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Alice and Bob take turns in rolling a fair dice. Whoever gets "6" first wins the game. Alice starts the game. What are the chances that Alice wins?

Jan 14, 2020

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The probability that Alice wins the first time is  1/6.

If Alice doesn't win and Bob doesn't win and then Alice wins is  [(5/6)(5/6)](1/6)  =  (25/36)(1/6)

If Alice wins on the next round:  [(5/6)(5/6)][(5/6)(5/6)](1/6)  or (25/36)2(1/6)

If Alice wins on the next round:  [(5/6)(5/6)][(5/6)(5/6)][(5/6)(5/6)](1/6)  =  (25/36)3(1/6)

etc.

We have:                                                                  (1/6) + (25/36)1(1/6)+ (25/36)2(1/6)  + (25/36)3(1/6) + ...

If we replace (25/36) with x, we have:                      (1/6) + (1/6)x + (1/6)x2 + (1/6)x3 + ...

Factoring:                                                                  (1/6)[ 1 + x + x2 + x3 + ... ]

But, the sum of 1 + x + x2 + x3 + ...  is  1 / (1 - x),

So we have                                                                (1/6)[ 1 / (1 - x) ]

Since  x = 25/36                                                         (1/6)( 1 / (1 - 25/36)

(1/6)( 1 / (11/36) )

(1/6)(36/11)

=  6/11

Jan 15, 2020