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A Dicey Problem: Newsletter Challenge (October 2013)

Posted September 30, 2013 4:59 PM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from GlobalSpec:

A bag holds 10 fair dice. Some of the dice are 3-sided, some 6-sided, and some 7-sided. When all of the dice are rolled, the most probable result (sum of the dice) is 32. How many of each kind of dice are present in the bag?

And the answer is:

The bag contains three 3-sided dice, four 6-sided dice, and three 7-sided dice. The following equation for determining the most likely result for a given number of similarly-sided dice may be helpful:

Where n is the number of dice and s is the number of sides of each die. Using this formula we see,

Thus:

Thirty-two is the most probable result for three 3-sided dice, four 6-sided dice, and three 7-sided dice rolled at the same time.

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#1

Re: A Dicey Problem: Newsletter Challenge (October 2013)

09/30/2013 5:58 PM

3 ea 3 sided....5 ea 6 sided....2 ea 7 sided.....

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#55
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 11:57 AM

3 ea 3 sided = 6
5 ea 6 sided = 17.5
2 ea 7 sided = 8
total is 31.5

(sorry!)

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#2

Re: A Dicey Problem: Newsletter Challenge (October 2013)

09/30/2013 6:11 PM

3 - 3 sided

4 - 6 sided

3 - 7 sided

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#3
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

09/30/2013 6:20 PM

DOH!....I believe you are correct.....(insert excuse here)

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#4

Re: A Dicey Problem: Newsletter Challenge (October 2013)

09/30/2013 10:26 PM

What is a three- sided dice?

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#5
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

09/30/2013 10:39 PM

It's a..

Well, no, it's more like a...

Okay, picture a 4-sided die, and then... umm...

HEY LOOK - a Squirrel!

/GA. Another poorly framed question.

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#25
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 1:42 PM

So a three sided dice is just something that can only fall on 3 faces flat. Here we go with another steeeeep learning curve!

Thanks for sharing!

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#26
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 1:51 PM

A prism of triangular cross-section with rounded ends will, when rolled, come to rest on one of 3 faces.

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#29
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 3:44 PM

Don't try to sugar coat it....

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#37
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/02/2013 6:43 AM

a coin

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#43
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/15/2013 10:50 AM

You have to expand your thinking a little to get your head around it. Think of a cube with matching quarter circles on opposing sides. Cut off everything outside of the arcs and what remains forms a curved three sided solid. I found this explanation on the web and used AutoCad with extrusions to verify that it works (I was sceptical at first, too).

This is a wireframe of the solid I created while I was trying to figure it out.

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#6

Re: A Dicey Problem: Newsletter Challenge (October 2013)

09/30/2013 11:15 PM

0 3-sided dice (this solid cannot exist GA Reuven)

9 6-sided dice with only a 0 on every surface

1 6-sided dice with only the number 32 on every surface

0 7 sided dice (I cannot think of a topology that could make a 7 sided solid fair to roll)

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#7
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 12:44 AM

Yes, that's a five....

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#24
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 1:30 PM

I don't get an intuitive feel that this will be a fair die. It looks like the center of mass is closer to the "seven" and (presumably) "six" surfaces that will rest on the surface than the other five surfaces. Additionally the areas of contact will be different. It would be interesting to test.

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#30
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 3:50 PM

Luckily just such a study has been conducted....

https://www.youtube.com/watch?v=mmbapsJudG4

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#8

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 12:54 AM

The 3-sided die could be made of orange-peel sections like ()()() rolled up; no need to be flat faces. The 7-sided die can be SE's pic, adjusted so that the sides and ends are equally probable (i.e., somewhat longer than shown). Troy's solution works, but is it unique?

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#19
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 7:17 AM

If one relaxes the implicit assumption that the numbering on each die starts from 1, then the solution is not unique. On my 6-sided die the spots are from 2 to 7, allowing the solution 5 3-sided, 4 6-sided and 1 7-sided.

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#20
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 8:19 AM

when cylindrical 3-sided has come to rest, I will answer

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#28
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 2:11 PM

I'm curious why that was supposedly off topic.

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#9

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 1:53 AM

Thanks.

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#10

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 2:59 AM

Let the number of the three sets of dice be x, y and z. I will assume that each dice contains all the integers from 1 up to the size of the dice. In this case, and given that the dice are fair, the average number for each type of dice will be respectively: (1+2+3)/3 = 2, (1+2+3+4+5+6)/6 = 21/6 and (1+2+3+4+5+6+7)/7 = 4. This average value is the most probable. (I will check this later... but if I'm mistaken at this point, the following analysis is void!)

