PROBABILITY
(Casino Game Example)
The Game of CRAPS
How to Play:
The player rolls a pair of dice.
1. If the roll is a sum of 7 or 11, the player wins even money.
2. If the roll is a sum of 2, 3, or 12, the player loses even money.
3. If the roll is any other sum, then that sum is called the “point”.
The player continues to roll repeatedly.
1. If the player rolls a sum of 7, the player loses even money.
2. If the player rolls the point (before they roll a sum of 7), the player
wins even money.
Ex.
If the point was a sum of 9, the player would continue to
roll until either a sum of 9 (win) or a 7 (lose) comes up
Probability Analysis:
P(win on first roll with 7 or 11) =
8
(0.2222)
36
P(win on the point) = ?????
Let’s say that the player rolls a sum of 4 on the first roll – this becomes the point.
The probability of this happening is: P(sum of 4) =
3
36
The player must roll another sum of 4 to win or a sum of 7 to lose.
Any other sum is irrelevant.
The probability now becomes conditional: P(sum 4/sum of 4 or 7 is rolled) =
Therefore, the probability of winning on a point of 4 is:
P(win on point of 4) =
3 3
x
(0.0278)
36 9
3
9
A similar calculation can be done to determine winning on a point of 5, 6, 8, 9, or 10:
P(win on point of 5) =
4
4
(0.0444)
x
36 10
P(win on point of 6) =
5
5
(0.0631)
x
36 11
P(win on point of 8) =
5
5
(0.0631)
x
36 11
P(win on point of 9) =
4
4
(0.0444)
x
36 10
P(win on point of 10) =
3 3
(0.0278)
x
36 9
To determine the player’s overall probability of winning, all the cases must be added
together (mutually exclusive):
P(WIN even money) = 0.2222 + 2(0.0278) + 2(0.0444) + 2(0.0631)
= 0.493
To determine the player’s overall probability of losing (the complementary event):
P(LOSE even money) = 1 – 0.493
= 0.507
To determine the player’s expected value on a $10 bet:
x
$10 (win)
–$10 (lose)
P(x)
0.493
0.507
E(x) = 10(0.493) – 10(0.507)
= – 0.14
 The expected value for each $10 bet is a loss of 14¢.
NOTE:
The theoretical expected value should be a reasonable value for a Casino game!!
Otherwise, payouts should be adjusted to make it more reasonable.
A pay out log (similar to the one shown) should be kept for the game on the Casino Day:
Player #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
etc.
Pay Out
$10
$10
–$10
–$10
$10
$10
$10
$10
–$10
$10
–$10
–$10
$10
–$10
–$10
–$10
etc.
Player #
Pay Out
Player #
Pay Out
The experimental expected value can then be calculated as follows:
E(X) = ∑ (pay outs) ÷ (# of players)
This experimental expected value can then be compared to the theoretical value calculated
in the probability analysis. Comments should include any reasons for discrepancies!!