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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 $X_2$.2011-11-12
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    Is this related to quantum mechanics? If not, why does it matter that the dice are indistinguishable?2011-11-12
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    @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.)2011-11-12
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    @Dinesh: I know what convolution is, but does that work for indistinguishable dice as well as distinguishable dice?2011-11-12
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    @Mehrdad: I interpreted the word "indistinguishable" in your question to mean that the two dice are identical. Apart from this, I can't see how indistinguishability will change the solution. Distinguishable vs Indistinguishable makes a difference in Combinatorics when counting the number of ways of picking k objects out of a set of n objects. But if you have two dice tosses, you have two values whether or not the dice are indistinguishable.2011-11-12
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    @Dinesh: Yes, it means they are identical. Which means you can't distinguish them. So the phrase "the probability of the first one coming up 3" doesn't make sense because once they are rolled, you can't distinguish the "first" one from the "second" one. I *know* this makes a difference in the outcome, but what I would like to know is what the difference is, and why (I don't know how to calculate it properly).2011-11-12
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    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?2011-11-12
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    @DougChatham: +1, That's a brilliant observation, and... frankly, I don't know how to answer it. I feel like it starts getting into quantum mechanics (observation affects the outcome?) but I don't know enough about it to make a call. (Having read *The Quantum Zoo*, though, I know for a fact that distinguishability at least makes a difference in quantum mechanics.) That said, I'm also hesitant to rely on common-sense reasoning like this -- I'd rather see a nice description somewhere that explains mathematically how indistinguishability does (or does not) affect the outcome.2011-11-12
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    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.2011-11-12
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    Mehrdad, it would seem your best strategy is to print all this out, answers and all, then show that to your professor or teaching assistant and ask for comments. There is no way for you to tell us every word your professor said, hence a certain amount of mindreading in the answers. However, the reverse is readily accomplished with a printout.2011-11-12
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    @Will Jagy: Will definitely consider doing that, thanks for the suggestion. :)2011-11-12
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    At the same time, it would have helped if you had identified your textbook.2011-11-12
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    @WillJagy: It's a reader, not a textbook.2011-11-12
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    If you're interested in the quantum mechanics aspect of this, you might want to take a look at [this Wikipedia article](http://en.wikipedia.org/wiki/Identical_particles), particularly [this section](http://en.wikipedia.org/wiki/Identical_particles#Statistical_properties). Note also that what is usually described as statistical effects of indistinguishability can be less mysteriously and with fewer philosophical complications be described as statistical effects of the symmetry or anti-symmetry of states.2011-11-13
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    @joriki: That's a great read, thanks for providing the link. (I started reading about the [Gibbs paradox](http://en.wikipedia.org/wiki/Gibbs_paradox) afterward, which is also interesting.)2011-11-13
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    @Mehrdad: We will not delete this, as there are already two answers written by other people.2011-11-13

4 Answers 4

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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...2011-11-12
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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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    No, my professor was not screwing with me. Yes, I have indeed read the book (what made you think I hadn't?), which mentions **nothing** about this topic. And please, please, don't assume that I wasn't smart enough to apply "common sense". The entire reason I asked it is that it's *not common sense*. It is *not* "obviously sensible" to draw a tiny little dot, because doing so violates the problem. Unless you have a credible reference supporting your claim, please don't assume the professor was making a mistake. His answer makes complete sense to me; it's just the calculation I'm asking about.2011-11-12
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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?2011-11-12
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    @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.2011-11-12
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    @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?2011-11-12
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    @Dinesh: I don't know. (That's why I'm asking!) What are they, and why?2011-11-12
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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.