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

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 $\sum \left({p}_{i} {r}_{i}\right)$. and if their are $n$trials wins are $n \times \sum \left({p}_{i} {r}_{i}\right)$Coming to the problem, probability of getting an ace, other than club, is $\frac{3}{52}$and return on it is $5

probability of getting a club, other than ace, is $\frac{12}{52}$ and return on it is $1 probability of getting ace of club is $\frac{15}{52}$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 $\left(\frac{3}{52} \times 5 + \frac{12}{52} \times 1 + \frac{1}{52} \times 15\right)$= (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 $\left(\frac{3}{52} \times 5 + \frac{12}{52} \times 1 + \frac{1}{52} \times 11\right)$= (15+12+11)/52=$38/52=$0.73 and in $100$games, it is 100xx38/52=$73.08