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Okay, so I am curious about a problem.

Say that you are given the opportunity to make a certain bet in which $P(Win)=.52$ and $P(Lose)=.48$ and that their payouts are equal but opposite. You will repeat the bet $10000$ times. Assuming that you start with $1000$ units, and you are allowed to bet $p$ percent of your total wealth after each round. For example, if you choose $p=.1$ you would bet $100$ units the first round. If you win the first round, you would bet $110$ units the next time. Conversely, if you lost the first round, you would bet $90$ units the second round. Notice that you can never go completely bankrupt since you are only betting a predetermined percent of your wealth each time.

Assuming that you can only, control $p$, what percent of your wealth would you choose to bet? I.e. what value of $p$ would give you the highest expected wealth after $10000$ trials?

My attempt: When I saw this problem, I initially thought that a larger $p$ would surely increase the expected value at the end of $10000$ rounds since you had a greater than $.5$ probability of winning.

After struggling to come up with a formula, I built a simulation and found out that this was clearly not the case. (Any simulation with $p\gt.1$ resulted in a near-zero end value).

This got me thinking and I thought about this problem as $(1+p)^{5200}*(1-p)^{4800}$ since there would be expected $5200$ wins and expected $4200$ losses. I then differentiated and looked for a $0$ over the interval $(0,1)$ to find a maximum. I ended up getting a derivate of $5200(1-x)^{4800}(1+x)^{5199} -4800(1-x)^{4799}(1+x)^{5200}$ and end up with a value of $.04$

When I plug in $.04$, it does indeed yield a higher end value, but it is so variable from time to time it is difficult to tell if my answer is correct.

I am hoping that someone can provide guidance on whether my general approach to this problem is correct, or if there is some other way to solve such a problem?

user345
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  • What is the objective? Is it to maximize expected final wealth? If so, one should bet $100$% of current wealth each time (i.e., $p=1$). – quasi Jul 13 '17 at 03:26
  • Yes, objective is to maximize expected final wealth. If one were to bet $100$% of wealth, just one loss would result in running out of money and at least one loss is all but guaranteed in $10000$ trials – user345 Jul 13 '17 at 03:32
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    That has nothing to do with expected wealth. The expected winnings are .02 units for every unit bet. So the more units bet, the higher the expected winnings. With the $100$% strategy, bankruptcy is almost certain, but the expected wealth is maximized. – quasi Jul 13 '17 at 03:33
  • @quasi maybe I am missing something, but if you were to bet $100$% of wealth, one loss would use all of you wealth and you would be unable to continue on. Thus, you would be almost certain to end up with $100$% loss. I agree that $100$% would give highest expected value after $1$ round, but I don't see that it would be possible to maximizing for $10000$ rounds – user345 Jul 13 '17 at 03:36
  • Expected values add. – quasi Jul 13 '17 at 03:37
  • You need to distinguish between maximizing the probability of achieving some particular amount of wealth, and maximizing expected wealth. Once you specify the objective, the optimal strategy is with respect to that objective. – quasi Jul 13 '17 at 03:38
  • Correcting my numerical error, the expected winnings per unit bet is $.04$ units, not $.02$ units – quasi Jul 13 '17 at 03:43
  • @ap2010 It's worth pointing out the asymmetry between going bankrupt and winning. While there's a very high probability of going bankrupt, you can only lose $1000$ units if you do. If you win all $10000$ times, you'd have $1000\cdot 2^{10000}$ units, even if this happens with extremely low probability. – RideTheWavelet Jul 13 '17 at 03:45
  • Same question, different odds https://math.stackexchange.com/questions/240619/coin-tossing-game-optimal-strategy Yes, the optimal strategy to maximize expected wealth is bet all your money. That gives zero most of the time, but the times you win you win a lot and enough to make up for it. – Ross Millikan Jul 13 '17 at 04:25

1 Answers1

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Combining my comments into an answer . . .

Assuming the objective is to maximize expected final wealth, one should bet $100$% of current wealth each time (i.e., $p=1$).

The expected winnings are $.52(1) + .48(-1) = .04$ units for every unit bet, so the more units bet, the higher the expected winnings.

Note: With the $100$% strategy, bankruptcy is almost certain, but the expected wealth is maximized.

Using the $100$% strategy, the expected wealth after $n$ bets is $$(.52)^n\left(2^n\right)(1000)$$

For small $n$, say $n \le 10$, you can easily verify this with a simulation.

For a fixed number of bets, you need to distinguish between maximizing the probability of achieving some particular amount of wealth (e.g., maximizing the probability of doubling the initial wealth) and maximizing expected wealth. Once you specify the objective, an optimal strategy is with respect to that objective.

quasi
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