Then it will hold:

x + y + z = 10 and 2.x + 21/6.y + 4.z = 32 => 12.x + 21.y + 24.z = 192. After some trivial math we conclude that:

z = 10 - x - y (1)

x + y ≤ 10 (2) (as a consequence of the above)

4.x + y = 16 (3)

With inspection we see that (3) - taking into account (2) - is satisfied with the values:

(x,y) = (2,8) or (3,4) or (4,0). By using (1) we eventually get the table:

x = 2 | 3 | 4

y = 8 | 4 | 0

z = 0 | 3 | 6

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#12
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 3:24 AM

...Of course, if by saying "some of the dice" it really means "a non-zero number of dice" then the solution x,y,z = 3,4,3 is the only valid. Mathematically speaking though, I cannot see how the word "some" excludes the possibility to have no die of a certain type. In this case, the problem has the above three solutions.

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#41
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/07/2013 4:43 AM

I'd like to consider the word "some" as "at least one exists". Otherwise, I'd like to assume that none of these 3 sided, 6 sided or 7 sided dice exists, just one ball with the number 32 shall be ok.

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#13
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 3:43 AM

tkot, equation (2) should be x + y < 10, since all three dice are present (at least one of each) and a total of 10 dice.

Based on this, the only solution to equation (3) is (3,4), which leaves z as 3.

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#14
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 4:01 AM

Hi Tkot. How the [4.x+y=16] came form? Also, I'm not sure about your claim that "the average value is the most probable". Does this mean that after many trials the (e.g.) 3-sided dice will give "2" as result, for most of the times? Because this is wrong. Could you clarify that? And what's the meaning of "most probable" in this situation, anyway?

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#15
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 4:21 AM

If you have two three sided dice, the possible combinations are 1,1 1,2 1,3 2,1 2,2 2,3 3,1 3,2 3,3. The most probable value is 4, which occurs 3 of the 9 times. The next is 3 or 5 which occur 2 of the 9 times (each). Finally 2 or 6 occur only 1 time of the 9 possibilities. Two dice divided by the most probable outcome of 4 gives a 2 for each three sided dice. You can do the same analysis for the 6 sided dice (which we know the most probable outcome is 7). Ditto for the seven sided dice.

Or you can mathematically solve it as Tkot did.

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#18
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 6:47 AM

I still don't get it. Your claim is that -in the case of a 3-sided dice- the "2" is the most probable number. This is wrong. The possibility that the result is "2" is the same as for being "1" or "3", i.e. 33,33%. Unless the dice is not "fair"...

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#23
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 12:15 PM

You are correct for a single die. However, when it comes to 2 3-sided dice the most likely combined result is 4. Assigning half of that result to each individual die gives the 2.

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#36
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/02/2013 3:05 AM

Hi both

I spent some minutes to derive some formula for the sum frequency when summing the result of two dice. Let two dice with n and m number of faces respectively. If we consider the sum s = i + j (i: the outcome of the first die, and j of the second) then this sum can also be acquired with the throws:

(i-1) + (j+1) , (i-2) + (j+2) , ... (min(i-1, m-j) times) plus

(i+1) + (j-1) , (i+2) + (j-2) , ... (min(j-1, n-i) times)

therefore the number of possible appearances of this sum is:

occurrence of s = counts = 1 + min(i-1, m-j) + min(j-1, n-i)

With some little extra work we eventually get to:

| s - 1 (if s < n+1)

counts = | n (if n+1 ≤ s ≤ m+1)

| m+n - (s - 1) (if s > m+1)

This is a symmetrical function with two linear parts and a maximum at n. This maximum appears for the middle sum(s), i.e. those lying between n+1 and m+1. Per instance, if we have a 3-sided and a 4-sided die, then we get:

sum | occurences

2 | 1

3 | 2

4 | 3

5 | 3

6 | 2

7 | 1

From this point one needs to generalize when we have more than two dice, maybe 10, like in our case. What is interesting in the problem at hand is to investigate if the most probable sum of 32 can be produced by only one configuration.

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#21
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 9:49 AM

I think you ought to remove a few terms since 1,2 is the same as 2,1 and so forth.

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#22
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 11:23 AM

No, that's fine. In the first case die A rolled a 1 in the second die B rolled a one. They're different scenarios that just happened to have the same numbers.

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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 2:01 PM

ooops, I guess.

Smart as a post strikes again lol.

