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BGonline.org Forums
Roulette puzzle
Posted By: Ian Shaw In Response To: Roulette puzzle (Timothy Chow)
Date: Wednesday, 28 April 2010, at 9:42 a.m.
My gut instinct is that the answer is D; the probability you will be poorer is 50% to 75%. Casinos want you to lose money, but slowly enough for you to have a good time doing so.
To come out ahead you need to win $35 four times or more to recoup the $108 laid out. To calculate the probability of success, I started with the chance of winning four straight games and built up from there.
The probability of success in the first four rolls is simply p4, where p is 1/38.
To succeed on the fifth roll you must have won three times in the first four rolls, then win the fifth roll. The probability of three wins and one loss is p3(1-p) and there are 4 combinations of these (lwww, wlww, wwlw, wwwl). So the probability of success on the fifth roll is 4p3(1-p)p = 4p4(1-p).
To succeed on the sixth roll you must have won three times in the first five rolls, then win the sixth roll. The probability of three wins and two losses is p3(1-p)2 and there are C(6,3) = 10 ways of this occurring. So the probability of success on the fifth roll is 10p3(1-p)2p = 10p4(1-p)2.
The overall probability of success is given by the formula: Σ(t=4 to 108) C(n-1,w-1).pw.(1-p)(t-w)
Where:
p =probability of winning $35 = 1/38
w = minimum number of wins to make a profit = 4
t = number of turns played.
C(x,y) be the Combination Function which gives the number of ways of choosing y items out of x items.Therefore the probability of coming out ahead is 31.7% and the probability of losing money is 68.3%, so the answer is indeed D.
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