Wage x dollars in biased coin flip You have 100 dollars. You are playing a game where you wager x dollars on a biased coin flip with a 90 percent probability of heads. You make 2x if it’s heads and lose the x dollars if it’s tails. How much do you think you should bet on each flip if you are going to play for 100 flips?
My approach:
If I wage x dollars then, 90% of them time (heads), I would win 2x - x = x dollars and 10% of the time (tails) I would lose x dollars.
Hence my Expected value E[x] = 0.9(x) + 0.1(-x) = 0.8x which is lesser than how much I wage.
 A: If you just want to maximize your expected winnings, then the best strategy is to bet your entire stake every time. This is because when you bet $x$, your expected gain is $0.9\cdot x+0.1\cdot (-x)$, which is monotonically increasing in $x$. This strategy works like the lottery; you lose your initial $\$100$ with overwhelming probability, but with the small chance of $(0.9)^{100}$, you win a huge prize of $\$(2^{100}-1)\cdot 100$.
Most people would agree this strategy is stupid. Instead, we should try to maximize our "worst case" winnings, so we can be confident we will win a lot of money. We cannot guarantee making money, but we can try to maximize the number $t$ such that $P(\text{winnings}\ge t)=0.99$. It turns out the best strategy in this case is given by the Kelly criterion; you should bet $80\%$ of your current wealth each time.
Let $X_0,X_1,\dots,X_{100}$ be your current wealth at each stage of the game, so $X_0=100$, while $X_{100}$ is your final wealth. Furthermore, let $r_1,r_2,\dots,r_{100}$ be the proportion of your current wealth that you place on each bet, so you bet $r_iX_{i-1}$ on the $i^\text{th}$ bet. Note that each $0\le r_i\le 1$, and
$$
X_{i}/X_{i-1}=\begin{cases}
1+r_i & \text{with probability }0.9\\
1-r_i & \text{with probability }0.1
\end{cases}
$$
That is, the ratio of our wealth is a simple, bounded random variable. We can then write
$$
X_{100}=100\cdot (X_1/X_0)\cdot (X_2/X_1)\cdot \cdots (X_{100}/X_{99})
$$
so our wealth is a product of these independent random variables. We now take $\log$'s of both sides, since addition is more nice than multiplication.
$$
\ln X_{100}=\ln 100+\sum_{i=1}^{100}\ln(X_{i}/X_{i-1})
$$
We have written $\ln X_{100}$ as a sum of independent random variables. The central limit theorem then implies that $\ln X_{100}$ is approximately normally distributed*, with mean equal to $\ln 100$ plus the sum of the means of $\ln (X_i/X_{i-1})$, and variance equal to the sum of the variances of $\ln(X_i/X_{i-1})$. This normal variable is pretty tightly concentrated around its mean, because as you add variables, the mean grows linearly, while the standard deviation only with the square root. This implies, in order to maximize the $99\%$ security level of your overall profit, you should maximize the expected value of $\ln (X_i/X_{i-1})$ for each $i$, which entails maximizing
$$
0.9\ln (1+r_i)+0.1\cdot \ln(1-r_i)
$$
Using calculus, you can show this is maximized when $r_i=0.8$, which is exactly the Kelly criterion.
* Since the variables are not identically distributed in the case where the $r_i$ are unequal, you need something like the Lyapunov central limit theorem here. Basically, since each $\ln(X_i/X_{i-1})$ contributes a small amount to the total variance, we still get a normal distribution.
A: A slightly different approach: I want the chance of not coming out with a million dollars to be tiny, but apart from that I want to make lots. Because my goals are different, my results are different.
We can estimate the chances of losing k times. The chances of losing 27 times and winning 73 times are less than one in a million.
If we start with $100 and bet 45% of our money each time, and lose 27 times (with a chance of less than one in a million), we end up with about 5.9 million dollars. If we lose 10 times and win 90 times, we end up with about 88 trillion dollars.
Now if I add that I don’t care about making more than a billion, and adapt my bets during the game, I should be able to improve the “worst” case outcome further at the expense of less chance to win a ridiculous amount of money.
Betting 80% each time is most likely to make more, but in the one-in-a-million bad case losing 27 times, we actually lose money.
So to solve the problem, we need human psychology and how much money is actually valued. With different numbers in the problem I would use a different strategy. With the given numbers, I’d be very happy with betting 45% each time. If it was for real with that much money at stake, I’d do some find tuning.