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#16
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 5:46 AM

First of all, the claim that the "average value is the most probable" is not proven, it's just a hunch. It comes from the feeling that the middle values can be combined with more ways than the extreme ones, therefore the sum of the "middles" will be more probable. Of course there is a chance that more than one sum can tie as the most probable ones! In this case we may have more than one solutions. I will try to investigate it if I get any free time; so far I admit defeat.

Meanwhile Autobroker gave a proof by example, but I have the feeling that is not easily generalizable for all cases.

As for the 4.x+y=16, it easily yields from the two conditions:

x + y + z = 10 and 12.x + 21.y + 24.z = 192

by eliminating z

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#17
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 6:38 AM

Thanks, for the clarification. Although I'm not convinced about the "average value is the most probable" issue, you get a GA from me.

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#31
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 5:14 PM

First of all, the claim that the "average value is the most probable" is not proven, it's just a hunch.

No, it can be proven by induction. Take 2 dice with spots from 1 to 7 and write out the table of sums, with the spots of 1 die represented by the row headers and the spots of the other represented by the column headers. The value 8 occurs once in each row. No other value occurs in each row, so all other values occur with lesser frequency. Now the 8s are in a row representing the base of a triangle, so throw away the bottom right of the square and look at the triangle. 8 is the sum of the maximum and minimum values of the triangle from 1 to 7. By inspection the base of the row for the 1 to 6 triangle is a set of 7s. We have shown that a statement (the most common value is the sum of lowest and highest values) is true for up to 7 spots and up to 6 spots. It must also be true for up to 5 spots and up to 8 spots. It is therefore true for n spots, where there is no upper limit for n.
Because there are 2 dice, the most common result for a single die is the average, obtained by dividing the value at the base of the result triangle by 2.

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#45
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/16/2013 4:17 PM

You're absolutely correct. This is the basics of probability (remember back in high school physics and math). Good job explaining this.

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#49
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/22/2013 11:43 AM

Good proof, although if you explore the concept of 'fair dice' you'll see that you can simplify your equations to determine the average to (x+1)/2. For a die of x faces, numbered 1 to x, this gives you the same result as adding all the faces and dividing by the number of faces.

This 'shortcut' has been discovered by gamers since the first release of Dungeons and Dragons, as well as the discovery that there is no such thing as truly 'fair dice.'

All dice have a bias towards a particular result, but most are so minor that they're considered 'fair enough.' Even Vegas understands the bias of dice, which is why the craps dice there are swapped out for a fresh pair as soon as the 'dealer' (don't know what they call the guy who runs the craps table) sees any wear on the so-sharp-they-hurt corners of the dice, the shooter decides he doesn't like the way they roll, or the 'dealer' just gets that something's-not-right feeling. Those 'dead dice' are still 'fair enough' for most people, they've just gone from 99.9999% unbiased to 99.999% unbiased.

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#50
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/22/2013 12:11 PM

I wonder, how many times should they throw a die, in order to ascertain that it is 99.9999% unbiased?

It is a typical statistical problem, which I have no time to solve now. I would be interested to know if one has got the answer!

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#59
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 12:42 PM

I remember coming across a program that would calculate the bias of a die after you plugged in the results of a certain (large) number of rolls.

What I found amusing was that the number required was so large (something on the order of 1,000 rolls per face, so 4,000 rolls for a d4, 6,000 rolls for a d6) that the wear and tear on the die was likely to make it biased during the evaluation, unless you rolled it on a padded felt table, but then the evenness of the padding becomes a factor in the die results.

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#11

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 3:02 AM

Too bad Tom and Ray retired; this one could be on Car Talk.

Even though there are only two equations for three unknowns, the Diophantine aspect may narrow things down to make Troy's solution unique.

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#32

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 6:42 PM

With the: 3 sided dices: average would tend to be 1.5 if we would throw it high number of times.

6 sided dice: average 3

7 sided dices: average 3.5

Answer is : 1x 3 sided for 1.5 average

+ 6x 6 sided for 18 average

+ 3x 7 sided for 10.5 average

for a total of 30

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#33
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 7:12 PM

Sorry, but your averages are incorrect. They should be 2, 3.5, and 4 respectively, as already shown in prior posts.

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#34
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Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/01/2013 7:44 PM

Yes i realized after my post when i read that prior one.

