Optimal bet according to the probability of win

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Suppose the following game:

  • You start with $5000;
  • You will roll a dice 100 times;
  • You have to choose a percentage of your accumulated money to bet all the 100 times. You will choose it only once and cannot change it during the game;
  • If you roll 1 or 2 (33,33% chance) you lose 12% of the money that you have bet;
  • If you roll 3,4,5,6 (66,66% chance) you win 10% of the money that you have bet;

Also, you have the following objectives:

  • You want to win the maximum money as possible, but...
  • You really don't want to lose more then 50% of your initial money. It is impossible, because theoretically you could get bad numbers all the 100 times that you would roll the dice. So, let's determine that you accept a 20% risk of losing more than 50% of your initial money.

Given that, what would be the optimal percentage of the accumulated money to bet each time? How to solve this kind of problem?

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If you bet all your money every time, then using the binomial distribution with parameters $n=100$ and $p=\frac23$, there is a $0.1876887$ probability that you will win $62$ times or fewer and lose $38$ times or more.

So you have over $80\%$ chance of winning at least $63$ times and losing no more than $37$ times, in which case you will have at least $5000 \times 1.1^{63} \times 0.88^{37} \approx 17888$ and probably much more.

To lose half your money after betting everything every time, you need to lose at least $46$ times, and that has a probability of about $0.57\%$

To lose any money overall after betting everything every time, you need to lose at least $43$ times, and that has a probability of about $2.76\%$