You are playing roulette, each time betting on “odd,” which occurs with probabil
ID: 3326840 • Letter: Y
Question
You are playing roulette, each time betting on “odd,” which occurs with probability 18/38 and gives you even money back. You start with $10 and decide to play until you have either doubled your fortune or gone broke.
Compute the probability that you manage to double your fortune if in each round you bet $10, $5, $2, and $1 dollar, respectively.
After you have found the best strategy, give an intuitive explanation of why it is the best and why it is called “bold play.”
even money back meaning getting the same amount back.
Explanation / Answer
The best strategy is to bet everything. This lets you win with probability 18/38, and you go broke immediately if it doesn't work. It is relatively easy to prove that this is optimal.
Let X be the amount of money you have at the end. E[X] = $20 * P(win).
This is also the initial amount minus your expected loss, which is the house advantage fo 1/19 times the amount you wager.
To reach $0 or $20, you have to wager a total of at least $10, so the maximum expected value you can have at the end is 10 - 10/19 = $10(1-1/19).
So, 20P(win)<=10(11/19),P(win)1/2(11/19)=18/38.
That means you can't win more than 18/38 of the time, so betting $10 on one color is optimal.The best strategy is to bet everything. This lets you win with probability 18/38, and you go broke immediately if it doesn't work. It is relatively easy to prove that this is optimal.
Let X be the amount of money you have at the end. E[X] = $20 * P(win).
This is also the initial amount minus your expected loss, which is the house advantage fo 1/19 times the amount you wager.
To reach $0 or $20, you have to wager a total of at least $10, so the maximum expected value you can have at the end is 10 - 10/19 = $10(1-1/19).
So, 20P(win)<=10(11/19),P(win)1/2(11/19)=18/38.
That means you can't win more than 18/38 of the time, so betting $10 on one color is optimal.
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