Probability Problem Trouble

Question

Looking for some guidance on this problem:

You are playing a betting game and there are 2 outcomes: you either win $\$1$ with probability $\frac {999}{1000}$ or you lose $\$200$ with probability $\frac 1{1000}$

It takes $1$ minute to play each round. You can play $60$ rounds/hour and you can play as long as you want.

What are the expectation and standard deviation of playing this game once? Of playing for 1/hr?

Would you play?

I found the $E = .799$ and the standard deviation to $=6.353$. I am unsure how to account for multiple tries so I am unsure where to go from here. Any guidance would be appreciated.

Answer 1

Each round of the game is independent of the last one. This means you can use linearity of expectation to work out the mean and standard deviation.

For two variables $X$ and $Y$, it's always true that $E[X+Y] = E[X] + E[Y]$. If $X$ and $Y$ are independent then it is also true that $E[XY] = E[X]E[Y]$. These are powerful results.

For the mean, it's fairly straightforward. The expectation over $N$ rounds is just the expected value of your total winnings from $N$ rounds, which is equal to the sum of the expected winnings from each of the rounds.

Since each round you have the same expected value, this leads to an expected value of $.799N$ over $N$ rounds.

The standard deviation is just the square root of the variance, which is given by $\mathrm{Var}(X) = E[X^{2}] - (E[X])^{2}$. If you play another round and score $Y$, your new variance is $\mathrm{Var}(X+Y) = E[(X+Y)^{2}] - (E[X+Y])^{2}$. Can you expand that out and see what you get?

You should get $E[X^{2}]+2E[XY]+E[Y^{2}] - (E[X]^{2} + 2E[X]E[Y] + E[Y]^{2})$, expanding the sums outside the expectation operator.
Since the rounds are each independent, $X$ and $Y$ are independent so we can use $E[XY] = E[X]E[Y]$.
Cancelling the terms should leave you with $E[X^{2}] - (E[X]^{2}) + E[Y^{2}] - (E[Y]^{2}) = \mathrm{Var}(X) + \mathrm{Var}(Y)$ - the variances just add!
This means your variance also scales with $N$ so your standard deviation scales with $\sqrt{N}$. For $N$ rounds, your standard deviation is $6.353\sqrt{N}$. So as $N$ gets larger, the standard deviation shrinks relative to the mean.

As to whether you'd want to play this game, even though the expected value of each round is positive, it's offset by the alarmingly large standard deviation (a consequence of the small chance of losing hard). However, the ratio between the mean and standard deviation improves with the number of rounds you play.

If you play enough rounds, the central limit theorem means you can approximate the distribution of your score as a normal distribution - so if your expectation is $4\sigma$ for example, you have a $0.003\%$ chance of a net loss. As lulu's shown, this takes a long time to build to though, and you could easily run out of money before reaching this point.

So how many rounds you play would probably depend on a number of factors - how much money you have as a safety net, how much time you have to play, how much of a risk-taker you are, whether you have a job that pays more than $\$.799\times 60 = \$47.94$ an hour...