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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.

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    $\begingroup$ The way the question is phrased seems like losses actually don't matter. In other words, if Robert "wins" 25 and "loses" 50, then by the phrasing he has still won 25 even though he has lost money overall. $\endgroup$ – NotMe Oct 10 '14 at 18:59
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    $\begingroup$ Did anyone consider that the title is a subtle cry for help, and that "Robert" is actually the OP, unable to abstain from gambling hundreds of times? See gamblersanonymous.org/ga . $\endgroup$ – beep-boop Oct 10 '14 at 22:33
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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

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

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  • $\begingroup$ Give me time to consider your answer. Your answer seems promising. +1, I really appreciate it for your response. Thank you... :-) $\endgroup$ – Venus Oct 10 '14 at 17:48
  • $\begingroup$ @Venus: This is just the standard normal approximation to the multinomial distribution, justified by the Central Limit Theorem. For n = 100 the error will be very small. $\endgroup$ – RRL Oct 10 '14 at 18:00
  • $\begingroup$ @Venus This is the best answer. You could try explicit computation by using multinomial coefficient, but you'll end up with a ugly double sum (say, all $X<20\$$ and some other free parameter) for which there is no known identity. $\endgroup$ – jgyou Oct 10 '14 at 19:55
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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).

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  • $\begingroup$ Give me time to consider your answer. +1, I really appreciate it for your response. Thank you... :-) $\endgroup$ – Venus Oct 10 '14 at 17:45
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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.

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  • $\begingroup$ Ya, I know 100% is impossible answer. It seems I can't get an exact answer to this question. +1 anyway, I really appreciate it for your response. Thank you... :-) $\endgroup$ – Venus Oct 10 '14 at 17:44
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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 $$\sum_{2a+3b\le 120, a,b\ge 0} \frac{100!}{a!b!(100-a-b)!}(.25)^{a+b}(.5)^{100-a-b} $$

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  • $\begingroup$ Give me time to consider your answer. Your answer seems promising. +1, I really appreciate it for your response. Thank you... :-) $\endgroup$ – Venus Oct 10 '14 at 17:46

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