Gambling problem

Question

Question

Robert will win $\$1$ with probability $\frac{1}{4}$, win $\$2$ with probability $\frac{1}{4}$, and lose $\$1$ with probability $\frac{1}{2}$ in a bet. Each bet is independent. Determine the probability that Robert will win at most $\$20$ after betting $100$ times.

Attempt

Let $X$ be the amount of money that Robert wins. I know how to determine expected money that Robert would get each bet i.e. $E[X]=\$1\cdot\frac{1}{4}+\$2\cdot\frac{1}{4}-\$1\cdot\frac{1}{2}=\$0.25$. I think the probability would be $100\%$ because after betting $100$ times the expected money that he would get is $\$0.25\cdot100=\$25$, but it's definitely wrong. What is the correct approach to solve this problem? Any help would be appreciated. Thanks in advance.

Answer 1

Let $S_{100} = \sum_{k=1}^{100}X_k$ denote the winnings after 100 bets.

Then

$$E(S_{100})= 100E(X)= 25$$

and

$$var(S_{100})= \sum_{k=1}^{100}var(X)= 100(27/16)\approx 168.75.$$

As a sum of $100$ independent random variables, $S_{100}$ is approximately normally distributed with mean $25$ and standard deviation $\sqrt{168.75} \approx 12.99$.

The probability of at most $\$20$ in gains is obtained using the standard normal distribution function

$$P(S_{100} \leq 20)= N\left(\frac{20-25}{12.99}\right)\approx 35\% $$

Answer 2

No, it can't be $100\%$ that he wins at least $20$, as he could lose all the bets. You are correct that his expectation is $25$. If he wins $2$ five times he has $20$ more bets of which he must win $15$ or more. The chance of winning specifically five bets at $2$ and $15$ bets at $1$ is ${25 \choose 5}{20 \choose 15}\frac 1{4^5}\cdot \frac 1{4^{15}}\cdot \frac 1{2^5}$ There are a lot of possiblities to add up.

Answer 3

A simple way to get a correct solution is using brute force and a spreadsheet, and calculating the following values:

After 1 game, what are the probabilities of having -1, 0, 1, 2 dollars? After 2 games, what are the probabilities of having -2, -1, 0, 1, 2, 3, 4 dollars? ... After 100 games, what are the probabilities of having -100, -99, ..., 199, 200 dollars?

Each can be calculated quite easily. If we define $P (n, d)$ = "probability of having d dollars after n games", then $P (n, d) = P (n-1, d+1)/2 + P (n-1, d-1)/4 + P (n-1, d-2) / 4$. Then you just add up the numbers P (100, -100) to P (100, 20).

You can check that the spreadsheet is correct by taking the probabilities that you calculated, use them to calculate the expected amount of money, and check that it is indeed $25.

(For a reasonably accurate result with fewer calculations, you might assume that the money won will follow a normal distribution, calculate average and standard deviation, and use a table to calculate that such a distribution would give a value of at most 20.5. This will not give the correct result, but it shouldn't be too far off).

Answer 4

I don't know if this helps, but here's a reformulation of the problem: Let $a$ be the number of \$1 wins; $b$ be the number of \$2 wins, and let $c$ be the number of \$1 losses.

We have $a+b+c=100$, and we want $a+2b-c\le 20$. (Of course, $a,b,c \ge 0$.)

From the equation, we have $c=100-a-b$, which we plug into the inequality, obtaining $a+2b-(100-a-b)\le 20$,

which simplifies to $2a+3b\le 120$.

So we want $P(2a+3b\le 120)$.

I guess this probability would be given by $$\sum_{2a+3b\le 120, a,b\ge 0} \frac{100!}{a!b!(100-a-b)!}(.25)^{a+b}(.5)^{100-a-b} $$