Craps Appendix
Introduction
Welcome to the craps appendix. This is where I
derivate the player's edge for all the major bets in craps. Outside of this
appendix I usually speak about the house edge, which is just the produect of the
player's edge and -1. To avoid multiplying by -1 for every bet I shall speak of
everything in term's of the player's edge, which you can expect to be negative
since the house ultimately has the edge on all bets except the free odss. Please
stay a while and work through some of the bets yourself. Not only will this give
you a deeper understanding of the odds but hopefully motivate you to refresh or
improve your math skills.
Before going on you must have an understanding of the probability of throwing
each total in one roll. This is explained in depth in my dice probability basics
page. If you didn't know or can't figure out that the probability rolling a 6 is
5/36 then a visit to that page is a prerequisite for this page.
The general probability for the expected return of a bet is:
(probability of event i)*(return of event i) over all events
i.
The player's edge is the expected return divided by the initial bet. For
example when betting against the line on a sporting event you have to bet $11 to
win $10. Assuming a 50% chance of winning the expected return would be 0.5*(10)
+ 0.5*(-11) = -0.5 . The player's edge would be -0.5/11 = -1/22 =~ -4.545%.
An exception to the house edge rule is when a tie is possible. In general
ties are ignorred in house edge calculations. To adjust for this, when a tie is
possible, divide the expected return by the average bet resolved. The "average
bet resolved" is the product of the initial wager and the probability that the
bet was resolved. In craps the only bets with a tie are the don't pass and the
don't come.
Many of the bets in craps win if one particular event happens before another.
These bets can take several rolls or more to resolve. If a wager wins with
probability p, loses with probability q, and stays active with probability 1-p-q
then the probability of winning eventually is:
p*(1-p-q)i (for i=0 to infinity) =
p *
(1/(1-(1-p-q))) = p * (1/(p+q)) = p/(p+q).
Throughout this page you will see a lot of expressions of the form p/(p+q).
To save space I do not derive the expression each time since it is worked out
above.
Pass/Come
The probability of winning on the come out roll is
pr(7)+pr(11) = 6/36 + 2/36 = 8/36.
The probability of establishing a point and then winning is pr(4)*pr(4 before
7) + pr(5)*pr(5 before 7) + pr(6)*pr(6 before 7) + pr(8)*pr(8 before 7) +
pr(9)*pr(9 before 7) + pr(10)*pr(10 before 7) =
(3/36)*(3/9) + (4/36)*(4/10) + (5/36)*(5/11) + (5/36)*(5/11) + (4/36)*(4/10)
+ (3/36)*(3/9) =
(2/36) * (9/9 + 16/10 + 25/11) =
(2/36) * (990/990 +
1584/990 + 2250/990) =
(2/36) * (4824/990) = 9648/35640
The overall
probability of winning is 8/36 + 9648/35640 = 17568/35640 = 244/495
The
probability of losing is obviously 1-(244/495) = 251/495
The player's edge
is thus (244/495)*(+1) + (251/495)*(-1) = -7/495 =~ -1.414%.
Don't Pass/Don't Come
The probability of winning on the come
out roll is pr(2)+pr(3) = 1/36 + 2/36 = 3/36.
The probability of pushing on
the come out roll is pr(12) = 1/36.
The probability of establishing a point
and then winning is pr(4)*pr(7 before 4) + pr(5)*pr(7 before 5) + pr(6)*pr(7
before 6) + pr(8)*pr(7 before 8) + pr(9)*pr(7 before 9) + pr(10)*pr(7 before 10)
=
(3/36)*(6/9) + (4/36)*(6/10) + (5/36)*(6/11) + (5/36)*(6/11) +
(4/36)*(6/10) + (3/36)*(6/9) =
(2/36) * (18/9 + 24/10 + 30/11) =
(2/36)
* (1980/990 + 2376/990 + 2700/990) =
(2/36) * (7056/990) = 14112/35640
The total probability of winning is 3/36 + 14112/35640 = 17082/35640 =
2847/5940
The probability of losing is 1-(2847/5940 + 1/36) = 1-(3012/5940)
= 2928/5940
The expected return is 2847/5940*(+1) + 2928/5940*(-1) =
-81/5940 = -3/220 =~ 1.364%
Most other sources on craps will claim that the house edge on the don't pass
bet is 1.403%. The source of the discrepancy lies is whether or not to count
ties. I prefer to count ties as money bet and others do not. I'm not saying that
one side is right or wrong, just that I prefer counting them. If you don't count
ties as money bet then you should divide by figure above by the probability that
the bet will be resolved in a win or loss (35/36). So 1.364%/(35/36) =~ -1.403%.
