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# Deck of Cards: Newsletter Challenge (02/06/07)

Posted February 04, 2007 5:01 PM
Pathfinder Tags: challenge questions
User-tagged by 1 user

The question as it appears in the 02/06 edition of Specs & Techs from GlobalSpec:

You have an incomplete deck of cards and are asked to randomly select five percent of the set. You follow a pattern for the selection - select the first card, miss one, select the next, miss two, select the next, miss three, etc. Following this method, the last card you selected is also the last card in the deck. And the number of selected cards is exactly five percent of the total cards in the deck. How many cards were selected?

Update (02/13/07 8:52 AM): And the Answer is....
Let n be the number of cards picked. Because n is five percent of the total, then the number of cards in the deck is 20n. The cards were picked following the following pattern:

So the number of the final card picked is , but this is precisely the last card in the deck. Therefore,

The non-zero solution to this equation is:

Therefore, you picked 39 cards that make up five percent of the deck.

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Guru

Join Date: Aug 2005
Posts: 5426
#1

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/05/2007 12:10 PM

I suspect a missprint here, but, the answer to the question as it stands is 39.

One heck of a deck: maybe a stack of punched cards?

|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|

|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|

1+ 2.. + 3.... + 4....... + 5.................................

[n(n+1)/2]/n = 20

(n+1)/2 = 20

n = 39

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Guru

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#2

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 6:58 AM

I couldn't see how Randall constructed his equation...until I spotted a "new to me" relationship between n and ∑n!

What I came up with, before reading Randall's entry is this:

.....1-1--1---1----1-----1-----1 etc

n ...1.2...3....4......5.......6.......7

∑n..1.3...6...10.....15.....21......28

where:

n = number of cards chosen

∑n = total number of cards

required: n = 0.05∑n or ∑n = 20n

Since ∑n = (n+1)/2 x n (the bit I hadn't previously spotted)

(n + 1)/ = 20

n = 39 and the deck has 780 cards.

If, however, this is a real deck of cards of 52 cards, then picking nine cards out of 45 would give you one-fifth (20%) of the available deck, using the equations above.

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Anonymous Poster
#70

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/08/2007 10:46 AM

No sums of simple sequences - English high-school maths is clearly not what it was.

Guru

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#94

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/14/2007 1:57 AM

Hi Rose,

I preferred your guess at the "missprint" (one fifth as opposed to 5%). Just for completeness: to derive the sum of an arithmetic progression.

S = ---1---+---2---+---3---+---4-------------+---n........(1)........Then reversing

2S = (n+1) + (n+1) + (n+1) + (n+1)------------+ (n+1)..................n times (n+1)

2S = n*(n+1)

and S = [n*(n+1)]/2

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Anonymous Poster
#3

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 10:54 AM

this equation cannot be completed within the constraints of <52 cards, following this method of selection.

Associate

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#5

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 12:00 PM

The method suggested also doesn't meet the requirements of the original request to "randomly select" the cards, but that wasn't really the question.

I suspect that the original asker meant 20% or 1/5 of the set, but perhaps he or she could clarify for us.

Associate

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#10

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 3:34 PM

i believe the that the "random selection" as written was just how it was reported to occur in this proposed case....but agree with everyone that there is something wrong with this challenge as an incomplete deck is always < 52 but the math just won't work for 5% sampling....must be a typo.

Associate

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#12

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 4:39 PM

I disagree that the 5% is a "typo," since it is written out as "five percent" and is written twice the same way.

The author of the question has made two big errors of conception: that following a pattern is "random selection," and that "five percent" means "one-fifth." These are thought errors, not typos.

I suspect, as do many other posters, that the bar for posting questions is too low. This one never should have made it to publication.

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#7

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 2:41 PM

Proof that the problem has no solution:

n = number of cards chosen, a positive integer

N = total number of cards in the short deck

N = 20n

N < 52 => N = 20 or N = 40 => n=1 or n=2

Since clearly neither n=1 nor n=2 is correct, we have a problem with no solution.

Anonymous Poster
#20

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 10:57 PM

Quicker proof:-

5% of 52 is 2.6 so as a whole card is required, 2 is max that can be drawn. Following the procedure, 1st card and third card in the deck would be the max that can be taken from a full deck as drawing card 3 would exceed the 5%.

Guru

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#4

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 11:46 AM

I agree with English Rose's final suggestion of 9 cards.

1. Top card - then miss no.2
2. Card 3 - then miss nos4&5
3. Card 6 - miss 7,8,9
4. Card 10 - miss 11,12,13,14
5. Card 15 - miss 16,17,18,19,20
6. Card 21 - miss 22,23,24,25,26,27
7. Card 28 - miss 29,30,31,32,33,34,35
8. Card 36 - miss 37,38,39,40,41,42,43,44
9. Card 45, last card.
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#6

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 2:11 PM

The number of selected cards is 9.

(That's because the total number of cards in the incomplete deck is 45.)

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#8

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 2:57 PM

But, but, but . . . 9 is 20% of 45!

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#9

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 3:15 PM

I selected 5 cards!

See: The answer will be announced on 02/13, right here on CR4...

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Anonymous Poster
#11

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 4:24 PM

The only problem with the answer 39 is that this is 5% of 780 and 780 is exactly 15 complete decks of cards.

Anonymous Poster
#58

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 12:49 PM

Not if you use 15 "incomplete decks" of 51 cards and 1 "incomplete deck" of 15 cards for a total of 780 cards.

Note: All 16 of which "incomplete decks" of cards can be found in my kids toy box.

Commentator

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#13

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 4:42 PM

I picked 2 cards cause there was only 40?

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#22

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 2:31 AM

I think the answer is 40 cards. you choose 2 cards out of 40 . The first card you select is the 38 th card (randomely selected) and the next card is 40 th card which is also the last card.

In no other way the 5% can be maintained in an incomplete deck of cards!

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Anonymous Poster
#33

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 8:33 AM

I agree with Nat. Although this problem identifies a pattern to be used it never suggests to what extent it is used. The answer is 40 cards or there is a typo. I would think a typo would have been addressed by now.

Anonymous Poster
#56

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 12:41 PM

Thirty-nine, or two, or five, depending on how you interpret "the first card" and "follow the pattern", and whether the largest deck you use is a basic pack with no jokers, or something a bit larger. (The question was the number of cards selected, not the number in deck). Not really the sort of question I'd like to see here - unless someone can come up with a more satisfactory answer than we've seen so far

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#55

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 12:25 PM

So you think it's a word game? That "first card" does not mean "first card in the deck"?