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#35

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/02/2013 12:07 AM

there are a dices 3-sided, b dices 6-sided and c dices 7-sided.

the most probable results of the dices are 2 for the 3 sided, 3.5 for the 6 sided and 4 for the 7 sided dice.

so 32= a*2 + b*3.5 + c*4 and 10=a+b+c!

the solution of these equations is/are the solution of the question (12=1.5*b + 2*c)!

Because the numbers of dices can't be broken, there are the following combinations:

b c a sum

0 6 4 32

4 3 3 32

8 0 2 32

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#38

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/03/2013 10:18 PM

A fairly easy solution to the original problem.

So what is the probability of actually getting a result of 32 when throwing all ten dice in a single throw?

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#39
In reply to #38

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/03/2013 11:54 PM

with the 10 dices in a single throw there are totally 6.001128 possibilities of combinations;with 3 times the 7 on the 3 7-sied dices there are still 11 points to throw!

There are a lot of possible throw's

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#40
In reply to #38

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/04/2013 3:51 PM

.00315317356

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#42

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/08/2013 8:32 PM

Assumptions and there are a few!!!

First, the three sided dice is a given so i think it fair to assume the practicality of building such a beast is not relevant to this discussion (although i'm sure we will find a way of doing it in the future!!).

Not being a gambler i assume "fair" means an equal chance of landing any number. Not stated but assumed is that all dice number from one.

Additionally, i assume we are counting in base 10 (or perhaps something higher than base 7).

Midpoint of 3 is 2, midpoint of 6 is 3.5 and midpoint of 7 is 4 so by taking various integer combinations of 2, 3.5 and 4 we need to get to 32. By my calcs i agree with poster number 2 i.e. (3 x 2) + (4 x 3.5) + (3 x 4) = 32.

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#44

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/16/2013 4:01 PM

Assuming you can get a fair 7 sided (perhaps similarly to how you get a fair 3 sided) and assuming numbering starts at one for each. A number of correct answers appear but at least one is missing:

Sides 3 6 7

Num 1 4 4

3 4 3 (prev, etc.)

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#46
In reply to #44

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/16/2013 11:47 PM

10 dices?

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#47

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/22/2013 11:09 AM

I had a full solution spelled out, but the blog is saying It can't process the form, so I'm going to give the short version here.

to find the 'most probable number' on a die, you add the values of the faces up and divide by the number of faces, or for a fair die of x faces, numbered 1 to x, you can shortcut it with the equation avg = (X+1)/2

So d3 = 2, d6 = 3.5, d7 = 4

Now it's just a simple factor of solving the equations: x+y+z=10 and 2x+3.5y+4z=32

And I need to rework my equations because I just realized I had used 2.5y instead of 3.5y, so my answer is off.

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#48
In reply to #47

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/22/2013 11:26 AM

It still doesn't like my proof, guess it's too long.

Short answer, three sets of dice satisfy the equations:

4d3, 0d6, 6d7

3d3, 4d6, 3d7 <- 'most correct' as it is the only one that does not allow zero to be in the set of 'some'

2d3, 8d6, 0d7

The trick to doing it fast is to eliminate all xd6 combinations that don't result in an even integer for the d6's, leaving only 0d6, 4d6, and 8d6, then it's just number crunching.

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#51
In reply to #47

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/22/2013 4:45 PM

I had a full solution spelled out, but the blog is saying It can't process the form,

This has happened to me so many times that I now routinely copy the entire response (ctrl-A, ctrl-C) and then use the browser back button to return to the blog. I then click on the reply button again, and copy the entire response (ctrl-V) into the empty box. It then gets accepted.

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#52

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 10:16 AM

Whoops - I made a big boo boo

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#53
In reply to #52

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 10:33 AM

(I had to rethink this, but I am right now)
1. A 3-sided dice can't exist, but I'm going to pretend it can.
2. The average roll of a 3-sided die is 2
The average roll of a 6-sided die is 3.5
The average roll of a 7-sided die (eh??) is 4
3. For an average of 32, an even number, the total quantity of 6-sided dice must be a multiple of 4.
4. 4 6-sided dice would have an average of 14, and 8 would have an average of 28, which is too large a number, so 4 it is.

5. With 4 6-sided dice and an average of 14, there remains an average of 18 among 6 dice. That's an average of 3 each. Half are 3-sided and half a 7-sided.

The key is calculating the average of each type of die.

