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Two identical dice are thrown simultaneously. Find the probability of getting a $3$ on one of the dice while a $2$ on the other.

I'm pretty sure the answer is $\frac{2}{36}$ but my friend says the answer should be $\frac{1}{21}$.

He argues that as the dice are identical, so sample space comprises only $21$ combinations.

I'm unable to explain it to him but I don't agree with his answer of $\frac{1}{21}$.

Can someone tell me the correct answer and provide a short explanation?

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  • $\begingroup$ The dice don't know they're identical, so the sample space isn't smaller because of that. There are still two ways for 2 and 3 to appear... $\endgroup$ – colormegone Jun 17 '15 at 5:06
  • $\begingroup$ The dice are identical - presumably they are also standard unbiased six-sided dice with faces numbered $1$ to $6$. $\endgroup$ – Mark Bennet Jun 17 '15 at 6:16
  • $\begingroup$ No two dice are ever "identical", in the sense that they're always in different locations. Name the die that started off over here "A" and name the one that started over here "B"… Now, it's easy to see that "A gets a 3 and B gets a 2" is different than "B gets a 3 and A gets a 2". $\endgroup$ – Akiva Weinberger Jun 17 '15 at 7:54
  • $\begingroup$ Unfortunately, it is easy to be misled by the thought that the two dice are "the same". But the situation is not any different in terms of outcomes from throwing one die, and then throwing the same one later. One can think of the two dice (or any number of dice, coins, etc.) as using the same random number generator "twice in the same time interval", rather than in subsequent intervals. $\endgroup$ – colormegone Jun 17 '15 at 16:09
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There might be 21 possible outcomes ( (1,1), (1,2),....(6,6) ) , but they don't have equal probability, 1+1 can only occur in one way, but 2+3 can occur two different ways. There are 6 ways to throw (1,1), (2,2), (3,3), etc, and 15 different pairs (1,2)/(2,1), (1,3)/(3,1), etc, making 36 possible combinations. We can get a 2 and 3 in two different ways, so the probability is 2/36 as you say.

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There are 36 possible outcomes. If it doesn't matter the order of the die, then the probability is $2* (1/36)$ ways it can happen. If it has to be 2 first, then 3 next, then it is $1/36$.

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The odds of getting a 3 on a die is 1/6 The odds of getting a 2 on a die is 1/6

There are two possibilities for a 3 on one and a 2 on the other: 1. Die #1 is 3 and Die #2 is 2 2. Die #1 is 2 and Die #2 is 3

2 possibilities * 1 in 6 * 1 in 6 = 2/6/6=1/18 (or 2/36)

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As the outcomes of dices should be $2$,$3$. So $2$ on one dices and $3$ on another as dices are identical we can't distinguish them so only one way that is $(2,3)$. Now, the sample space for this will be $21$ not $36$ because the dices are identical, i.e. $(1,2)$ is not different from $(2,1)$ etc. So sample space = $21$ and probability = $1/21$.

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    $\begingroup$ Please remember that clarity is important when attempting to explain a concept to someone. Consider revising your post to use cleaner punctuation and formatting. $\endgroup$ – Chantry Cargill Oct 8 '16 at 19:34

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