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Wins 2 - wins all (Posted on 2010-05-06) Difficulty: 3 of 5
Three backgammon players of equal game skills compete for a prize.
The prize will be awarded to the winner of two games in a row.
A and B, following a drawing, play the first game, then the winner will face C.
Next game, if needed, will be by C and the player who lost the 1st game and so on.
Determine a priori chances of winning for each of the 3 players, assuming Lady Luck treats them without discrimination.


Rem: There are no draws in backgammon.

No Solution Yet Submitted by Ady TZIDON    
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analytical solution | Comment 1 of 5
listing who can win in each round:
round 1: nobody
round 2: a,b
round 3: c (2 ways)
round 4: a,b
round 5: a,b
round 6: c (2 ways)
round 7: a,b
round 8: a,b
round 9: c (2 ways)
....
so it can be seen that A and B have equal chances of winning so we need only compute C's chances.  The probability of C winning is equal to
2*(1/2)^3 + 2*(1/2)^6 + 2*(1/3)^9 + .....
which is equal to 2/7
thus the odds for A and B are (1- (2/7))/2 = 5/14
which can also be confirmed by computing the infinite sum
(1/4) + (1/2)^4 + (1/2)^5 + (1/2)^7 + (1/2)^8 + ....
thus confirming that somebody will eventually win
thus the odds are
A: 5/14 or approximately 35.7%
B: 5/14 or approximately 35.7%
C: 2/7 or approximately 28.6%


  Posted by Daniel on 2010-05-06 12:46:18
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