Question

What is the expected value to gain in following game and how much one is expected to win if one plays `100` games?

In a card game return on getting an ace, other than club, is `$5` but if it is a club he gets extra `$10` and getting a club, other than ace, gets you `$1`.

Answer 1

He is expected to win `$0.81` per game and

in `100` games, he is expected to win `$80.77`

Explanation:

When we have to find expected win or loss, what we have to do is identify probability of the event `p` and multiply it with expected return `r`.

If more than one events are there, which are mutually exclusive and their probabilities are `p_1,p_2,p_3,---` and corresponding returns are `r_1,r_2,r_3,---` and aggregate expectation is `sum(p_ir_i)`.

and if their are `n` trials wins are `nxxsum(p_ir_i)`

Coming to the problem,

probability of getting an ace, other than club, is `3/52` and return on it is `$5`

probability of getting a club, other than ace, is `12/52` and return on it is `$1`

probability of getting ace of club is `15/52` and return on it is `$15`, (here as he is expected to win `$5` for an ace and extra `$10` for club, he wins `$15`. If it is, however, `$1` for a club and extra `$10` for an ace, he wins `$11`).

So for one game (considering `$15` for ace of club), one wins

`(3/52xx5+12/52xx1+1/52xx15)`

= `(15+12+15)/52=$42/52=$0.81`

and in `100` games, it is `100xx42/52=$80.77`

Considering `$11` for ace of club, one wins

`(3/52xx5+12/52xx1+1/52xx11)`

= `(15+12+11)/52=$38/52=$0.73`

and in `100` games, it is `100xx38/52=$73.08`