- #1
jamcc09
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This question comes from the "Introduction to Probability" book (Blitzstein & Hwang).
Martin has just heard about a gambling strategy: bet 1 dollar that a fair coin will land heads. If it does, stop. If it lands tails, then double the bet for the next toss, now betting 2 dollars on heads. If it does, stop. Otherwise, double the bet to 4 dollars for the next toss. This strategy is continued, i.e. doubling the bet each flip and stopping after winning a bet. Assume each individual bet is fair. The basic idea is that he will be 1 dollar ahead when he wins a bet.
Martin decides to try out the strategy, but he has only 31 dollars, so he could end up bankrupt. On average, how much money will Martin win?
My solution: (Martin can bet a maximum of 5 times because of having only 31 dollars)
The assumption that each individual bet is fair means that if Martin is on flip $n$, he is betting $2^{n-1}$ dollars. He has equal probability of winning or losing on any given bet, because of the fair coin, so the expected winnings on any bet is 0. If $X_n$ is the amount of money he wins on the $n^{th}$ flip then $E(X_n) = p*2^{n-1} - (1 - p)*2^{n-1}$ and since $p = 1/2$, $E(X_n) = 0$ and since $p = 1/2$, $E(X_n) = 0$. I would say, by linearity of expectation, that if $X$ is the amount of money he wins using this strategy, then $X = X_1 + X_2 + X_3 + X_4 + X_5$ and $E(X) = E(X_1) + E(X_2) + E(X_3) + E(X_4) + E(X_5) = 0$.
Another way that I solved it what to basically consider what is the probability that he goes bankrupt. Since the coin is fair, it is just the probability of getting 5 tails in a row, which is $1/32$. That means the probability of winning is $31/32$. In the case of bankruptcy he loses his 31 dollars, if he wins he gets 1 dollar. These are the only two possibilities, since he stops after winning a bet or losing all his money. So, the expected winnings are $E(X) = \frac{31}{32}*1 + \frac{1}{32}*-31 = 0$.
I actually came up with the first solution while writing the question and get the same as the last solution, so I feel a bit more confident about it now, but still would like to know if this is actually a correct way to reason about this. This is the first time I feel like the linearity property of the expected value could be an intuitive way to solve it (even though it wasn't my first path to a solution). Thanks to anyone who can confirm or provide insight into where I went wrong :)
Martin has just heard about a gambling strategy: bet 1 dollar that a fair coin will land heads. If it does, stop. If it lands tails, then double the bet for the next toss, now betting 2 dollars on heads. If it does, stop. Otherwise, double the bet to 4 dollars for the next toss. This strategy is continued, i.e. doubling the bet each flip and stopping after winning a bet. Assume each individual bet is fair. The basic idea is that he will be 1 dollar ahead when he wins a bet.
Martin decides to try out the strategy, but he has only 31 dollars, so he could end up bankrupt. On average, how much money will Martin win?
My solution: (Martin can bet a maximum of 5 times because of having only 31 dollars)
The assumption that each individual bet is fair means that if Martin is on flip $n$, he is betting $2^{n-1}$ dollars. He has equal probability of winning or losing on any given bet, because of the fair coin, so the expected winnings on any bet is 0. If $X_n$ is the amount of money he wins on the $n^{th}$ flip then $E(X_n) = p*2^{n-1} - (1 - p)*2^{n-1}$ and since $p = 1/2$, $E(X_n) = 0$ and since $p = 1/2$, $E(X_n) = 0$. I would say, by linearity of expectation, that if $X$ is the amount of money he wins using this strategy, then $X = X_1 + X_2 + X_3 + X_4 + X_5$ and $E(X) = E(X_1) + E(X_2) + E(X_3) + E(X_4) + E(X_5) = 0$.
Another way that I solved it what to basically consider what is the probability that he goes bankrupt. Since the coin is fair, it is just the probability of getting 5 tails in a row, which is $1/32$. That means the probability of winning is $31/32$. In the case of bankruptcy he loses his 31 dollars, if he wins he gets 1 dollar. These are the only two possibilities, since he stops after winning a bet or losing all his money. So, the expected winnings are $E(X) = \frac{31}{32}*1 + \frac{1}{32}*-31 = 0$.
I actually came up with the first solution while writing the question and get the same as the last solution, so I feel a bit more confident about it now, but still would like to know if this is actually a correct way to reason about this. This is the first time I feel like the linearity property of the expected value could be an intuitive way to solve it (even though it wasn't my first path to a solution). Thanks to anyone who can confirm or provide insight into where I went wrong :)