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Am I doing this right? I split the problem up into the cases of 2 same, 3 same, 4 same, but I feel like something special has to be done for 2 of the same, because what if there are 2 pairs (like two 3's and two 4's)?

This is what I have:

For 2 of the same: $5\times 5\times 6\times {4\choose 2}=900$

For 3 of the same: $5\times 6\times {4\choose 3}=120$

For 4 of the same: $6\times {4\choose 4}=6$

Combined: $900+120+6=1026$

Total possibilities: $6^4=1296$

Probability of at least 2 die the same: $\frac {1026}{1296}\approx 79.17$%

Confirmation that I'm right, or pointing out where I went wrong would be appreciated. Thanks!

Sorry if the formatting could use work, still getting the hang of it.

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    $\begingroup$ It's much easier to calculate the probability that they are all different, then subtract from $1$. $\endgroup$ – Thomas Andrews May 1 '15 at 0:46
  • $\begingroup$ You have overcounted, because you haven't considered the case where you get two pairs. $\endgroup$ – Thomas Andrews May 1 '15 at 0:48
  • $\begingroup$ I should've thought to do the 1-chance of all different, that was stupid of me. Can you show me how to deal with the over-counting? That was my concern and reason for posting it. But even having you confirm it, I'm not sure how it's over-counting, because 2 pairs is still a case where at least 2 dice show the same number. Thanks again! $\endgroup$ – Bob May 1 '15 at 0:50
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You answer for "exactly two the same" counts some cases twice - when you get two pairs ($4545$, for example.)

The case of $2$ the same the others different counts to $6\cdot 5\cdot 4\cdot\binom{4}{2}=720$.

The case of two pair is $\binom{6}{2}\cdot\binom{4}{2}=90$.

Those two values add up to $810$, and you over-counted by $90$ - that is, you counted each "two pair" result twice.

This gives a total of $720+90+120+6=936=1296-360$.

This sort of problem is much easier to do by calculating the probability of the opposite (that they are all different) and subtract that from $1$. The probability that they are all different is $\frac{6\cdot5\cdot 4\cdot 3}{6^4} = \frac{360}{1296}$.

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  • $\begingroup$ I originally had $6\cdot 5\cdot 4\cdot\binom{4}{2}=720$, but wasn't sure if the change to $6\cdot5\cdot5\cdot\binom{4}{2}$ instead would work for pairs. I couldn't figure out how to represent the case of 2 pair, or figure out if it was counting more than once. $\binom{6}{2}\cdot\binom{4}{2}$. Couldn't figure that out. Thanks. $\endgroup$ – Bob May 1 '15 at 1:10
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I split the problem up into the cases of 2 same, 3 same, 4 same, but I feel like something special has to be done for 2 of the same, because what if there are 2 pairs (like two 3's and two 4's)?

Yes. Your cases are

  • 1 quadruplet: $\binom{4}{4}\times \binom{6}{1}$ arrangements.
  • 1 triplet, 1 singleton: $\binom{4}{3,1}\times \binom{6}{1}\times \binom{5}{2}$ arrangements.
  • 1 doublet, 2 singletons: $\binom{4}{2,1,1}\times \binom{6}{1}\times \binom{5}{2}$ arrangements.
  • 2 doublets: $\binom{4}{2,2}\times \binom{6}{2}$ arrangements.

And the complement is the remaining case of 4 singletons: $\binom{6}{4}$ arrangements.

Finally, the total space is, of course, of $6^4$ arrangements.


NB: $\dbinom{4}{2,1,1} = \dfrac{4!}{2!\,1!\,1!}$ is a multinomial coefficient.

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  • $\begingroup$ Think you messed up somewhere, because I'm pretty sure you have the same over-counting that I did (maybe even more). Not sure if I'm right, or where the mistake is if I am. Edit: Pretty sure it's on the 1 doublet, 2 singleton. Because what you have for that part comes out to 1440, which is more than even the total number of possibilities. Am I mistaken? $\endgroup$ – Bob May 1 '15 at 2:00
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The probability that at least two are the same is one minus the probability that all four are different. This is \begin{align} & \Pr(\text{2nd differs from 1st}) \\[8pt] \times {} & \Pr(\text{3rd differs from 1st and 2nd} \mid \text{2nd differs from 1st}) \\[8pt] \times {} & \Pr(\text{4th differs from 1st, 2nd, and 3rd}\mid \text{first three differ}) \\[10pt] = {} & \frac 5 6 \cdot \frac 4 6 \cdot \frac 3 6 = \frac 5 {18}. \end{align}

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  • $\begingroup$ That's definitely the easiest, and Thomas noted that. It was easy to solve once he reminded me of it. I guess what I wanted after that was an explanation for why what I was doing (long-ass method or not) wasn't working. Thanks for simple and clean explanation of the easy method though :) $\endgroup$ – Bob May 1 '15 at 3:10
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Indeed, when you count the combinations with pairs, you counted twice each combination that has two pairs.

There are 3 * 5 * 6 = 90 such combinations of 2 pairs. So, for 90 combinations, you counted the combination twice in your total of 900.

The number of combinations having one or two pairs is then 900 - 90 = 810

The total number of combination having at least two identical dice is then (810 + 120 + 6) / 1296 = 72.2222%

Numerical check: the easy method gives: 1 - (5/6 * 4/6 * 3/6) also equals to 72.2222%

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