# Expected Value of Gambling Strategy - Martin's Winnings

• MHB
• jamcc09
In summary, Martin has a gambling strategy of betting 1 dollar on heads that a fair coin will land. 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. Assuming each individual bet is fair, the expected winnings on any bet is 0. If $X_n$ is the amount of money he wins on the $n^{th}$ flip then $E(X_n) = p*2^{n-1} - (1 - jamcc09 This question comes from the "Introduction to Probability" book (Blitzstein & Hwang). Martin has just heard about a gambling strategy: bet 1 dollar that a fair coin will land heads. If it does, stop. If it lands tails, then double the bet for the next toss, now betting 2 dollars on heads. If it does, stop. Otherwise, double the bet to 4 dollars for the next toss. This strategy is continued, i.e. doubling the bet each flip and stopping after winning a bet. Assume each individual bet is fair. The basic idea is that he will be 1 dollar ahead when he wins a bet. Martin decides to try out the strategy, but he has only 31 dollars, so he could end up bankrupt. On average, how much money will Martin win? My solution: (Martin can bet a maximum of 5 times because of having only 31 dollars) The assumption that each individual bet is fair means that if Martin is on flip$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 :) Hi jamcc09, Welcome to MHB! It's fitting that the player's name is Martin as his strategy is known as the Martingale strategy. I have to run right now but I would look at the example shown in this link on the specific case of a 50/50 game: Stack Exchange. Ah ha, I see! I will try to consider the true meaning of the names a bit more next time ;) It seems from the wikipedia article that if each bet is fair the expected profit is 0 and is negative for$p < 0.5\$. This seems to come from a similar argument to mine that the expected value of 1 application of this strategy is equal to 0. Is that correct?

Your suggested stack exchange article was a little confusing for me, because there was a lot of debate. However in the top of that article there is a link to another rather hilarious post that is a bit easier to understand and a bit more congruent to my problem, see this post. In short, someone's friend came up with this strategy for a casino in a "world of warcraft"-like video game (however, if the person won, he or she would start with a bet of 1 dollar again), declared it genius (famous last words?), let his/her strategy run over night and woke up bankrupt ;)

The first reply to the post shows why the strategy will almost always end up with a bankruptcy and although I don't completely understand everything in the mathematical part of the post yet, it does seem intuitively clear to be the case :) Thanks!

## 1. What is the expected value of gambling strategy?

The expected value of gambling strategy is a mathematical concept that calculates the average amount of money a player can expect to win or lose over time. It takes into account the probability of winning and the amount of money won or lost in each outcome.

## 2. How does the expected value of gambling strategy affect my chances of winning?

The expected value of gambling strategy can give you an idea of the long-term outcome of your gambling. It can help you determine if a particular strategy is worth pursuing or if it is more likely to result in a loss.

## 3. How is the expected value of gambling strategy calculated?

The expected value of gambling strategy is calculated by multiplying the probability of each outcome by the amount won or lost in that outcome, and then summing up all the values. This gives you an idea of the average amount you can expect to win or lose over time.

## 4. Can the expected value of gambling strategy guarantee a profit?

No, the expected value of gambling strategy is not a guarantee of profit. It is simply a mathematical concept that helps you make informed decisions about your gambling strategy. It does not take into account factors such as luck or unforeseen circumstances.

## 5. How can I use the expected value of gambling strategy to improve my chances of winning?

You can use the expected value of gambling strategy to compare different strategies and determine which one has a higher expected value. This can help you make more informed decisions about your gambling and potentially improve your chances of winning over time.

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