Expected Value

BH Statistics

March 30, 2000

Expected Value

Let’s say that you are playing a game.

You pay $1 each time you play.
The game is played by flipping two coins.
If they both land heads, you receive 50 cents (for a 50 cent loss).
If they land one head, one tail, you a dollar (for no loss).
If they land on both tails, you earn back 2 dollars (for a 1 dollar gain).

The idea of a game’s expected value is "The amount per play that you expect, more ore less, to win or lose over the long run".

One way to look at "over the long run" is to imagine we repeat the game a large number of times (e.g., 1000 times).

So, let’s pretend we’re going to play the game 1000 times. We’ll see how much we can expect to win or lose in total for those 1000 games, and then compute an average win or loss per game.

Cost of playing

How much will we pay to play 1000 games? $1 per game, or $1000.

Winnings from getting two heads

What fraction of the time can we expect to get both heads? 1/4. So, on approximately 1/4 of the plays, or 250 plays, we will receive 50 cents per play. On those plays we will receive a total of $125.

Winnings from getting two tails

Approximately what fraction of the time will we get two tails when we toss two heads? 1/4. So 1/4 of the time you will receive 2 dollars, so on 1/4 of 1000 plays, or 250 plays, you will receive $2 per play, for a total of $500.

Winnings from getting one head and one tail

About what fraction of the time will you get one head, and one tail? 1/2. So about 1/2 of the plays you will receive1 dollar, so on those 500 plays you will receive 500 dollars.

In total

So you have paid 1000 dollars to play the game 1000 times, and you received 125, 500 and 500 dollars, or 1125 dollars after playing all 1000 games. You can expect to earn a net of 125 dollars per 1000 games, or $.125 per game.

Generalize

Suppose you play n games. In approximately 1/4 of those n games you will get two heads, in 1/4 of n games you will get two tails, and in 1/2 of n games you will get one head and one tail.

Let X be the net amount you receive on a given play of this game (taking into account that you must pay a dollar to play).

So "winning" 50 cents means that you actually receive a net amount of -.50 dollars.

Then the total amount we receive on those plays in which we receive a net amount of -.50 dollars per play is the product of the number of such games times -.50, or

By the same logic, the total amount we net on those plays in which we gain $0.00 ($1 returned minus $1 paid to play), over the long run, will be approximately

The total amount we net on those plays in which we gain $1.00 ($2 returned minus $1 paid to play), over the long run, will be approximately

So, the total amount net over n plays of the game will be approximately

+ + .

This says simply that the total amount you expect to net is (the number of times you net -.50 dollars) times -.50 dollars, plus (the number of times you win 0 dollars) times 0 dollars, plus (the number of times you net 1 dollar) times 1 dollar.

The net amount per play over n plays is:

This simplifies to

which is P[X=^-.5](^-.5) + P[X=0](0) + P[X=1](1).

Let X be "the amount you net on a play". Suppose you make a list of the amounts you can net on any play as x₁, x₂, x₃ and so on (meaning, x₁is the first amount in the list, x₂ is the second amount in the list, and so on.) Let x_i stand for the i^th amount in the list. Then, the expected value of the game is:

P[X=x_i] is the fraction of the time that the experiment (game) produces a value of x_i_,. So, (P[X=x_i]n) is the number of times the experiment produced a value of x_i , Thus, the expression P[X=x_i](n)(x_i) represents the total gain for those times that the experiment (game) produces a value of x_i.

The sum is the total net gain over n trials. The quotient is the average net gain per play for n plays.

Using a little algebra we see that the n cancels in numerator and denominator of . This means that the expected (EXPECTED) average net gain per play will be the same regardless of how many times you play the game.