This is the house edge assuming that the player never rolls a 12 on the come out
roll.
Place Bets to Win
Place bet on 6 or 8: [(5/11)*7 +
(6/11)*(-6)]/6 = (-1/11)/6 = -1/66 =~ -1.515%
Place bet on 5 or 9: [(4/10)*7
+ (6/10)*(-5)]/5 = (-2/10)/5 = -1/25 = -4.000%
Place bet on 4 or 10: [(3/9)*9
+ (6/9)*(-5)]/5 = (-3/9)/5 = -1/15 =~ -6.667%
Place Bets to Lose
Place bet to lose on 6 or 8: [(6/11)*4 +
(5/11)*(-5)]/5 = (-1/11)/5 = -1/44 =~ -1.818%
Place bet to lose on 5 or 9:
[(6/10)*5 + (4/10)*(-8)]/8 = (-2/10)/8 = -1/40 = -2.500%
Place bet to lose on
4 or 10: [(6/9)*5 + (3/9)*(-11)]/11 = (-3/9)/11 = -1/33 =~ -3.030%
Buy
Buy bet on 6 or 8: [(5/11)*23 + (6/11)*(-21)]/21 =
(-11/11)/21 = -1/21 =~ -4.762%
Buy bet on 5 or 9: [(4/10)*29 +
(6/10)*(-21)]/21 = (-10/10)/21 = -1/21 = -4.762%
Buy bet on 4 or 10:
[(3/9)*39 + (6/9)*(-21)]/21 = (-9/9)/21 = -1/21 =~ -4.762%
Lay
Lay bet to lose on 6 or 8: [(6/11)*19 + (5/11)*(-25)]/25 =
(-11/11)/25 = -1/25 =~ -4.000%
Lay bet to lose on 5 or 9: [(6/10)*19 +
(4/10)*(-31)]/31 = (-10/10)/31 = -1/31 = -3.226%
Lay bet to lose on 4 or 10:
[(6/9)*19 + (3/9)*(-41)]/41 = (-9/9)/41 = -1/41 =~ -2.439%
Big 6/Big 8
[(5/11)*1 + (6/11)*(-1)]/1 = -1/11 =~ 9.091%
Hard 4/Hard 10
The probability of a hard 4 on any given roll is
1/36.
The probability of a 7 on any given roll is 6/36.
The probability of
a soft 4 on any given roll is 2/36 (1+3 and 3+1).
The probability of winning
on any given roll is 1/36.
The probability of losing on any given roll is
6/36 + 2/36 = 8/36.
The probability of winning the bet is p/(p+q) (see above)
= (1/36)/(9/36) = 1/9
The expected return is (1/9)*7 + (8/9)*(-1) = -1/9 =~
11.111%.
The player's edge is also -1/9 since the bet is 1 unit.
The odds
are the same for a hard 10.
Hard 6/Hard 8
The probability of a hard 4 on any given roll is
1/36.
The probability of a 7 on any given roll is 6/36.
The probability of
a soft 6 on any given roll is 4/36 (1+4, 2+3, 3+2, and 4+1).
The probability
of winning on any given roll is 1/36.
The probability of losing on any given
roll is 6/36 + 4/36 = 10/36.
The probability of winning the bet is p/(p+q)
(see above) = (1/36)/(11/36) = 1/11
The expected return is (1/11)*9 +
(10/11)*(-1) = -1/11 =~ 9.091%.
The player's edge is also -1/11 since the
bet is 1 unit.
The odds are the same for a hard 8.