Could be. If so, it's still a pretty lame question, especially considering that it explicitly states that you "pick one, skip three, etc."

Anonymous Poster
#32

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 8:22 AM

Or one card, and there were 20.

Anonymous Poster
#14

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 4:52 PM

I also would say 2 cards, based on a possible incomplete deck of 51.

Anonymous Poster
#15

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 5:01 PM

Well, there must be an answer since it will be given on 02/13. But at first glance and using simple math and using the method given on can not come up with 5% of the deck with the last card selected being the last card in the incomplete deck.

My question is, what does the challenger mean by "incomplete". If he means each card is a single piece and not cut/torn into multiple pieces and the deck has less than 52, then there is no answer based on the method he provides of deriving the answer. If "incomplete" means that one or more cards are cut/torn into multiple units then that is too much hassle to figure out while at work and I pass on it.

Anyway, if all the cards are single pieces and there is less than 52 cards in the deck, then there is no way to select 5% of the deck using the method given and end up on the last card of the deck. How do we know this? Simple math.

5% of 100 cards would be 5 cards, correct? 100 x .05 = 5. Even using the method given, you would get more cards than 5. You would end up with 13 cards thus ending on the 91st card of a 100 card deck and not able to proceed further since you can not continue the same method of counting beyond that card and not go past 100 for the next count to work out. 13 is definitely more than 5%.

That means that 5% of a full deck of 52 (a full deck w/out jokers) would be about 2.6 cards. If each card is a whole piece, then this number is not possible. If you followed the method all the way through as far as you could go with 52 cards you would end up with 9 cards with the 9th card being the 45th card. You can not go beyond that card with the method given or you would have to go beyond 52 for the next count to work out.

So in order to pick a "whole number" of cards that equal to 5% of an incomplete deck (all cards a single piece and none are cut/torn into multiple pieces), then you can either only have 40 cards which gives you 2 cards to choose or 20 cards which give you only one card to choose. It would be impossible to get 5% of 20 or 40 using the method of counting given. Plus, you would not end on the 20th or 40th card. The numbers you can end on using the method given are 1, 3, 6, 10, 15, 21, 28, 36, & 45 if I figured it all out correctly.

So in order to get the answer the challenger is proposing, one would definitely have to think outside the card box in regards to the terminology used and maybe use some fuzzy math. Of course, I am sure when I see the answer, it will be so simple that I will slap my forehead and go "I could have had a V-8!".

Vikki

Anonymous Poster
#17

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 5:54 PM

Thinking outside of the box, what makes you think that this is a subset of a 52 card deck. It just said a deck of cards not a 52 card deck.

Anonymous Poster
#16

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 5:51 PM

This prob & stat stuff is getting rather boring. Lets have a good MacGuyver (sp?) challenge next week.

Like you have two tin cans, a really good swiss army knife (think 150 functions), bubble gum, and a balaclava. Now, how do you disable(or redirect) an ICBM and free 4 captured British special forces in Teheran after [unintentional] food poisioning.

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#18

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 6:38 PM

It is such a simple problem that either a child or ??? can arrive at FIVE percent of the set easily.

"Lick that calf over again!"

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Anonymous Poster
#19

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/06/2007 7:30 PM

If the answer is 5 then there must be 100 cards (5% of 100 is 5).

But based on the sequence given, the cards you select are

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 106

So you can never choose card 100 -> Don't satisfy criteria that final card selected is final card of set.

Also, card 96 is the 13th card choosen -> much greater than 5% of set (~13.5%)

If stop after the 5th card -> 15 cards in set-> 33% of set choosen

As mentioned in other posts, it is quite obvious there is a mistake in the question and if so the answer would be 9 cards selected out of a set of 45 (1/5 of set)

Anonymous Poster
#21

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 12:37 AM

I agree with Randall & English replies that the answer should be 39.

If you have been to any casino you will notice that their "deck" of cards is not just a single deck of 52 cards but multiple sets of cards.

Anonymous Poster
#24

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 3:43 AM

five to be exact. So you're saying the hypothetical you will select cards from a shoe? (pun intended)

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#57

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 12:45 PM

If you read up on casinos' card stocks, used, for instance, for blackjack, you'll notice that the whole stock is referred to as a "five-deck shoe." A deck is still 52 cards.

Any other definition of "deck" is non-standard and must be explicitly defined in order to have a well-formulated question.

Also, to reach 39 cards you'd need three 5-deck shoes. Since a standard new deck is about 1/2" thick, this stack would be 7.5" - hardly something you'd call a "deck."

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#23

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 3:29 AM

If the choice of five cards is correct, then there remain 95 cards in the deck after selection.

What has not been said in the question is that the reference to "miss one" is one block of the same size as the final tally of chosen cards.

The sequence given is too short to be assumed to be linear - it is infact the fibonachi sequence, so:

Pick 1 card

Miss one block of five (5)

Pick 1 card

Miss two blocks of five (10)

Pick 1 card

Miss three blocks of five (15)

Pick 1 card

Miss five blocks of five (25)

Pick 1 card

Miss eight blocks of five (40)

Oh S\$%@ all the cards are gone and I can't take the back one!

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Anonymous Poster
#25

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 5:12 AM

Think that the answer is 37...

let n= no. of cards collected

Total no of cards= 20n (as n= 5% of total cards)

Sequence:

1(1st collection), 1 (missed), 1(2nd collection), 2 (missed), 1(3rd collection), 3 (missed)... , n-1 (missed), 1(nth collection)

therefore, total no. of cards=

1+1 +1+2 +1+3 +...+1+(n-1) +1= 20n

n +1+2+...+(n-1) =20n

1+2+3+...+(n-1) =19n

Note that LHS is an AP, whose sum= n[(2(1)+(n-1)] /2

=[n^2+n]/2

Equating and solving,

n=37

Anonymous Poster
#35

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 9:13 AM

It is 37 and 740 as 37 is exactly 5% of 740

Anonymous Poster
#47

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 10:33 AM

You have a little mistake at the end. You could solve:

n + 1 + 2 + ... + (n-1) = 20n

1 + 2 + ... + (n-1) + n = 20n

n(n+1)/2 = 20n

n = 39

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#50

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 10:52 AM

or
1+2+....+n-1 = 19n
(n-1)n/2 =19n
n-1=38
n=39

Anonymous Poster
#26

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 5:55 AM

Clearly the '5%' should be one fifth. The problem setter made a hames of this.

