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Casino Games - Craps
Author: Yui-Bin Chen, Ph.D.
The Copyright of this article is held by the author.
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Table Layout
_______________________________________________________
| | | | | | | DON'T | DON'T |
| 4 | 5 | SIX | 8 | NINE | 10 | COME | PASS |
_____|_____|_____|_____|_____|______|______| bar 6-6| bar 6-6|
| |________|________|
| C O M E | |
|_____________________________________| |
| | |
| (2) 3 4 9 10 11 (12) | |
| F I E L D | |
|_____________________________________| |
| /
| P A S S /
_____|____________________________________________________/
First, we need to look at the probability of each number being shown:
Pr(2) = Pr(12) = 1/36
Pr(3) = Pr(11) = 2/36
Pr(4) = Pr(10) = 3/36
Pr(5) = Pr(9) = 4/36
Pr(6) = Pr(8) = 5/36
Pr(7) = 6/36
The expected return of bet X is
E(X) = sum of (Pr(Xi) × V(Xi)) for all i,
where V(Xi) is the return value of Xi case.
In the rest of this article,
we will discuss Pass, Don't Pass, and Field bets as follows.
First Roll
The roller shoots a pair of dice. If the sum of numbers on the dice
is 7 or 11, it is a "Pass" roll; if the sum is 2, 3 or 12, it is a
"Craps" roll; other numbers make points. After a point is made, the
roller continues to shoot dice until either the point is shown (pass)
or 7 is shown (don't pass, 7-out). The roller gives the dice to the
next player when he/she shoots a "Craps" or "7-out."
Pass Bet
| Case 1 |
When 7 or 11 is shown, you win
1 unit and keep your original bet.
So, Pr(X1) = (6+2)/36 = 8/36 and V(X1) = 2. |
| Case 2 |
When 2, 3, or 12 is shown,
you lose your bet.
So, Pr(X2) = 4/36 and V(X2) = 0. Indeed, you don't need to count
this case, because V(X2) = 0. But to make sure we do not miss
any count, we want sum of all Pr(Xi) = 1. |
| Case 3 |
When 4 or 10 is shown
(Pr(X3) = 6/36), your winning chance is
(3/36) / (3/36 + 6/36) = 1/3 (only 7 and point counts).
Therefore, V(X3) = 2 × 1/3 = 2/3. |
| Case 4 |
When 5 or 9 is shown
(Pr(X4) = 8/36), your winning chance is
4 / (4+6) = 2/5 and V(X4) = 4/5. |
| Case 5 |
When 6 or 8 is shown
(Pr(X5) = 10/36), your winning chance is
5 / (5+6) = 5/11 and V(X5) = 10/11. |
Overall, E(X) = 8/36 × 2 + 0 + 6/36 × 2/3 + 8/36 × 4/5
+ 10/36 × 10/11
= 488/495 = 0.98586,
which means you lose 7/495 = 1.414% edge.
Don't Pass Bet
| Case 1 |
When 7 or 11 is shown,
you lose your bet and V(Y1) = 0. |
| Case 2 |
When 2 or 3 is shown,
you win your bet. So, Pr(Y2) = 3/36 and V(Y2) = 2. |
| Case 3 |
When 12 is shown, it is a push
and Pr(Y3) = 1/36, V(Y3) = 1. |
| Case 4 |
When 4 or 10 is shown
(Pr(Y4) = 6/36), your winning chance is
(6/36) / (3/36 + 6/36) = 2/3 (only 7 and point counts).
Therefore, V(Y4) = 2 × 2/3 = 4/3. |
| Case 5 |
When 5 or 9 is shown
(Pr(Y5) = 8/36), your winning chance is
6 / (4+6) = 3/5 and V(Y5) = 6/5. |
| Case 6 |
When 6 or 8 is shown
(Pr(Y6) = 10/36), your winning chance is
6 / (5+6) = 6/11 and V(Y6) = 12/11. |
Overall, E(Y) = 0 + 3/36 × 2 + 1/36 × 1 + 6/36 × 4/3
+ 8/36 × 6/5 + 10/36 × 12/11
= 217/220 = 0.98636,
which means you lose 3/220 = 1.364% edge.
Indeed, E(Pass) + E(Don't Pass) = 2 - 1/36 (because of the push on 12).
This means that if you bet $1 on Pass and $1 on Don't Pass, 1/36 of chance
that you will lose $1 (on your Pass Line).
Odds Bet
Now, let's consider the odds bet. After a point is made, players can
put Odds on their Pass and Don't Pass (and Come, Don't Come) bets. The
Odds can be up to K times of the original bets determined by the house.
All Odds are paid fairly. That is, the payout of the Odds on Pass is
Pr(7)/Pr(point) to 1; the payout of the Odds on Don't Pass is Pr(point)/Pr(7)
to 1.
Since it get paid fairly, the expected return will be 1. Assume the casino
allows players to place a K times odds. The total expected return is now
E(Pass) + K*Pr(a point is made) for the Pass Line, but the total bet is
1 + K*Pr(a point is made) units. Since Pr(a point is made) = Pr(4)+Pr(5)
+Pr(6)+Pr(8)+Pr(9)+Pr(10) = 2/3. So, the house edge is 1 - (E(Pass) + 2K/3)
/ (1 + 2K/3). On Pass Line, with double odds, the house edge is 0.606%;
with single odds, 0.848%. On Don't Pass Line, with double odds, the house
edge is 0.585%; with single odds, 0.818%. With Frontier's 10X odds, the
house edges are reduced to 0.184% (Pass) and 0.178% (Don't Pass).
Recently, in Las Vegas, there are casinos (Binion's Horseshoe &
Stratosphere) which offer a 100X odds game. This will reduce
the house edge to 0.021% (Pass) and 0.020% (Don't Pass).
Note that although the odds bet is a fair bet,
it indeed increases the risk. The odds bet amount may not be
counted for comp.
For a long term, you will lose
TOTAL-BETS-ON-PASS × E(Pass), if you only
bet on Pass Line and Odds. There is no known strategy to avoid this loss.
Come and Don't Come are similar to Pass and Don't Pass, and I think the
house edges are the same (because each bet can be considered as an
independent bet).
Field Bet
(win on 2, 3, 4, 9, 10, 11, and 12; lose on 5, 6, 7, and 8)
It may pay differently when 2 or 12 is shown. So, let's break it into
3 groups: 2, 12, and others for all winning numbers. Then the expected
return can be calculated by
E(Field) = Pr(2) * V(2) + Pr(12) * V(12)
+ Pr(Others) * V(Others)
= (V(2) + V(12) + 14*2) / 36
If the payoffs of 2 and 12 are 2:1, then V(2)=V(12)=3 and E(Field) = 34/36
(a losing edge of 5.56%); if the payoff of 2 is 2:1 and the payoff of 12
is 3:1, E(Field) = 35/36 (a losing edge of 2.78%).
Place Bet
You can bet whether an indicated number (4, 5, 6, 8, 9, or 10) comes
before 7. If you bet "to win", it means you bet that the number comes
before 7; if you bet "to lose", it means you bet that the number comes
after 7. The payoff is varied depending on which number you bet.
The payoff and casino edge of each number are listed below:
| To Win |
|---|
| Place Number |
Winning Chance |
Payoff |
Casino Edge |
|---|
| 4 or 10 |
3/9 |
9:5 |
6.667% |
4 or 10
(Bought) |
3/9 |
2:1
(-5% commission) |
5.000% |
| 5 or 9 |
4/10 |
7:5 |
4.000% |
| 6 or 8 |
5/11 |
7:6 |
1.515% |
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