A Dicey Problem: Newsletter Challenge (October 2013)

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

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

(sorry!)

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

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.}

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!

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

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

a coin

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)

Yes, that's a five....

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?

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.

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

I'm curious why that was supposedly off topic.

Thanks.

...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.

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.

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.

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?

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.

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"...

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.

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 = count_s = 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)

count_s = | 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.

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

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.

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

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

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.

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

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.

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.

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

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

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

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

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?

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

.00315317356

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.

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.)

10 dices?

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.

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.

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.

Whoops - I made a big boo boo

(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.

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).

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

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

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.)

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

try an electronic random number creator/calculator

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.

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!

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.

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.

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