Craps 2/Craps 12
[(1/36)*30 + (35/36)*(-1)]/1 = -5/36 =~
-13.889%
Craps 3/Craps 11
[(2/36)*15 + (34/36)*(-1)]/1 = -4/36 =~
-11.111%
Any Craps
[(4/36)*7 + (32/36)*(-1)]/1 = -4/36 =~ -11.111%
Any 7
[(6/36)*4 + (30/36)*(-1)]/1 = -6/36 =~ -16.667%
Horn
The probability of rolling either a 2 or 12 is 1/36 + 1/36
= 2/36.
The probability of rolling either a 3 or 11 is 2/36 + 2/36 =
4/36.
The probability of roling anything else is 1-2/36-4/36 =
30/36.
Remember that the horn bet is like all four craps bets in one. Even if
one wins the other three still lose. The house edge is:
[(2/36)*27 +
(4/36)*12 + (30/36)*(-4)]/4 = (-18/36)/4 = 12.500%
Field
When the 12 pays 2:1 the expected return
is:
2*(pr(2)+pr(12)) + 1*(pr(3)+pr(4)+pr(5)+pr(10)+pr(11)) +
-1*(pr(6)+pr(7)+pr(8)+pr(9)) =
2*(1/36 + 1/36) + 1*(2/36 + 3/36+ 4/36 + 3/36
+ 2/36) + -1*(5/36 + 6/36 + 5/36+ 4/36) =
2*(2/36) + 1*(14/36) + -1*(20/36)
= -2/36 = -1/18 =~ 5.556%.
When the 12 pays 3:1 the expected return is:
3*pr(2) + 2*pr(12)) +
1*(pr(3)+pr(4)+pr(5)+pr(10)+pr(11)) + -1*(pr(6)+pr(7)+pr(8)+pr(9)) =
3*(1/36) + 2*(1/36) + 1*(2/36 + 3/36+ 4/36 + 3/36 + 2/36) + -1*(5/36 + 6/36
+ 5/36+ 4/36) =
3*(1/36) + 2*(1/36) + 1*(14/36) + -1*(20/36) = -1/36 =~
2.778%.
Buying Odds
4 and 10: [(3/9)*2 + (6/9)*(-1)]/1 = 0.000%
5
and 9: [(4/10)*3 + (6/10)*(-2)]/2 = 0.000%
6 and 8: [(5/11)*6 +
(6/11)*(-5)]/5 = 0.000%
Laying Odds
4 and 10: [(6/9)*1 + (3/9)*(-2)]/1 = 0.000%
5
and 9: [(6/10)*2 + (4/10)*(-3)]/2 = 0.000%
6 and 8: [(6/11)*5 +
(5/11)*(-6)]/5 = 0.000%
Combined pass and buying odds
The player edge on the combined
pass and buying odds is the average player gain divided by the average player
bet. The gain on the pass line is always -7/495 and the gain on the odds is
always 0. The expected bet depends on what multiple of odds you are allowed.
Lets assume full double odds, or that the pass line bet is $2, the odds bet on a
4, 5, 9, and 10 is $4, and the odds on a 6 or 8 is $5.
The average gain is -2*(7/495) = -14/495.
The average bet is 2 + (3/36)*4 + (4/36)*4 + (5/36)*5 + (5/36)*5 + (4/36)*4 +
(3/36)*4] =
2 + 106/36 = 178/36
The player edge is (-14/495)/(178/36) = -0.572%.
The general formula if you can take x times odds on the 6 and 8, 6 times on
the 5 and 9, and z times on the 4 and 10 is (-7 / 495) / [ 1 + ((5x + 4y + 3z) /
18) ]
Combined don't pass and laying odds
The player edge on the
combined don't pass and laying odds is the average player gain divided by the
average player bet. The gain on the don't pass is always -3/220 and the gain on
the odds is always 0. The expected bet depends on what multiple of odds you are
allowed. Lets assume double odds and a don't pass bet of $10. Then the player
can lay odds of $40 for a win of $20 on the 4 and 10, $30 for a win of $20 on
the 5 and 9, and $24 on the 6 and 8 for a win of $20. The average gain is
-10*(3/220) = -30/220.
The average bet is 10 + 2*[(3/36)*40 + (4/36)*30 + (5/36)*24] = 30.
The player edge is (-30/220)/30 = -0.455%.
The general formula if you can buy x times odds then the house edge on the
combined don't pass and laying odds is (3/220)/(1+x).
Go back to craps.
