What is a fair game? One in which you have just as much chance of winning as your opponent. If money were involved we would say it was a fair game if neither of the players lost any money, at least if they played long enough.
If a gambler offered us the following game what would he expect us to pay before each turn (ante up) to make it a fair game? Here is the game. We have one die. If we throw a one the gambler pays us one dollar, if we throw a two the gambler pays us two dollars and so forth upto six dollars. Obviously the gambler will not do this for nothing, and we will not do it if there is no hope of us making any money. If he charged six or seven dollars we would pay and have no chance of a profit.
Let us use probability to look at the game. A die has six faces each numbered one through six. So P(1),P(2),P(3),P(4),P(5),P(6) are all 1/6. So one sixth of the time we can expect a one to turn up, and the gambler pays one dollar. One sixth of the time a two will show up and the gambler pays two dollars and so forth. If we were to pretend the gambler had to pay the same amount every time then he would expect to pay
1 of $1 + 1 of $2 + 1 of $3 + 1 of $4 + 1 of $5 + 1 of $6.
Using the distributive law we factor the 1/6 getting
1 ( 1$ + $2 + $3 + $4 + $5 + $6) = 1($21) = $3.50.
So if we pay him $3.50 as our ante and he expects to pay out $3.50 each time, everything will balance and it will be a fair game. If a 1,2,or 3 comes up we lose from 50¢ to 2.50 if a 4, 5, or 6 comes up we win from 50¢ to $2.50. Our wins balance off our losses so it is a fair game.
Here are some vocabulary words to go with the ideas above. The $3.50 is also called the
mathematical expectation
of the game. We get the mathematical expectation of a game by multiplying the probability of each payoff by the value of each payoff and then adding up all the products. If we were to drop the dollar signs, we could think of the numbers on the dice as scores and the 3.5 would be the expected score on one die. It is impossible to get 3.5 on one die. Think about two die, 3.5 is expected on the first 3.5 is expected on the second so 7 is expected on both of them together. Does that make sense?
Let us use the rule for mathematical expectation and see if it gives seven too. From our sample space for two dice we know the probabilities: P(2) = 1/36; P(3) = 2/36; P(4) = 3/36; P(5) = 4/36; P(6) = 5/36; P(7) = 6/36; P(8) = 5/36; P(9) = 4/36; P(10) = 3/36; P(11) = 2/36; P(12) =1/36. To find the expected score multiply the score by its probability:
2/36; 6/36; 12/36; 20/36; 42/36; 40/36; 36/36; 30/36; 22/36; 12/36.
Add them up getting: 252/36 which is seven. So we get the same mathematical expectation two different ways. Furthermore, we know from our sample space for two dice that seven is the most probable score. So the mathematical expectation is the most probable score or result.
There is another way to look at the mathematical expectation. When we found $3.50 as the value of the game, what did we do? We added the payoffs or scores and then divided the answer by 6 the number of scores. That is called the
mean average
. So the mean average is the most probable score.
There is still more to say about the mean average. Let us look at the one die game where the value of the game, the mean average of the points, the ante, was 3.50. Here is the payoff chart:
Point
|
Payoff
|
Ante
|
Net
|
1
|
$1
|
$3.50
|
-$2.50
|
2
|
$2
|
$3.50
|
-$1.50
|
Negative numbers
|
3
|
$3
|
$3.50
|
.50
|
mean we lose.
|
4
|
$4
|
$3.50
|
.50
|
Positive numbers
|
5
|
$5
|
$3.50
|
$1.50
|
mean we win.
|
6
|
$6
|
$3.50
|
$2.50
|
Add up our net we get zero. For every gain there was a balancing loss. The numbers in the Net column were calculated by subtracting the $3.50 from the payoff. The $3.50 is the mean the net numbers are known as the
deviations from the mean
and the sum of the deviations from the mean is zero.
Let us use some algebra. We have a set of numbers we will subtract an unknown number a from each of the given numbers add up all the differences and set it equal to zero. What is a when we solve for it?
(figure available in print form)
If you know about sigma notation for sums, you could do the same proof for any sat of numbers not just the 3 numbers 2,6,and 7.
So we have seen two properties of the mean it is the expected or most probable score. The mean is the number the deviations from which sum to zero. There is a third property. If we squared the deviations from the mean and added the squares up the answer would be smaller than if we used any number other than the mean.
Case
|
Numbers
|
Deviations
|
Squares
|
I
|
2
|
25 = 3
|
9
|
|
6
|
6-5 = 1
|
1
|
|
7
|
7-5 = 2
|
4
|
|
|
|
14
|
Case
II
|
2
|
2-6 = -4
|
16
|
|
6
|
6-6 = 0
|
0
|
|
7
|
7-6 = 1
|
1
|
|
|
|
|
17
|