Substituting 20% for 5%, the answer is 45 cards from which 9 are chosen

1+1+1+2+1+3+1+4+1+5+1+6+1+7+1+8+1=45

Anonymous Poster
#27

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 7:11 AM

The Guest is right. Clearly 5% was a mistake. The questiont should read 1/5th.

Anonymous Poster
#28

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 7:12 AM

39 cards were selected. 780 cards in the deck, that's one big deck!

Anonymous Poster
#29

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 7:39 AM

The question to answer is "are the cards skipped over discarded or returned to the deck?" Because assuming that the cards are discarded means a deck of 780 cards the solution must be either a really big deck or that the cards are returned to the deck after skipping. I do not have time to work this out.

Anonymous Poster
#30

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 7:52 AM

The answer has to be 2 cards taken and 40 cards in the deck or 1 card taken and 20 cards. The other assumption is that a pass was taken through the whole deck which would lead to a really big deck.

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#31

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 8:05 AM

Assuming the percentage was stated incorrectly and we are looking for 20%, I believe the answer would be 7 cards selected.

S1, D1 total cards=2, S1, D2 total cards=5, S1,D3 total cards=9, S1,D4 total cards=14, S1,D5 total cards=20, S1, D6 total cards=27, S1,D7 total cards=35.

7/35= 20%.

MechTech

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#34

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 9:01 AM

I'm assuming that Stirling Stan knows his stuff here, and the issue is how we can interpret the question - some things are precise, others vary - even according to the version of English that you use.

One ambiguity is "total cards in the deck". This could mean the sum of the face values.
As pointed out by NatNAt, there is nothing to say that the "first card" selected is the first card in the deck - this itself could be classed to "randomise" the selection, in spite of the existence of a subsequent pattern. But NatNat's answer ignores the fact that the minimum number selected can be five, because the "etc." in the selection pattern requires this.
Then, as pointed out by GM1964, another is in regard to what we miss each time - a single card, the whole deck (but two misses gets you back to where you started), groups of N. He also pointed out that the counting could be a Fibonacci sequence.

Given the reference to the child, I'll go with NatNat's simple insight. This means that, five cards being the minimum that satisfy the question, we have a deck of at least 100 cards. I know of several decks that use more than 100 cards (double packs for poker, Tarot decks, some "royal" games that use a double deck including Jokers of 106 cards...). I haven't come across a game that uses more than 110 cards, though they probably exist.

But this does give an answer of 5 cards (=minimum from picking schedule, maximum from deck size), in an incomplete deck of 100 cards. Dubious, etc., but could be seen by a child.

I also tried "total cards in pack" = sum of values, but this allowed anything between five and seven cards with an incomplete deck from a standard 52-card pack.

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#59

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 12:50 PM

What reference to a "child?"

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#63

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 1:18 PM

Stirling Stan - posts ## 9 and 18. In one of these he implies that he has the official answer in front of him, in another he refers to a child.

Fyz

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#64

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 1:42 PM

I see. I think you went wrong here:

"I'm assuming that Stirling Stan knows his stuff here..."

Stirling Stan is incoherent.

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#65

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 5:56 PM

I had read Stirling Stan's incoherence in this thread to be of the form of someone dropping clues as to where to look - on the basis that he implied knowledge of the official answer. In other threads where he has contributed, he tends to be a bit dogmatic and also cryptic, and is (like the rest of us) frequently unaware of some of the implications (sorry S-S), but not incoherent as he appears here.

So far, I have seen one mathematically good answer to the question, but the size of the deck makes it improbable that this was what was intended. So it seemed worth following S-S's hints to see where they lead. So far, the possible answers are either half-baked (ambiguity between 1 and 2) or exceed the size of deck I'd wish to see.

So we either need another insight, or continue to believe that the question was misguided - which wouldn't be the first time. I've also noticed that some decent questions have had 'official' answers that were incomplete, or wrong, or even answered a different question to the one posed.

Fyz

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#66

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 8:59 PM

You may well be right about me and/or the statement of the problem as presented.

I honestly don't know the proper interpretation and/or solution. If I had I would not divulge any real help but perhaps did a bit of needling to stimulate the discussion.

I realized my error not too long after making the first post.

On the other hand I expect to have a LOT of company at the CROW FEAST come Tuesday February 13th 2007!

The problem with the problem seems to be that it can be picked apart in too many ways and arrive at more and different solutions than the poser intended.

It's OK Physicist, I'm just an ole injinear but have a few friends who are physicists, I'll add you to the list.

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#67

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/08/2007 4:43 AM

Hi Stirling Stan

I'm flattered, if unworthy, as I'm just a superannuated ... myself, and even my moniker of "Physicist" came out wrong - it should have been "Physicist?", which is quite different, but some-how the punctuation got lost - but I couldn't be .... to start again. Story of my life?

Fyz

Anonymous Poster
#36

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 9:19 AM

2 cards

Anonymous Poster
#39

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 9:53 AM

But you haven't answered the question, which says that you "select the first card, missed one, select the next, missed two, select the next, missed three, etc.", as this gives a minimum of five cards.

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#37

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 9:33 AM

The simplest answer is one. To clarify - You have an incomplete deck of 20 cards. You randomly select the first card on the bottom. You now have 5 percent of the deck, you've stayed within the pattern and the last card you selected is also the last card in the deck. The Challenge says "select the first card". It doesn't say "select the first card on the top of the deck".

Anonymous Poster
#40

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 9:54 AM

But you haven't answered the question, which says that you "select the first card, missed one, select the next, missed two, select the next, missed three, etc.", as this gives a minimum of five cards.

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#51

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 11:12 AM

The Challenge says " you follow a pattern for the selection - select the first card, miss one, select the next, miss two, select the next, miss three, etc.". It doesn't say that you have to draw this many cards as a minimum. It's telling you what the pattern is. You would follow this pattern until you selected the last card in the deck and had five percent of the cards, then you and the pattern would stop. If you draw the first card from the bottom of a deck of twenty and then stop, you have fulfilled the requirements of The Challenge and you have not deviated from the pattern. You have only followed the pattern as far as necessary.

