MHB Expected value and variance of profit

[Yankel]

Hello all, I have this question, which I think I partially knows how to solve, but need some completion.

"A man is playing versus a machine in the following way: The machine chooses 2 numbers randomly from the set of numbers 1,2,3,4,5, where a number can be chosen twice (with replacement). If the multiplication of the 2 chosen numbers is even, the man gets 5 dollars Calculate the expected value (mean) and variance of the profit after 100 games, if for every game he pays 2 dollars to play."

What I did to start with, is to calculate the probability of having an even multiplication and I got p=16/25. Now I know I can calculate the profit for a single game, get a probability function with 2 values, and find E(X), and multiply it by 100. To be more specific, if X is the profit in 1 game, then it can get the values 3 and -2 only. The matching probabilities are 16/25 and 9/25 respectively. Therefore it's easy to find E(X) and V(X). However I need the expected value and variance after 100 games. I know that for E it's 100*E(X), what about V ?

Thank you

 

[Siron]

Your calculations are correct in my opinion if you assume that the order of picking two random numbers in the set $\{1,2,3,4,5\}$ does matter.

If you denote $\{X_1,\ldots,X_{100}\}$ as the sequence of profits of the first $100$ games then I think you can assume this random variables are independent and Bernoulli distributed. Hence their sum $S_{100} = \sum_{j=1}^{100} X_j$ represents the pay-off after the $100$th game which has a Binomial distribution.

 

[chisigma]

Hello all, I have this question, which I think I partially knows how to solve, but need some completion.

"A man is playing versus a machine in the following way: The machine chooses 2 numbers randomly from the set of numbers 1,2,3,4,5, where a number can be chosen twice (with replacement). If the multiplication of the 2 chosen numbers is even, the man gets 5 dollars Calculate the expected value (mean) and variance of the profit after 100 games, if for every game he pays 2 dollars to play."

What I did to start with, is to calculate the probability of having an even multiplication and I got p=16/25. Now I know I can calculate the profit for a single game, get a probability function with 2 values, and find E(X), and multiply it by 100. To be more specific, if X is the profit in 1 game, then it can get the values 3 and -2 only. The matching probabilities are 16/25 and 9/25 respectively. Therefore it's easy to find E(X) and V(X). However I need the expected value and variance after 100 games. I know that for E it's 100*E(X), what about V ?

Thank you

At each game the probability of winning is...

$\displaystyle p = \frac{1}{5}\ \frac{2}{5} + \frac{1}{5} + \frac{1}{5}\ \frac{2}{5} + \frac{1}{5} + \frac{1}{5}\ \frac{2}{5} = \frac{8}{25}\ (1)$

For a binomial distribution the expected number of successes in n games is...

$\displaystyle \mu = n\ p\ (2)$

... and the variance...

$\displaystyle \sigma= n\ p\ (1-p)\ (3)$

In your case is $n=100$ and $p = \frac{8}{25}$, so that the expected profit is $E = 32\ 5 = 160$ with variance $V= 32\ \frac{17}{25}\ 5 = \frac{1088}{5} = 108.8$ ... taking into account that if You want to play 100 games You have to pay 200 dollars, the global expected profit is negative...

Kind regards

$\chi$ $\sigma$