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For dices that we cannot distinguish we have learned in class, that the correct sample space is $\Omega _1 = \{ \{a,b\}|a,b\in \{1,\ldots,6\} \}$, whereas for dices that we can distinguish we have $\Omega _2 = \{ (a,b)|a,b\in \{1,\ldots,6\} \}$.

Now here's the apparent paradox: Suppose we have initially two identical dices. We want to evaluate the event that the sum of the faces of the two dices is $4$. Since $ 4=1+3=2+2$, we have $P_1(\mbox{Faces}=4)=\frac{2}{|\Omega_1|}=\frac{2}{21}$. So far so good. But if we now make a scratch in one dice, we can distinguish them, so suddenly the probability changes and we get $P_2(\mbox{Faces}=4)=\frac{3}{|\Omega_2|}=\frac{3}{36}=\frac{1}{12}$ (we get $3$ in the numerator since $(3,1) \neq (1,3$)).

Why does a single scratch change the probability of the sum of the faces being $4$ ?

(My guess would be that either these mathematical models, $\Omega _1,\Omega _2$, don't describe the reality - meaning rolling two dices - or they do, but in the first case, although the dices are identical we can still distinguish them, if we, say, always distinguish between the left dice and the right, so applying the first model was actually wrong. But then what about closing the eyes during the experiment ?)

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The error is in the first solution: if one uses $\Omega_1$ as sample space then the weight of $\{1,3\}$ must be twice the weight of $\{2\}$ since $\{1,3\}$ corresponds to two elementary events while $\{2\}$ corresponds to a unique elementary event. In short: there are other probability measures on a finite set than the uniform one. –  Did Mar 6 '12 at 17:28
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The first sample space is a perfectly valid one, whether the dice are distinguishable or not. It happens to be not particularly useful for computing our probabilities, since the outcomes in it are not all equally likely. Oddly enough, a sample space of this type in which the outcomes are equally likely turns up in Physics (Bose-Einstein statistics). –  André Nicolas Mar 6 '12 at 17:31
    
@AndréNicolas But shouldn't change the fact that we are using indistinguishable dice (vs. distinguishable ones) somehow change the probability of the outcome ? –  temo Mar 6 '12 at 18:05
    
@If you are also changing the definition of outcome, sure. But suppose that you are blindfolded. Does that change the probability that when you toss two dice, distinguishable or not, we get a sum of $3$? Often, to correctly compute probabilities, it is useful to imagine that you have made identifying scratches on the dice. –  André Nicolas Mar 6 '12 at 18:14
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Suppose that the dice are colored red and green respectively and thus are distinguishable to you but not to me since I am red-green colorblind. Should the probabilities that we compute be different or the same? What if we try an empirical test of rolling the dice $1680$ times to see how well our model approximates the real world? Will we have rolled $4$ approximately $1680\times\frac{2}{21}=160$ times as per your model or $1680\times\frac{1}{12}=140$ times as per my model? –  Dilip Sarwate Mar 6 '12 at 19:19

2 Answers 2

up vote 4 down vote accepted

In the case where you don't distinguish between the dice, it is fine to use a sample space that consists of unordered pairs. But the price you pay for that is that the elements in $\Omega_1$ not equally probable.

In particular, it is not valid to compute probabilities simply by counting relevant elements of $\Omega_1$ and divide by its total cardinality $|\Omega_1|$.

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But what happens if I imagine the outcome of rolling my two indistinguishable two dices as a ball with two numbers written on it the opposite ends of an axis through the sphere, corresponding to the numbers of the faces. Shouldn't each rolling of two indistinguishable dices correspond to pulling on ball out of an urn with $|\Omega_1|$ balls ? If that would be the case the probability would still be $\frac{2}{21}$. (Sorry if I'm insisting, but I didn't quite get why the elements in $\Omega_1$ aren't equally probable). –  temo Mar 6 '12 at 17:58
    
Does your ball produce outcomes with the same probabilities as a pair of dice? That's an empirical question, not a mathematical one. The probabilities for dice are different because they are, as an experimental fact, different. One way to express this is that Nature can tell the dice apart even if you decide not to. –  Henning Makholm Mar 6 '12 at 18:04
    
So this last comment implies that there's nothing logically inconsistent about the first model, so the second model is taught because it is backed up empirically not because it's any better logically? –  user9352 Jun 10 '12 at 17:39
    
@user9352: Correct -- there's nothing logically inconsistent about the idea of dice that behave in such "strange" ways. They just don't seem to occur in reality. –  Henning Makholm Jun 10 '12 at 22:38

The correct probability distribution for dice treats them as distinguishable. If you insist on using the sample space for indistinguishable dice, the outcomes are not equally likely. However, if you are doing quantum mechanics and the "numbers" become individual quantum states, indistinguishable dice must be treated using either Fermi or Bose statistics, depending on whether they have half-integer or integer spin.

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