Anonymous Poster
#53

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 11:26 AM

That is the difference between defining a pattern (as you state) and following the pattern (as requested). We're probably suffering from an international linguistic problem. The other issue is that your interpretation allows two different solutions with a standard 52-card deck, 5 solutions with a Tarot or poker deck, or 39 solutions with an unrestricted deck.

Anonymous Poster
#38

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 9:51 AM

The question is flawed. The maximum number of cards that can be selected in a full deck of 52 is 2 cards (5% of 52 = 2.6). 5% of 40 is 2 cards. 5% of 20 is one card. The selection strategy cannot acheive a result of 1 card in a total of 20 cards, or 2 cards in a total of 40 cards.

Anonymous Poster
#44

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 9:56 AM

Assumptions, assumptions. Even most single bridge/whist/poker packs as bought have 54 cards, and there are decks out there with at least 106 cards.

Anonymous Poster
#41

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 9:54 AM

"Stop making sense" David Byrne.

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#42

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 9:54 AM

There are two acceptable solutions. The first is that the 5% figure in the challenge is incorrect and should read one fifth. The second is that 5% is correct and the deck consists of a shoe of at least 16 standard decks (noting that the guest responder determined that 780 is a shoe of 15 complete decks).

The first solution is sigma n = x. n = x/5. Sigma n = n(n+1)/2 = 5n. n+1 = 10. n = 9. x = 45. Thus a standard deck missing 7 cards.

The second solution is sigma n = x. n = x/20. Sigma n = n(n+1)/2 = 20n. n+1 = 40. n = 39. x = 780. Thus a 16 deck shoe missing 52 cards.

Just as a side note, the 16 deck shoe may consist of 12 decks each missing 3 cards and 4 decks each missing 4 cards.

Re. randomness of selection: Assuming that the "deck" has been shuffled and is therefore random, any selection method will result in a random set of cards. Or were you looking for a second order random set?

I don't think that even the Las Vegas mavens can handle a 16 deck shoe. Is there such a thing?

Anonymous Poster
#43

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 9:56 AM

The question makes no statement about the type of deck of cards. It could easily be a Yugioh or Pokemon deck of cards. I would think that there are more than 780 which would make it incomplete. I think the correct answer has already been given.

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#45

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 10:08 AM

Yes, of those given so far, I too prefer the Randall/English_Rose solution. But half the fun is looking for an acceptable alternative answer. We've come up with at least two families of alternative answers so far - none of them at all satisfactory, in my opinion. My fear is that the desired answer is as I described - five beginning from part-way down a pack of 100; but I'm still hoping someone can come up with something better.

(If the question had said the smallest number, I think the candidate I described would become more competitive)

Fyz

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#46

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 10:30 AM

The solution according to my Excel spreadsheet is:

n = 37

The total number of cards is 740.

Anonymous Poster
#49

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 10:45 AM

37 doesn't comply with the rule of choosing the last one. Instead, you are choosing the 37th and missing 37.

Anonymous Poster
#48

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 10:36 AM

The answer is 39, easy to calculate using Excel.....

Anonymous Poster
#60

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 1:08 PM

I agree the answer is 39 cards chosen from a deck of 760.

The formula to solve for this is:

n/(((n-1)^2+(n-1))/2+n) = x, where n is the number of cards chosen,

((n-1)^2+(n-1))/2 is the number of cards skipped

x is the ratio of cards chosen to total cards, and we want to set to 0.05 in our case.

Take for instance you choose 5 cards, n = 5

you have done: 1-1--1---1----1 note that you don't skip 5 cards after chosing the last card.

There are 5 chosen cards and 1+2+3+4 = 10 skipped cards, the percentage chosen = 5/(10+5) = 0.333 or 33.3%

By using the formula, n = 5, (5)/((4^2+4)/2+(5)) = 5/((20/2)+5) = 5/15 = 0.3333

For 39 cards, n = 39, 39/((38^2+38)/2+39) = 39/(741+39) = 0.05

Total number of cards = 741+39 = 760

I believe many people are including an additional set of "skipped" cards after the last card has been chosen.

If you do that then you would falsely get (by taking the "n-1" from the equation and repace it with "n") a solution of

37 cards, n = 37, 37/((37^2+37)/2+37) = 37/(703+37) = 0.05

Assuming you must choose the first and last cards from a deck using the prescribed method, the deck must be 760 cards!

Anonymous Poster
#52

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 11:21 AM

Perhaps we have overlooked an obvious answer. Algebra doesn't lie:

Let:

N = number of cards chosen

X = Total Cards in the incomplete deck (assume X <= 52)

Y = Cards skipped.

N = 5/100 * X or X = 20N

Y is the summation of integers from 1 to N-1 (Not 1 to N because you don't skip on the last draw since it's the last card) so

Y = (N-1) (N) / 2

So since N + Y = X you get N + (N-1)*N/2 = 20N or N(N-1) = 38N or N(N-39) = 0 which yields N = 39 which, as pointed out, is outside the range of solutions assuming a 52 card deck max.

But... N = 0 is also a valid solution. If X = 0 (a very incomplete deck to say the least) then 5% of 0 is certainly 0. So 0 is the only practical solution, albeit not a very exciting one.

If you change the problem to the proposed correction of N being 1/5 of X you get

N(N-9) = 0 which yields N = 9 or N = 0 both solutions within the permissable range.

David Johns

Anonymous Poster
#54

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 11:34 AM

Algebra absolutely does produce solutions that do not fit the problem, and it's a standard pitfall that you need to check the answer really does fit the constraints. This isn't algebra lying - it's the user introducing disallowed solutions when setting up the algebra or when manipulating the algebra. As in this case, the problem states that there is a deck of cards, and also that you follow the pattern - not merely that you define a pattern. You'd be pretty surprised if someone accused you of following them when all you'd done was stand still. The same absolutely applies to the single-card solution, though the situation for the proposed two-card solution is less clear-cut.

Anonymous Poster
#61

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 1:13 PM

Sorry...780, typo...

Anonymous Poster
#62

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/07/2007 1:15 PM

Nat Nat #22 is right, the answer is 2. The convoluted offerings are typical of the engineering trained mind. The answer and solution must be more complicated than first flush because, well, that should be the basis for the next question. The best questions are the ones that simply allow the questioned to follow their own meme.

