# Tossing two indistinguishable dice/coins/whatever

If I toss two indistinguishable dice, what is the distribution for the sum of the two numbers I get?

Why?

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Let $X_1,X_2$ be the random variables denoting the outcome of your tosses. Then, $X_1$ and $X_2$ are independent and identically distributed and you want the distribution of $X_1+X_2$. The key idea you want to look up is "convolution" of the distributions of $X_1$ and $X_2$. – Dinesh Nov 12 '11 at 6:19
Is this related to quantum mechanics? If not, why does it matter that the dice are indistinguishable? – joriki Nov 12 '11 at 6:20
@joriki: Not necessarily (but maybe?), but my professor (discrete math) said it matters whether the dice are distinguishable or not, so I'd like to understand why. (He never got around to explaining it.) – Mehrdad Nov 12 '11 at 6:21
So a pair of dice, on red and one green, will work differently for a colorblind person that they would for someone with normal color vision? – Doug Chatham Nov 12 '11 at 11:35
It is becoming increasingly obvious that the OP is not after a mathematical answer (which has been provided several times on this page) to the (trivial) question asked but rather after an answer to some worn-out paradoxes associated to indistinguishability in quantum mechanics. Two remarks are in order. First, QM applies to the behaviour of matter and energy at the subatomic level, thus, not to actual dice or coins as we know them (but maybe to whatever). Second, the QM question is not in the scope of this forum. – Did Nov 12 '11 at 13:37

The probaility of throwing a total of n with two dice is $\frac{n-1}{36}$ if $n\leq7$, $\frac{13-n}{36}$ if $n>7$ and 0 otherwise. The probability is proportional to the number of ways of expressing n as the sum of 2 numbers in the range 0-6.

With x dice, the proability of a total of n if $x\leq n \leq \lfloor{ \frac{7x}{2}}\rfloor$, if the entry a is in the n-x+1 th row and xth diagonal of Pascal's triangle, is $\frac{a}{6^{x}}$. For $\lceil{ \frac{7x}{2}}\rceil \leq n \leq 6x$, the probability is the same as for $7x-n$.

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That's certainly true for distinguishable dice, but are you sure it applies to indistinguishable ones? My prof. was saying they are different... – Mehrdad Nov 12 '11 at 6:54

Mehrdad, Angela is correct. Your professor was screwing with you, possibly in an effort to get you to think for yourself. And please, please, read the book. Carefully. I have rarely let a false idea hold past the end of a class, but some people do it.

If two dice cannot be distinguished, throw them, say 1000 times, taking careful notes of what happens, how many times each total holds.

Now, get some paint, put a tiny green dot on one die, a tiny red dot on the other. Throw them another 1000 times. The proportions of 2,3,4,5,6,7,8,9,10,11,12, are very similar to the first 1000 times. However, it is obviously sensible to draw a square, one row each for the red die, (1-6), one column each for the green die (1-6). In the square you have outlined, put the sum in each box. One may now simply count, there are 36 things that can happen, 6 of them total to 7, so the probability of throwing a 7 is 1/6. The probability of throwing an 8 is 5/36, and so on.

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@Mehrdad: The point Will is trying to make about the red/green dots is a thought experiment to help you see the equivalence between the case with the indistinguishable dice and the distinguishable dice. To put the point another way: is there a difference (probabilistically) between rolling the same die twice and computing the sum of the rolls and rolling two distinguishable (with red/green dots) dice once each and computing their sum? – Dinesh Nov 12 '11 at 7:46
@Dinesh: Yes, there is a difference, because the sample spaces are different. If you roll the same die twice, your sample space is {(H, H), (H, T), (T, H), (T, T)}, while if you roll two identical dice simultaneously, your sample space is {HH, HT, TT}. (HT and TH don't count twice in the set, because there is no difference between them -- they cannot be distinguished in the outcome.) The trouble I'm having is how to go about calculating the result -- does that mean they all occur with the same probability? I don't know. – Mehrdad Nov 12 '11 at 7:49
@Mehrdad: Lets say T=0, H = 1. You are right that the sample spaces of the rolls are different in the two cases. But the question says we are interested in only the sum of the rolls in which case the sample space is {0,1,2} in both cases. What are their probabilities in the two cases? – Dinesh Nov 12 '11 at 7:58
@Dinesh: I don't know. (That's why I'm asking!) What are they, and why? – Mehrdad Nov 12 '11 at 8:00

You can construct a table of the probability of throwing a particular pair of numbers like this

    Larger  1       2       3       4       5       6
Smaller
1           1/36    2/36    2/36    2/36    2/36    2/36
2           0/36    1/36    2/36    2/36    2/36    2/36
3           0/36    0/36    1/36    2/36    2/36    2/36
4           0/36    0/36    0/36    1/36    2/36    2/36
5           0/36    0/36    0/36    0/36    1/36    2/36
6           0/36    0/36    0/36    0/36    0/36    1/36


Then the probability of throwing a particular total is the sum of the probabilities on a diagonal corresponding to that total. For example the probability of a total of 4 is $\frac{0}{36}+\frac{1}{36}+\frac{2}{36} = \frac{1}{12}$. This will give Angela's results.

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I think that the professor uses the term indistinguishable because he may think that a roll of (4,3) is different from a roll of (3,4) when the dice are distinguishable but not so when they are indistinguishable. But it should affect the distribution of the sum of the numbers on the dice.

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