1. ## Wagering problem

I have a question I have been trying to challenge myself with, but unfortunately I don't have a method for obtaining the exact solution. The question is this:

You have 1,000 dollars. If you wager any amount, you have a 90% chance of receiving double the amount of the wager and a 10% chance of losing the amount. You want at least a 50% probability of reaching X dollars through multiple wagers. How do you wager so as to maximize the value of X?

I have determined that I can, with at least 50% probability of success, attain the value of 3,402,823,669 dollars by wagering 60% of whatever I have each time. However, I am sure that this is not the optimal solution; it is merely "close" and I'm not sure how to arrive at the exact answer.

2. Originally Posted by icemanfan
I have a question I have been trying to challenge myself with, but unfortunately I don't have a method for obtaining the exact solution. The question is this:

You have 1,000 dollars. If you wager any amount, you have a 90% chance of receiving double the amount of the wager and a 10% chance of losing the amount. You want at least a 50% probability of reaching X dollars through multiple wagers. How do you wager so as to maximize the value of X?

I have determined that I can, with at least 50% probability of success, attain the value of 3,402,823,669 dollars by wagering 60% of whatever I have each time. However, I am sure that this is not the optimal solution; it is merely "close" and I'm not sure how to arrive at the exact answer.
Guess what the maximum value of $X$ is: $+\infty$. In fact, this game is so advantageous for the player that there is a strategy for which you can win as much as you want with probability greater than 99.99%...

And the strategy is almost the simpliest: wage repeatedly $1. Either you win$1 (probability 90%) or you lose $1 (probability 10%). A result from probability theory tells you that, starting from$1000, your amount of money with this strategy will go to $+\infty$ if you play repeatedly, and you will get ruined (i.e. have $0 left at some time) with probability $\left(\frac{10}{90}\right)^{1000}$ which is a really really tiny probability. That's not surprising: it is far more likely to win$1 than to lose \$1, and being broke happens if your number of losses exceeds the number of wins by 1000 at some time.