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• Hard Eight Craps Payout Payouts
• Hard Eight Craps Payout Rules
• Hard Eight Craps Payouts
• Hard Eight Craps Payout Odds

Say you turned the hard 8 bet into a lay the hard 8 bet. You win on 7 or easy 8 before the next hard 8. You’re essentially acting like the casino at that point and the payouts go from 9:1 to 1:9. Shouldn’t the house edge on this new hypothetical bet be -9.09% since the house edge on hard 8 is 9.09%? Winning Payout: House Advantage: Hardways- The idea is that the combination you bet on will come up before craps OR a 'soft' version of the number (i.e., you make '4' by a 1 and 3 instead of 2 and 2)- Hard Four (2 and 2) or Hard Ten (5 and 5) 8:1: 7:1: 11.11%.

fair odds are 10:1 ... I'd guess if they ever let you play darkside you'd put up something like $15 to win $1
if you only had to put up $9 I agree that the player would get the casino's exact edge for himself
in any case that has to be the most unheard of bet in a casino - to take the large end of the bet. I can't think of anything like that for the moment
I take it back, the most unheard of thing is to intentionally allow a bet with a player edgeRS

I'm sorry but I really don't understand what you mean. Are you saying that this bet really would only be an advantage of 1% if the player could make it? I don't see how that could be.
If the casino has a 9% advantage on a hard 8, surely the player would have a 9% advantage if they could book the same exact bet as the casino, right? Putting my $9 up against the casino's $1 (which is the same scenario that plays out on a standard hard 8 bet, just reversed) is the same as putting my $1 up against the casino's $(1/9).

'House Edge' is actually a bit of a misnomer.
The casino isn't betting $1 against your $1. The casino is betting $9 against your $1. You just have a 9% disadvantage.
The casino has a 1% or so advantage on their $9 bet. You have a 9% (or so) disadvantage on your $1 bet.
IE: Player $1 bet: has 10 ways to lose $1 and 1 way to win $9. The total money wagered is $11. The sum of outcomes (10*-1 + 1*9) is -1.
-1/11 ~ -9%
The casino, on the other hand, is betting $9. 1 way to lose $9 and 10 ways to win $1. Total money wagered is $99. The sum of outcomes (-9*1 + 1*10) is +1.
+1/99 ~ +1%.ThatDonGuy

'House Edge' is actually a bit of a misnomer.
The casino isn't betting $1 against your $1. The casino is betting $9 against your $1. You just have a 9% disadvantage.
The casino has a 1% or so advantage on their $9 bet. You have a 9% (or so) disadvantage on your $1 bet.

This. +1. Whatever the term for 'exactly' is this week.
The amount of the house edge is the same on both sides of the place bet because you are putting up 9x what the casino is.
The percentage, on the other hand, is different because each side is betting a different amount.Thank you both for the explanation. It's much clearer now. I looked over the code 10x and knew it was making the bet and payouts in the exact opposite way so figured it must have just been my lack of understanding the math.

in any case that has to be the most unheard of bet in a casino - to take the large end of the bet. I can't think of anything like that for the moment
I take it back, the most unheard of thing is to intentionally allow a bet with a player edge

Hard Eight Craps Payout Payouts

When it comes table games I think Craps has the largest lay bet available at 19 to 41 when laying the 4 or 10 (when factoring in the vig). I can't think of anything with a smaller payout compared to the original bet.
You can lay much higher odds on sports. The infamous No Safety SB bet comes to mind.
This is just a thought experiment for me, not a serious idea that I'm going to try to sell to casinos or anything. They wouldn't want this bet. Obviously if a casino did implement this it would not be at 1 to 9. It would be at 1 to 10 or 1 to 11. Much like how 99% of sports betters don't lay -900 on No Safety in the SB, 99% of players would never lay 1 to 9 on a bet at a Craps table even if they had an advantage so no players would ever lay 1 to 10 or 1 to 11 on a bet where the casino had an advantage so it would just never be implemented.
I'm just messing around trying to intersect my interest in (basic) programming with my interest in gambling.

Introduction

Introduction

Welcome to the craps appendix. This is where I derive 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 product 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 odds. 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 formula for the expected return of a bet is:

∑ (probability of event i) × (return of event i) over all possible outcomes.

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 ignored 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 = -1/55 ≈ -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%

Note: These bets are not allowed in land casinos. They can only be found in some Internet casinos.

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

Note: The hard 4 and hard 10 pay 7 to 1, or 8for 1. In craps the odds on the cloth are listed on a for 1 basis, including the graphic above.

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

Note: The hard 4 and hard 10 pay 9 to 1, or 10for 1. In craps the odds on the cloth are listed on a for 1 basis, including the graphic above.

The probability of a hard 6 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+5, 2+3, 3+2, and 5+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, y 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).

Net Gain/Loss per Session

The chart below shows the net gain or loss you can expect over 100 trials, or come out rolls. For purposes of creating the chart the player would bet $1 on the pass line and take full double odds.

Hard Eight Craps Payout Rules

Here are some actual numbers that show the probability of falling into various intervals.

Session Win/Loss

Interval	Probability
loss of over $100	0.0422%
loss of $76-$100	0.6499%
loss of $51-$75	4.6414%
loss of $26-$50	16.3560%
loss of $1-$25	30.0583%
break even	0.6743%
win of $1-$25	28.6368%
win of $26-$50	14.4257%
win of $51-$75	3.9097%
win of $76-$100	0.5639%
win of over $100	0.0418%

The graph and table were created by simulating 1,000,000 sessions of 100 trials, or come out rolls, and tabulating the results of each session.

Internal Links

Hard Eight Craps Payouts

• How the house edge for each bet is derived, in brief.
• The house edge of all the major bets on both a per-bet made and per-roll basis
• Dice Control Experiments. The results of two experiments on skillful dice throwing.
• Dice Control Advantage. The player advantage, assuming he can influence the dice.
• Craps variants. Alternative rules and bets such as the Fire Bet, Crapless Craps, and Card Craps.
• California craps. How craps is played in California using playing cards.
• Play Craps. Craps game using cards at the Viejas casino in San Diego.
• Number of Rolls Table. Probability of a shooter lasting 1 to 200 rolls before a seven-out.
• Ask the Wizard. See craps questions I've answered about:
  
• Simple Craps game. My simple Java craps game.

Hard Eight Craps Payout Odds

Written by: Michael Shackleford