Anonymous Poster
#68

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/08/2007 9:52 AM

39 out of 780

Someone had taken all the aces and left the jokers in 15 decks!

Don

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#71

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/08/2007 12:18 PM

I think its the one who set the question that's the joker!

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#72

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/08/2007 12:52 PM

I also find fault with the method of random selection

If a pattern is followed in RANDOM?? selection the cards can also be DESELECTED and the original order restored!

Don

Anonymous Poster
#73

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/08/2007 1:12 PM

Given a random starting point, and a randomly selected pattern, we can allow the selection to be sufficiently chaotic to count as random. Even starting at card1 would be possible - but should not form part of the criterion - so perhaps this forms a hole in the 39 steps solution. Divulging or repeating the pattern (or using patterns with similar characteristics) would invalidate this. (Observation of the locations of selections would also allow de-selection, however random the selection pattern).

Anonymous Poster
#69

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/08/2007 10:23 AM

The answere is 39 see details below. Pick skip Total Cards After Pick Total Picked % 1 1 1 100.00% 1 1 3 2 66.67% 2 1 6 3 50.00% 3 1 10 4 40.00% 4 1 15 5 33.33% 5 1 21 6 28.57% 6 1 28 7 25.00% 7 1 36 8 22.22% 8 1 45 9 20.00% 9 1 55 10 18.18% 10 1 66 11 16.67% 11 1 78 12 15.38% 12 1 91 13 14.29% 13 1 105 14 13.33% 14 1 120 15 12.50% 15 1 136 16 11.76% 16 1 153 17 11.11% 17 1 171 18 10.53% 18 1 190 19 10.00% 19 1 210 20 9.52% 20 1 231 21 9.09% 21 1 253 22 8.70% 22 1 276 23 8.33% 23 1 300 24 8.00% 24 1 325 25 7.69% 25 1 351 26 7.41% 26 1 378 27 7.14% 27 1 406 28 6.90% 28 1 435 29 6.67% 29 1 465 30 6.45% 30 1 496 31 6.25% 31 1 528 32 6.06% 32 1 561 33 5.88% 33 1 595 34 5.71% 34 1 630 35 5.56% 35 1 666 36 5.41% 36 1 703 37 5.26% 37 1 741 38 5.13% 38 1 780 39 5.00%

Anonymous Poster
#74

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/08/2007 4:14 PM

39

Ken m

Anonymous Poster
#75

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/08/2007 11:25 PM

A simple understanding of the english language is all that is needed to get the answer, and clear up all of the confusion. Are all of you ESLs? No, I guess not. They understand the language better than you guys. I suspect that we are dealing with a very serious problem here. Namely that you are a bunch of engineers. That's too bad. If you just stick to building what others tell you to build, and stay away from The Challenge, you will do much better. After all, what do you think you have managers for, and did you notice, they are not engineers? Do you now understand why?

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#77

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/09/2007 3:00 AM

Are all of you ESLs?

The simple answer to this is CRIMINAL.

A simple understanding of the english language is all that is needed to get the answer, and clear up all of the confusion.

OK lets look at the question and see what a simple understanding of the English language give us.

You have an incomplete deck of cards

From what I have been able to ascertain the term deck of cards could either refer to the song "The Deck of Cards" or a deck pf playing cards. Now I could be incorrect but I can't see the question being about a song so lets for now assume we are talking about a deck pf playing cards. Having quickly read the article I can see that a deck of cards could have 32, 36, 40, 48, 52, 54, 56, 78, 108 cards. Now the word incomplete implies that the deck of cards has something missing. Since we have seen that the deck of cards could contain up to 108 cards then the maximum number of cards in a single deck could be 107.

randomly select five percent of the set.

By selecting 5% we are talking about a ratio that means 5 out of every 100 or 1 of 20 nothing else. But not at this point the description has changed from a deck of cards to a set. A set could be anything so if we are talking about a literal interpretation we are talking of two different things. Initially we spoke about an incomplete deck of cards but we are asked to select 1 in 20 from a set.

You follow a pattern for the selection - select the first card, miss one, select the next, miss two, select the next, miss three, etc.

This is fairly self explanatory and describes the mathematical model use first by Randall and confirmed by English Rose.

Following this method, the last card you selected is also the last card in the deck.

This implies that the last card we select mist be the last card available. Notice however the cards are again described as a deck rather than a set as in the previous sentence.

And the number of selected cards is exactly five percent of the total cards in the deck.

This re-enforces the point that the number of cards selected must be 5% or one twentieth of the total number of cards.

How many cards were selected?

So what is the answer. Clearly to use the sequence and end with one card in twenty being selected the number of cards selected must be 39 and there must be a total of 780 cards.

However a deck of cards has various definitions none of which describe it as having more than 780 cards.

It is possible that the writer actually meat to say select one card in five rather than 5% but this mistake would need to be made twice. Also not that the number of cards to be selected is expressed as a number 5% and in words five percent.

So the question is did the writer mistakenly tell us to select 5% or the cards on two separate occasions of are we talking about a set of cards that contains multiple decks?

Mathematically the answer is 39 out of 780 cards and since mathematics is far less ambiguous then I would need to lean towards this being the answer. It is however possible we should be selecting one card in 5 and that there was a mistake in the original question that has been erroneously corrected. This also fits as the writer states 5% numerically initially but in words later which indicates to me that one reference was changed some time by a person other than the initial author.

Put simply the question contradicts itself.

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#78

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/09/2007 4:57 AM

Hi Masu

I think that the basis for this has to be to find an answer that turns the question into a consistent whole - or as near as possible. There are an irritating number of 'contributors' who simply state that it's easy if you read it correctly, but for the present I suspect them of merely being provocative - if they really knew the answer, they would surely explain it.

Your observation of the use of "set" was helpful, I think. If the cards are well shuffled, then you can select systematically from the deck, but are still selecting randomly from the set. This overcomes one of the problems.

