Question 267157
If North American casino roulette were fair, how much should it pay to a player who bets $10 on red and a red number comes up? The odds against winning a bet on a red in roulette are 1.01 to 1.
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Odds against "e" = P(not e)/P(e)
If odds against red are 1.01:1
P(not red) = 101/2.01
P(red) = 1/2.01
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Let "x" be the payoff for "red"
E(x) = winnings*P(win) + loss*P(lose)
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If the game is "fair" E(x) = zero
E(x) = (x-10)(1/2.01) - 10(1.01/2.01) = 0
Multiply thru by 2.01:
(x-10) - 10(1.01) = 0
x = 10(1+1.01)
x = 10(2.01
x = $20.10
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Cheers,
Stan H.
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If "odds for an event (e)" are a:b
then P(e) = a/(a+b) P(not e) = b/(a+b) 
P(winning chance) 1.01/(1.01+1) = 1.01/2.01 
P(winnings) 10*2.01 =2.01 
Total winning =$20.10 
Answer: $21.10