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#54
In reply to #53

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 11:03 AM

if you limit your definition of 'die' to mean 'a numbered regular platonic solid of n with numbers from 1 to n,' then the number of dice are limited to d4, d6, d8, d12, and d20, for easily determined solids. (This is also the set of dice that came with D&D.) Even without going into 'odd' die shapes (such as faceted cylinders which are rolled along their axis) we can produce many 'impossible' dice with simple math:

d2: Roll a d4, 1-2 = 1, 3-4 = 2.

d3: Roll a d6, 1-2 = 1, 3-4 = 4 5-6 = 3.

d4: We have a die for that.

d5: Roll a d20, 1-4 = 1, 5-8 = 2, 9-12 = 3, 13-16 = 4, 17-20 = 5

d7: Prime number, no multiple within first 5 platonic solids.

d8: We have a die for that.

d9: No multiple within first 5 platonic solids.

d10: Roll a d20, 1-10 = use rolled number, 11-20 = subtract 10 from the roll.

d36: Roll two d6's, one red and one white, subtract one from the red die and multiply the result by 6, then add the value of the white die. (Or just make a matrix of the results with the red die across the top and the white die down the side and call the roll a d66, like the gamers do).

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#56

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 12:06 PM

How will you add outcome of odd-numbered-faceted dice? They will not have TOP face.

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#57
In reply to #56

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 12:16 PM

All dice have a 'top face' even if it is not a true planar surface parallel to the table the die is resting on. For example, with a d4, the 'top face' is the point sticking up in the air; a d4 will traditionally have the 'face numbers' near the points, so if you look at one planar side you will see three different numbers, but if you look at one corner while rotating the die so you can see all planes that meet at that corner, you will see the same number no matter which plane you are looking at.

Seriously people, a lot of this confusion can be cleared up by going to your nearest gamer or gaming store and asking to see their dice. Gamers are a friendly bunch, they don't bite, and they'd love to share their knowledge of their hobby, and by extension, their vast knowledge of dice and probabilities, with you.

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#58
In reply to #56

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 12:35 PM

All dice have a 'top face' even if it is not a true planar surface parallel to the table the die is resting on. For example, with a d4, the 'top face' is the point sticking up in the air; a d4 will traditionally have the 'face numbers' near the points, so if you look at one planar side you will see three different numbers, but if you look at one corner while rotating the die so you can see all planes that meet at that corner, you will see the same number no matter which plane you are looking at.

Seriously people, a lot of this confusion can be cleared up by going to your nearest gamer or gaming store and asking to see their dice. Gamers are a friendly bunch, they don't bite, and they'd love to share their knowledge of their hobby, and by extension, their vast knowledge of dice and probabilities, with you.

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#60
In reply to #58

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 1:27 PM

If my memory is correct, many dice (particularly a ten sided dice) do not have a numerical sequence starting at 1. This allows for a weighting of a rolled die to permit for both a wider spectrum of outcomes than the number of surfaces and for non-uniform probabilities in the outcome result.

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#61
In reply to #60

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 1:55 PM

Yes, most d10's are numbered from 0-9, with 0 taken to mean 10 by gamers, unless it's rolled in pairs to make a 'd100, then 0 means zero, unless they're both zero, in which it means 100.

Die numbering can and does vary, but for the original puzzle, the stipulation was 'fair' dice, which means a dn is numbered from 1-n.

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#62
In reply to #56

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 2:06 PM

How will you add outcome of odd-numbered-faceted dice? They will not have TOP face.

I think you need at least 4 sides for a 3 dimensional object with flat sides, such as a die, so the 3 faced die cannot even exist. I suppose that if you had a die with no top side, such as a 4 sided one, then logically, you would have to use the bottom side.

Now here's my question that I hope someone will be kind enough to answer
. . . I see 4 blank posts near the end of this thread, and I am wondering if they are in fact blank or perhaps I can't see them because I am a newbie. I checked the FAQ, but I found nothing regarding this. Are people posting blank replies?

Thanks,
Asheville, North Carolina

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#63
In reply to #62

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 2:31 PM

I 'm new here myself, and I just discovered the reason:

The 'blank posts, are flagged as 'off-topic,' and therefore are 'hidden' from the main conversation.

At the top right corner of every post there is a little magnifying glass with a red minus sign on it if the post is 'expanded' a green plus sign if it is 'hidden.'

Clicking on the magnifying glass flips its state and conceals/reveals the post it is connected to.

Hope this helps.

(This is technically an 'off-topic' post, but if I flag it as such, then Ashenville won't know how to read it until after ze has read it, creating a logical paradox.)