Although you covered some of the possible interpretations, I think you missed (at least one of) the possibilities. Although use of "the first card" OUGHT to mean something more than "a first card", I don't think it absolutely MUST mean something more. So, "the first card" could either carry no additional meaning, or it could be something other than "the first card in the deck". (If there is any additional meaning other than this, I think it doesn't matter what it is)

Then there is ambiguity about the pattern - "miss one... miss two...". That could just mean that you discarded the card after taking it... But, once we have decided not to select the first card in the deck, the nature (if any) of the pattern doesn't actually matter. Come to that, if we assume that the deck can have up to 108 and no more, neither does is the location of the last card important, as the question states that you have followed* the pattern up to the fifth card, which is the maximum you could possibly take from a deck of <108).
* Not just defined it, actually followed it

I think that means that both 5 and 39 have claims as valid answers. But sloppiness in the meaning of "etc." could still allow an answer of four - and we could still be missing something.

Thanks for the insight on set vs deck

Fyz

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#79

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/09/2007 8:37 AM

amen masu...but i find the word "incomplete" a problem when applied to the answer of 39 out of 780 cards along with the comment of the 780 cards being 15 complete decks of playing cards. The math has been shown correct, but something is amiss, perhaps it is the assumption of standard deck of playing cards.

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#81

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/09/2007 11:53 AM

We shall just have to draw on our unbounded patience and endure the conundrum while waiting for the publication of the official answer.

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Anonymous Poster
#87

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/12/2007 5:09 PM

It looks like #58's answer in reply to #11 hold a possible explanation for "incomplete decks".

I believe I could also find the required 16 "incomplete decks" at the bottom of my kid's toy box as well.

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#76

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/09/2007 1:52 AM

Hi Chris

This is a little screwed up. If asked to randomly select five percent why did we decide to follow a pattern? Patterns are not random. Any way I did it your way under protest and selected 37 cards out of a deck of 740.

Anonymous Poster
#80

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/09/2007 11:47 AM

A shuffled DECK performs the randomisation of the SET. Therefore, the selection remains random w.r.t. the set, even though you follow a pattern within the shuffled deck. So the pattern was just for fun, and made it possible to pose the problem. But 37 cards?? - Randall got 39 in the very first post (also clearly explained by English Rose and others) - so perhaps you would you like either to explain your 37, or to check it.

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#82

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/09/2007 1:05 PM

there is no description of a "shuffled" deck in the problem.

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#83

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/09/2007 3:25 PM

It's not much of a challenge if they tell you relevant things that you could work out for yourself*. The wording doesn't say the pack is not to be shuffled, and shuffling (or something to serve an equivalent function) is needed if the "you" of the question is to be seen to select randomly as requested, while still following a selection pattern. However, it's also possible for the "you" to ignore this part of the request, so this additional action is not strictly necessary part of the solution (but it's rather satisfying, in that it shows the significance of some of the very specific wording).
*Additional information for distraction is another matter - that simulates real problem solving

Unfortunately, all the interpretation to date has come up with answers that are potentially problematic.
39 (too large?);
5 (requires two stages of interpretation i.e. it's not necessary to start at beginning of the deck, "follow the pattern" means actually take all the steps described; and one external constraint i.e. that the number of cards in the deck must be less than 108); and
1 or 2 (this requires that you don't start at the beginning of the deck, and also satisfies the constraint that the deck is a standard single pack - but so far it seems impossible to decide between the two options).

It will be interesting to see if the "official" answer resolves the residual issues, or if it just sweeps them under the carpet and asserts one of the existing suggestions to be the answer.

It would be even better if one of the CR4 contributors came up with a logical explanation as to why the question itself should uniquely and logically lead to an answer that can be met within a realistically-sized deck.

Over to you...

Fyz

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#89

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/12/2007 5:47 PM

fyz....actually the issue of a shuffled deck is not relevant as we do not care what is contained on the cards anyway for this problem...for that matter, we should assume they are blank, as we never look at them, etc. only pick cards to get 5% of total after selection is completed as per instructions....we must wait another day.

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#91

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/13/2007 6:19 AM

As I said elsewhere, it is not relevant to the numerical answer - merely to whether we actually followed the instructions; of we did not have this out, we should be looking for a way that allows the system we choose to be consistent with following the instructions.

Anonymous Poster
#84

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/10/2007 7:05 PM

The difficulty with this challenge is not to find a solution, but to "interpret" the question. Certainly, the problem is formulated beyond mathematical strictness.

It makes sense, to understand the deck request ("an incomplete deck") with minimal assumptions. In doing so, the statement only means that there are not exactly 52 cards in the deck (respectively any other conventional number of cards). In what follows, the deck size could be either less or greater than 52, nothing else. Particularly, the deck size is not restricted to less than 52 cards. I think, the purpose of the originator is to make clear, that non-standard card decks are explicitly allowed. Clearly, this hint is redundant, because 5% of 52 is not an integer.

To come to the solution, now let n be the number of cards subsequently selected, and let D be the total number of cards in the deck. Further, let N be the (ordinal) number of the first card selected by random minus 1 (N=0, if no cards are selected; the trivial case).

As pointed out, we are asked to "randomly select five percent of the set" and to "follow a (specific) pattern for the (process of) selection. This formulation is not quit clear and seems to be contradictory, at least it is ambiguous. Nevertheless, one reasonable understanding is as follows. Step 1: select one card of the deck by random. Step 2: follow the selection process as described. However, this comprehension has the disadvantage of leading to a number of 40 different solutions. Indeed, now each n=0, 1, 2,…, 39 is a solution. Following the problem description, this can be derived from

p*n=D=N+n*(n+1)/2 (where p=20=1/0.05).

This is a quadratic equation for n. By standard algebraic methods we get the two solutions

n=p-1/2+1/2*sqrt((2p-1)2-8N)

and

n=p-1/2-1/2*sqrt((2p-1)2-8N)

A closer look to these relations finally leads to fact, that exactly all the (40) pairs

(n, N)=(n, n*(39-n)/2))

are proper solutions of the problem (for n=0, 1, 2,…, 39).

There are no other solutions.

Examples:

n=39, N=0: D=N+n*(n+1)/2=780, 0.05*D=39

n=0, N=0: D=N+n*(n+1)/2=0, 0.05*D=0 (trivial)

n=1, N=19: D=N+n*(n+1)/2=20, 0.05*D=1

n=38, N=19: D=N+n*(n+1)/2=760, 0.05*D=38

n=6, N=99: D=N+n*(n+1)/2=120, 0.05*D=6

n=33, N=99: D=N+n*(n+1)/2=660, 0.05*D=33

n=19, N=190: D=N+n*(n+1)/2=380, 0.05*D=19

n=20, N=190: D=N+n*(n+1)/2=400, 0.05*D=20

The most simple non trivial solution is n=1 (with N=19, D=20, 0.05*D=n).

