MHB Calculating the Best Bet: Solving the Problem of Roulette Odds

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The discussion revolves around the mathematical analysis of roulette odds, specifically evaluating the expected value of bets. A player calculates that betting $1 on a standard roulette wheel with 38 spaces results in an average loss of $0.0526 per bet, leading to the decision not to play. In a hypothetical scenario, the player dreams of a roulette wheel where the average return is $0.44 per $1 bet, prompting a calculation to determine the number of spaces on this fantasy wheel. The solution reveals that this fantasy roulette wheel would contain 25 spaces, illustrating the concept of expected value in gambling. Overall, the thread emphasizes the importance of understanding odds and expected value when making betting decisions.
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You and your friends are going to Vegas. They find a roulette table and while watching your friends play, you ponder if this is mathematically a good decision in the long run. You choose to bet \$1 while you think.

There are 38 spaces on the wheel and you bet on one number of out of the 38 being chosen. If you choose correctly you will receive \$35, if not then you lose your bet. You are good at math and quickly calculate that your average return for this \$1 bet is -\$0.0526 so you decide not to play anymore.

While asleep you are dreaming about a wonderful fantasy roulette where you average \$0.44 on every \$1 bet. Assuming you still get \$35 for choosing the correct number and lose \$1 when not choosing the correct number, how many spaces does this fantasy roulette wheel contain?

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Congratulations to the following members for their correct solutions:

1) Sudharaka
2) soroban
3) Reckoner

Solution (from soroban):

[sp]Comments from Jameson: The roulette wheel is a great example of how the casino makes money. All possible bets have a negative expected value for the player in the long run, that is to say, if you play long enough you will lose your money even if you have big wins. If you didn't know before, after reading this solution you should now be able to make some useful calculations about roulette.

A main concept used here is expected value. If you haven't studied this before, I highly recommend doing so. It can be applied to investments and other financial decision, or even general decisions for that matter.

Solution below.

Let n = number of spaces.

\tfrac{1}{n} of the time you will win \$35.

The other \tfrac{n-1}{n} of the time you will lose \$1.

Your expected value is \$0.44.

Hence: .[/color]\frac{1}{n}(35) + \frac{n-1}{n}(-1) \:=\:0.44

Multiply by n\!:\;35 - (n-1) \:=\:0.44n

. . . . . . . . . . . . [/color]35 - n + 1 \:=\:0.44n

. . . . . . . . . . . . . . . . . . . [/color]36 \:=\:1.44n

. . . . . . . . . . . . . . . . . . . . [/color]n \:=\:\frac{36}{1.44} \:=\:25

This roulette wheel has 25 spaces.
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