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?
+23246 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
