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I was thinking about how a certain probability would be calculated and the answer is eluding me. It has been a number of years since I have taken a statistics and probabilities class so I am a little rusty.

Anyway, to jump right into my question: How would one figure out the probability of rolling 3 dice and getting a 3 6's. The twist is, dice rolled on a 6 remain the 6 for the next round.

So for example, if I roll 3 dice and get a 1/3/6 my next roll would be the first two dice as I already got a 6 and that is removed/remember by the game.

Also, how would one go about figuring out how many turns of the game one would expect before they attained 3 6's.

I know that for rolling 3 sixes at once, I could call the first roll X and the second roll x(1-x) such that on the nth round I would have x(1-x)^{(n-1)}

Knowing that I came up with [tex]E[/tex]ix(1-x)^{(i-1)}when i=1 to +[tex]\infty[/tex]

Then solve the limit to get the expectation of rounds =x(1/x^{2})

This wont work however for a game where a roll of 6 is removed, just the expectation of rolling all 6's. I am confused on how to start to setup the equation for this new game.

Any help would be much appreciated:rofl:.

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# A complex dice game (at least to me)

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