Depending on how strict the wording of the question is understood, n=1 could serve as the unique solution, because it is the simplest one. However, this interpretation is far away from being strong.

In my opinion, the following approach is the most reasonable one:

If we suppose that the problem has to be understood as to have a unique solution, we must take the problem formulation as it is. In doing so, we have to suppose N=0, because it is explicitly said "select the first card". Uniqueness can only be reached, when the first card selected is the first card in the deck. In any case, this is the simplest understanding of the challenge. Having this in mind, we get

n=39, N=0: D=N+n*(n+1)/2 =780, 0.05*D=39

as the unique (non trivial) solution. As described above, all other solutions need some additional assumptions.

This unique solution is also consistent with the requirement of a "random selection". Well understood "random" here means "arbitrary", nothing else.

(n,N)=(1,19) is not a proper solution, because the fist card selected is the 20-th card in the deck. Moreover, why should n=1 be a better solution than n=2 (with N=37)? In both cases a deck of less than 52 cards suffices, in what follows, that there is really no strong argument to characterize n=1 as to be a better solution than n=2. By the same reason we may realize that n=2 is no proper solution too.

n=3,4,5,…,38 (with N=54,70,85,99,112,…,184,187,189,190,190,189,187,…,54,37,19) are also no proper solutions, because in all these cases the fist card selected never is the first card in the deck. Furthermore: why should n=5 (with N=85) be a better solution than n=11 (with N=154), for example? In all these cases we need a deck of more than 52 cards. This means, we can't reasonable argue to indicate one of these solutions to be better than any other.

All in all I finally think the solution n=39 coincides the constraints at the best.

Greetings from Munich, Germany

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#85

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/11/2007 6:53 AM

Hi Guest from Munich

I'm rather afraid that your solution is the least-bad single-number solution available, but I feel that your statement about interpretation has missed some salient features that affect both the route and the validity of the (or any unique) answer.

Of course, these are purely semantic points, and I shan't list all features here, only those that impact your statements (others may be found in various contributions to the thread):

First - randomness: the question says you select randomly from the 'set' (of cards -assumed). There is nothing to say that you don't shuffle the deck - which is the standard way to 'randomise' a set of cards. So the selection of the first card in the deck can be as random as any other selection - not merely arbitrary. This removes one problem from the 39 answer.

Second - arbitrary size for the non-unique solutions: the thread says that you "follow the pattern", not merely that you define it. In addition, the description goes further than would be required usefully to define the progression - any simple sequence that starts 1, 2 ... (that I've identified*) has no solution that starts from the first card. This makes the minimum allowable number of cards 4, not zero. Unfortunately, this is insufficient to define a unique solution within the constraints of any 'standard' deck of cards (standards of 52 and 54 cards are disallowed), and decks that are incomplete subsets of 104 and 108 cards would not lead to a unique solution (4 and 5 are allowed).
*e.g. (1, 2, 3, 5, 8...) or (1, 2, 4, 8, 16...), or (1, 2, 2, 4, 8, 32...)

Third, you rightly comment that the statement "non-standard deck of cards" would be redundant. This leaves us with two possible interpretations - that the statement "incomplete deck of cards" does not in any way constrain the solution (just like the "randomly select" constraint), or that it means that the set is of a size that could not possibly comprise a complete deck. Now, I'd like to make that second interpretation, which would disallow the deck of 780 cards (=15 packs that don't include jokers); unfortunately, it doesn't lead to a unique answer.

Another potential (but in the end invalid) counter-argument to the deck being 780 cards it the proper meaning of the term "deck" - a stack of cards laid flat on a surface". I don't know if you've tried this with standard playing cards, but the resulting pile is not exactly stable; however, it's fine with oversized cards, and these are not explicitly excluded.

You rightly argue that the description of the selection process does not of itself demand that you start at the beginning of the deck. Therefore, the single justification for the solution 39 is that it corresponds to the only interpretation of the question (so far) that leads to a unique solution

So, I agree that 39 is the least-bad single answer. Unlike you however, I would tend to the viewpoint that the question as posed does not have a unique solution, and that the 'correct' answer is that any solution between 4 and 39 would be valid.

Hopefully, the "official" solution will find a semantic twist that we have all missed - and that leads to a satisfactory unique solution.

Fyz

Anonymous Poster
#88

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/12/2007 5:43 PM

Note: in posting #84 I erroneously typed "fist" instead of "first" in two of the last sections.

==>

As already pointed out, the problem is NOT formulated in mathematical strictness. Definitely! One consequence is that the discussion tends to become linguistic or philosophic in some way.

Virtually, the official solution can't ignore the fuzziness of the problem due to the rather vague wording. In what follows: whatever the official solution will be, the following statement remains true:

Depending on how strict the wording is being understood, the problem has an unique solution (n=39, 780 cards) or 40 different solutions (n=0,1,2...39,) where the solution n=0 is a trivial one which has been considered for the sake of completeness only).

The crux is that there is no serious and incontrovertible argument to decide between these perceptions. This lack of strictness and completeness is not satisfying.

@Physicist. Less is clear with the question but one thing is quite clear: it is totally irrelevant which cards are selected (the 8, the 4, the 9, for example). In what follows that shuffling the deck makes no sense and is not valid, virtually. Indeed, we are asked to follow the suggested procedure of selection what implies to do neither more nor less than described. If we were allowed to include some additional steps not mentioned explicitly, it would be no problem – trivially – to indicate any number (>1) of cards to be a valid solution. For example n=2 would be a very good solution: (step1) take the first card of the deck, (step 2) shuffle the deck until the last card becomes the second card of the remaining deck. (step 3) leave one card and take the second card of the deck. Ready! - And we selected the first and the last card of a deck of 40 cards. We may argue analogous to give reasons for any other number to be a good solution.

I agree with you that the thread says "follow the pattern" and I also agree that this not means "define the pattern". The relevant pattern is given by the triangle sequence 1, 3, 6, 10, 15, … in what follows that starting from the first card results in the solution n=39 as already shown. For this pattern n=39 is both the minimum and the maximum number of cards selected, provided the first card selected is the first card in the deck.

