Expected Number of Rolls to Throw Away Both Dice?

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SUMMARY

The expected number of rolls to throw away both dice when rolling two dice simultaneously is derived from the geometric distribution. Each die is rolled until a 6 appears, at which point it is discarded. A Monte Carlo simulation suggests the expected number of rolls is approximately 8.7. To analytically confirm this, one must compute the expected value of the maximum of the two independent geometric random variables representing the rolls until a 6 is obtained.

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Homework Statement


You roll two dice at the same time. Each time you get a 6 on a die you should throw it away and roll the other one. Otherwise, you keep rolling both of them. The game is over when you throw away both dice. What is the expected number of times you roll?


Homework Equations



Not sure.

The Attempt at a Solution



I tried defining a random variable called Y that has the form

<br /> Y = \mathbf{T} \cdot \mathbf{X}<br />

This is the dot product between random vectors X and T,
where X = (X1 X2 X3) and T = (T1 T2 T3). Here T1 has value 1 if there's no 6 and zero otherwise. T2 has value 1 if there's one 6 and zero otherwise. T3 has value 1 if there are two sixes and zero otherwise. I'm thinking the X_i should be Bernoulli random variables, but am not sure. A monte-carlo simulation gives me 8.7, and I think this is right, but I have to show it analytically.

Thanks.
 
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The number of rolls it takes each die to get a 6 is http://en.wikipedia.org/wiki/Geometric_distribution" . You just need to compute the expected value of the maximum of the two times, which is easy by conditioning. (It's actually easier than that.)
 
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