expected value

Question

in a game of chance, a player blindly selects (without placement) two bills from a box containing four bills ($1, $2, $5, and $10) and gets to keep the larer of the two bills selected. The game costs 8 dollars to play. Let X=the players net winnings. Find E(X) the players expected winnngs.

 

 

Please show the work, every step I really wanted to understand how to get the answer. I can't find an example expected value question where the person is paying or putting in money (such as 8 dollars)

Answer 1

First of all, order doesn't matter. There are six possible ways the player can draw:

(1,2), (1,5), (1,10), (2,5), (2,10), (5,10).

The corresponding prizes are 2, 5, 10, 5, 10, 10.

The net winnings are X = prize - wager, so since the wager is always $8, the corresponding winnings are -6, -3, 2, -3, 2, 2.

The expected value is the arithmetic mean of these:

<X> = ΣX / N = (-6 + -3 + 2 + -3 + 2 + 2) / 6 = -6/6 = -1.

So the player loses $1 on average. This is expected as it implies a house-edge.