Clearly, for any other sequences there are different solutions. If we indicate the pattern sequence by a(n) then the criterion for a n to be a solution is p*n=a(n) (with p=1/0.05). So, if a(0)=0 then n=0 is a valid solution – trivial but nevertheless valid.

For your "simple" sequences the minimum number of cards is 3 not 4. Consider the following sequence defined recursively by
a(1)=1, a(2)=2,
a(j)=2*(1+a(j-1))*(1+a(j-2))*(1+a(j-1)+1+a(j-2)).
It starts 1, 2, 60, 23424, … Based on this sequence the criterion is fulfilled for n=3 because of 0.05*a(3)=3.
Clearly, for n=1 or n=2 there is no solution. So the minimum allowable number of cards for those "simple" sequences is 3 actually.

In my opinion the "incomplete deck" statement is a sort of "positive constraint". It should help to find a solution. A deck less than 52 cards is incomplete but a deck of 60, 100 or 780 cards I consider as incomplete too. So the n=39/D=780 cards solution is allowed and even is unique provided the first card selected is the first card in the deck.

Nevertheless, because of the vague wording we all can't be absolutely certain about this interpretation. Anyhow, the assumption of uniqueness I consider to be not a weak argument.

Greetings from Munich

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Posts: 488
#90

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/13/2007 5:52 AM

Some of your points are fine - I'll not address those, or others that don't directly concern me

We are not asked to follow a pattern - we are asked to select randomly, and the question then states that we follow a pattern. Even so, the existence and following of instructions does not mean that we don't do anything else. If I ask someone to go to the shops, I do not expect to tell them to check for traffic before crossing the road (young children excepted). Thus, the question requests random selection, and then tells you some of the things you choose to do - nowhere does it state that you do nothing else. If we wish both to follow a pattern and select randomly, then we must shuffle the cards. Nothing in the question says we don't - and as it's essential to comply with the original request, I suggest we do. Not that it has any effect at all on the final answer.

And I still maintain that the range is 4 - 39, not 0 to 39. BTW, have you considered what happens if you replace the cards you have taken at the bottom of the deck?

Anonymous Poster
#86

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/12/2007 12:22 PM

I believe the partial deck size is a red herring.

Consider:

Partial Deck, Sample, Full Deck

1, 1, 20

3, 2, 40

6, 3, 60

10, 4, 80

15, 5, 100

21, 6, 120

28, 7, 140

36, 8, 160

45, 9, 180

55, 10, 200

etc.

The partial deck size is constrained by n(n+1)/2 to satisfy the "random" sequence given. You can adjust the partial deck size to yield any size sample you want. Because the full deck is 20 times the sample, you can make any selection rule you desire that will have no impact on the answer. The only relationship between the partial deck and the full deck is that it is smaller.

I like: 1 card from a full deck of 20 cards. In the 1800's poker was played with a deck of 20 cards in the U.S of A. and I feel a little nostalgic today.

Anyone know of any other decks?

BR

Anonymous Poster
#92

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/13/2007 3:55 PM

the answer is 20 cards in the deck

solution is in the sequence of skipping cards

take 1 skip 1

take 1 skip 2

take 1 skip 3 now the trick

take 1 skip 1 again

take 1 skip 2 etc....

5% of 20 is 1

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Location: Port Noarlunga, South Australia, AUSTRALIA (South of Adelaide)
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#93

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/14/2007 1:50 AM

I appears the description of an incomplete deck of cards is the red herring part as the official answer states that you select 39 cards from 780 cards which is exactly 15 decks of the standard 52 playing cards.

Clearly the phrase "incomplete deck of cards" dose not mean a single deck of playing cards that contains less than 52 cards.

Regardless the question is poorly stated and ambiguous with the word "incomplete" confusing the issue.

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#95

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/14/2007 4:08 AM

It could be that the reference to "incomplete deck of cards" was to what was selected, rather than what it was selected from.

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#96

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/14/2007 6:28 AM

"You have an incomplete deck of cards and are asked to randomly select five percent of the set."

There is no way around this the word "incomplete" is an adjective that is describing the "deck of cards" from which the five percent will be selected and not the other way round. Since the answer is 39 cards from a set that contains 780 cards then the complete set or deck must contain more that 780 cards.

So to what sort of set of cards could the author be referring? Well how about a set of computer punch cards or blank credit cards, either could easily contain more than the given number.

The mistake we were all making was to assume that with the phrase "incomplete deck of cards" the author is referring to playing cards. The term could easily be referring to a wide variety of cards any of which would normally be found in sets greater than the 780 of the answer.

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#98

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/14/2007 7:44 AM

masu....i agree. the answer was a disapointment as it was never stated what deck it was, especially with the incomplete statement. as noted before, i was bothered by the 780 card count with 15 complete decks.

Anonymous Poster
#97

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/14/2007 7:40 AM

Great question! A good solid seventh grade word problem that stumped how many adults?

Anonymous Poster
#99

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/14/2007 8:07 AM

What!!

The difficulty was only for contributors who were trying to find a sound interpretation with a unique answer, but in the end it turned out that the wording was full of fifth-grade errors. The most important of these was probably "select the first card" instead of "select the first card in the deck", particularly in the context of "select the second" clearly not meaning the second card in the deck. Of course, there were several things in there that did need intelligent interpretation to allow you to realise they could be neglected, but that does not excuse the basic flaws.

Power-User

Join Date: Mar 2005
Posts: 214
#100

### Re: Deck of Cards: Newsletter Challenge (02/06/07)

02/14/2007 8:48 AM

Why you want to make this challenge more complicated?

(1) A "deck of cards". Why some of you want to constraint the problem to a real deck of cards with 52 cards? Think outside the box. Think that this is a deck in a far away planet where the typical deck of cards contains 1000 cards. So, a deck with 780 cards is indeed incomplete.

(2) It is clear that once you start selecting the cards according to the stated pattern, you cannot re-shuffle the deck. The most important statements of the problem are the ones referring to the way you must select. These constitute the actual statements of the problem. The idea is to drive you to discover that the selection process constitute a sequence whose sum is the number of cards in the deck. Very simple. If you discover this pattern, then the rest is a matter of solving a high school algebra problem.

(3) The rest of the wording in the problem is nothing more than embellishments to confuse the reader in order to make the challenge more beautiful.

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