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July 22nd, 2010, 11:11 AM   #1
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Ways to roll 6 dice such that there are exactly 4 unique #s?

This one has me stumped. I've written a monte-carlo simulator, and probability of rolling such a combination is something like 50.1%

My thinking is that there are C(6, 4)=15 ways to have 4 unique numbers, and for each of these there are C(6+4-1, 6)=84 ways to have 6 dice land on these numbers. but this gives 1260 ways, which is completely unreasonable when there are C(6+6-1,6)=462 ways to roll 6 dice.

I just can't wrap my brain around this one. Any ideas?
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July 22nd, 2010, 01:12 PM   #2
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Re: Ways to roll 6 dice such that there are exactly 4 unique #s?

Hello, nice to meet you,

if we chose 4 numbers : C(6,4) to be unique, it remains to choose among 2 numbers, a number to be repeated, shouldn't we multiply by 2 ?
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July 22nd, 2010, 01:18 PM   #3
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Re: Ways to roll 6 dice such that there are exactly 4 unique #s?

There are C(6,4) ways to pick four numbers a,b,c,d. Then you either (i) pick one of the four and treble it to get "aaabcd", for which there are 6!/3! permutations; or (ii) pick two of the four and double them to get "aabbcd", for which there are 6!/(2!2!) permutations. Add your results for (i) and (ii), then divide by 6^6. You will get something just a bit over 50.1 %.
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