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15 fair 6-sided dice are rolled. What is the probability that all 6 faces show twice? That is: 11, 22, 33, 44, 55, and 66. The remaining 3 dice can be any number. Would appreciate any help. Thank you.

 Mar 2, 2020
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I will give this a try !

 

112233445566111 =15!/(5!*32)=340,540,200 x 6 =2,043,241,200......................(1)
112233445566112 =15!/(4!*3!*16)=567,567,000 x 30=17,027,010,000..............(2)
112233445566123 =15!/(3!3!3!2!2!2!)=756,756,000 x 20 =15,135,120,000........(3)

 

We add the 3 sums above =2,043,241,200 + 17,027,010,000 + 15,135,120,000 =34,205,371,200 / 6^15 =0.072748 x 100 =7.2748%

Note: somebody should check this, please. Thanks. [The multiplications by 6, 30, 20 comes from counting the number of combinations for the last 3 remaining dice, with repeats allowed, and was arrived at by: (6 + 3 -1) C  3 =8C3 =56.

 Mar 2, 2020

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