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 January 24th, 2012, 11:04 PM #1 Member   Joined: Jan 2012 Posts: 31 Thanks: 0 Roulette probability Hi I'm just doing some problems and this question doesn't have an answer. Can someone check these over please? Suppose I bet on red at roulette and you bet on black, both bets on the same spin of the wheel: a) What is the probability that we both lose? b) What is the probability that at least one of us win? c) What is the probability that at least one of use lose? a) $\frac{20}{38}\times \frac{20}{38}= \frac{400}{1444} = \frac{100}{361}$ b) $\frac{18}{38}\times \frac{20}{38} + \frac{20}{38}\times \frac{18}{38} + \frac{18}{38}\times \frac{18}{38}= \frac{1044}{1444} = \frac{261}{361}$ b) $\frac{18}{38}\times \frac{20}{38} + \frac{20}{38}\times \frac{18}{38} + \frac{20}{38}\times \frac{20}{38}= \frac{1120}{1444} = \frac{280}{361}$ Cheers David.
January 25th, 2012, 02:50 PM   #2
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Re: Roulette probability

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 Originally Posted by Bucephalus Hi I'm just doing some problems and this question doesn't have an answer. Can someone check these over please? Suppose I bet on red at roulette and you bet on black, both bets on the same spin of the wheel: a) What is the probability that we both lose? b) What is the probability that at least one of us win? c) What is the probability that at least one of use lose? a) $\frac{20}{38}\times \frac{20}{38}= \frac{400}{1444} = \frac{100}{361}$ b) $\frac{18}{38}\times \frac{20}{38} + \frac{20}{38}\times \frac{18}{38} + \frac{18}{38}\times \frac{18}{38}= \frac{1044}{1444} = \frac{261}{361}$ b) $\frac{18}{38}\times \frac{20}{38} + \frac{20}{38}\times \frac{18}{38} + \frac{20}{38}\times \frac{20}{38}= \frac{1120}{1444} = \frac{280}{361}$ Cheers David.
All your answers are wrong. You are making an assumption of independence that is wrong. You need to look at a Roulette wheel in its entirety. I'll assume a wheel with 38 spaces, 36 numbers (18 red and 18 black) plus 0 and 00.
a) Both lose only if it lands on 0 or 00. Odds are 2/38.
b) If it lands on any of the 36 numbers one of you will win. Odds are 36/38.
c) Since there is no slot which is both red and black, it is certain that at least one of you will lose.

 January 25th, 2012, 03:47 PM #3 Member   Joined: Jan 2012 Posts: 31 Thanks: 0 Re: Roulette probability Hahaha Yeah you're right. I don't know where my head was with that one. Thanks for your help. David.

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