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#.

1 Answer
Jan 10, 2017

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#