Question

Alice rolls a pair of fair six-sided dice until a sum of 6 appears for the first time. Independently, Bob rolls two fair six-sided dice until he rolls a sum 7 for the first time. Find the probability that the number of times Alice rolls her dice is equal to or within one of the number of times Bob rolls his dice.

Answer 1

Let X be the required number of rolls for alice and Y be the required number of rolls for Bob.

Each observation of a pair of dice has probability = 1/6 x 1/6 = 1/36 For sum of 6 required possibilities are (1,5) (5,1) (2,4) (4,2) (3,3)

So probability of success for alice is 5/36   [as 5 possibilities]

And clearly X follows Geometric distribution with parameter 5/36

P(X = k) = (1 - 5/36)k-1 x (5/36) [as we require k-1 failiures and then finally a success]

Now for sum of 7 required possibilities are (2,5) (5,2) (3,4) (4,3) (1,6) (6,1)

So probability of success for Bob is 6/36 [as 6 possibilities]

And clearly Y follows Geometric distribution with parameter 6/36

P(X = k) = (1 - 6/36)k-1 x (6/36) [as we require k-1 failiures and then finally a success]

Now

Becuase of independence. So,

By infinte sum of geometric progression,

Now,

And,

So the required probability is

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