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#

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#

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Answer 2

Anthony Amato

To determine the expected value of the game, we need to calculate the average amount one would win (or lose) per game.

Let's denote the probability of winning $1 as$ P($1) $, and the probability of losing$ 1 as $P(-\$ 1) $. From the given information, we know that:

$P(\$ 1) = 0.45 $and$ P(-$1) = 0.55 $

Now, we calculate the expected value (EV) using the formula:

$EV = (P(\$ 1) \times \text{amount won}) + (P(-$1) \times \text{amount lost}) $

$EV = (0.45 \times \$ 1) + (0.55 \times -$1) $

$EV = \$ 0.45 - $0.55 $

$EV = -\$ 0.10 $

This means that on average, a player can expect to lose $0.10 per game.

Now, to find out how much one is expected to win if one plays 100 games, we simply multiply the expected value per game by the number of games:

$\text{Expected winnings in 100 games} = EV \times \text{Number of games}$

$\text{Expected winnings in 100 games} = -\$ 0.10 \times 100 $

$\text{Expected winnings in 100 games} = -\$ 10 $

Therefore, if one plays 100 games, one is expected to win -$10 on average.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.