(And that wasn't a typo, 'ze is a neuter pronoun I learned from an old sci-fi author, I forget which one. He, him, his; she, her, hers' ze, zir, zis. It's designed to 'waffle' between sounding masculine, ze rhymes with he, zir rhymes with her, and zis is supposed to have a mangled vowel to sound like it's trying to rhyme both his and hers, but generally it's spoken to rhyme with his.)

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#64
In reply to #63

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 2:39 PM

Thank you, adreasler!

BTW, you kinda blew my mind with the Ze stuff. I like to understand what other people write, but I don't think I am going to use that particular pronoun.

It's funny, English is very gender non-specific compared to other languages. Those pronouns are almost all we have to differentiate the genders.

I s'pose I have to mark this as off topic now.

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#65
In reply to #64

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 3:20 PM

I figured the 'ze' pronoun would be better than the other 'gender neutral' pronoun I remembered from a George Carlin routine. It was a contraction of 'he, she, or it' which spelled out looked somewhat rude, and if pronounced sounded VERY rude.

(For those that don't remember the routine, drop the e's, replace then with apostrophes, and jam it all together, the second apostrophe is silent, the first is pronounced like 'or.')

George Carlin had a way with words, especially seven particular ones...

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#66

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/29/2013 6:30 PM

My thoughts are that a 3 sided dice would not be possible.

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#67
In reply to #66

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/30/2013 1:06 AM

try an electronic random number creator/calculator

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#68
In reply to #66

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/30/2013 1:23 AM

Only in the sense that "dice" is plural, making a single "dice" impossible. Please read the whole thread for a few pictures of how a 3-sided die can be done.

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#69

Re: A Dicey Problem: Newsletter Challenge (October 2013)

10/30/2013 10:58 AM

There are more than two 'equations' for the three unknowns.

a, b, and c are integers

a>0

b>0

c>0

2a + 3.5b + 4c = 32

a + b + c = 10

Thus when you interpret 'some' as one or more, you get a unique solution. This was a fun exercise, with an embarrassing number of mistakes along the way. Thanks!

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#70

Re: A Dicey Problem: Newsletter Challenge (October 2013)

11/04/2013 6:23 PM

The expected value of each die (assuming that each face is equally likely to come up) is the sum of the faces divided by the number of faces:

3 sided: E3=2 = (1+2+3)/3

6 sided: E6=3.5 = (1+2+3+4+5+6)/6

7 sided: E7=4 = (1+2+3+4+5+6+7)/7

Let N3, N6, and N7 be the number of 3 sided, 6 sided, and 7 sided die in the bag.

N3 + N6 + N7 = 10 (1)

The Expected Value of all 10 die is 32

E = 32

It is also equal to the sum of the expected values of each die times the number (N3, N6, or N7) of that die.

E = 32 = N3 * E3 + N6 * E6 + N7 * E7

or

2 * N3 + 3.5 * N6 + 4 * N7 = 32 (2)

So we have 2 equations and 3 unknowns and may have multiple solutions.

But only 0,1,2,...,10 are possible answers (no fractions, no negative values, and N6 must be even or 3.5 * N6 won't be an integer)

if we solve (1) for N3 and substitute into (2) and simplify we get:

N3 = 10 - N6 - N7

and

3 * N6 + 4 * N7 = 24

so

N6 = 8, N7 = 0, N3 = 2 is a solution

and

N6 = 0, N7 = 6, N3 = 4 is a solution

as well as the answer given

N3 = 3, N6 = 4, N7 = 3.

Looking again at

2 * N3 + 3.5 * N6 + 4 * N7 = 32 (2)

We already know solutions for N6 = 0 and N6 = 4, and that N6 = even, which leaves

N6 = 2 or 6 (N6 = 8 is a solution when N7 = 0)

N6 = 2 won't work because that would give us

2 * N3 + 4 * N7 = 25

or 2*(N3 + 2 * N4) = 25

and the left side must be even. Similar logic eliminates N6 = 6 as a solution.

So the only possible solutions are for N6 in {0,4,8}, and there is only one possible solution in each case because N6 = {0,4,8} is the third of our 3 equations with 3 unknowns.

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#71
In reply to #70

Re: A Dicey Problem: Newsletter Challenge (October 2013)

11/05/2013 9:26 AM

Very good, I had a similar solution to post, but I gave up when the forum kept refusing to accept tthe reply. I didn't know about the time limit to make an entry and thought it was being rejected for exceeding a character count limit.

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#72
In reply to #71

Re: A Dicey Problem: Newsletter Challenge (October 2013)

11/05/2013 9:43 AM

My reply was rejected at first as well, then I logged out and back in and resubmitted.

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