Hard Pass math - what's the edge?
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Poll
 | 1 vote (33.33%) | ||
| | 2 votes (66.66%) |
3 members have voted
1) Shooter comes out. If it's a natural winner or craps, bet gets no action.
2) If an easy point rolls, bet loses. That's easy 4, 6, 8, 10 or 5, 9.
3) If a hard point number rolls (hard 4, 6, 8, 10), the shooter keeps rolling.
4) If the shooter makes his point hard, the bet wins 50-to-1.
5) If the shooter 7s out or makes his point easy, the bet loses.
Calculation 1:
p(win) = 1.12%
p(push) = 33.33%
p(lose) = 65.54%
EV = 1.12%*50 +33.33%*0 + 65.54%*-1 = 9.43%
Calculation 2 (from boymimbo):
So you have a 1/6 x (1/4 x 1/9 x 2 + 1/4 x 1/11 x 2) = 1/6 x (2/36 + 2/44) = 1/108 + 1/132 = 1 / 59.4.
So you pay it at 50 to 1.
HA = (51 - 59.4) / 59.4 = 14.14 percent
Question: is the house edge 9.43% or 14.14%?
This is effectively the same question as "what's the house edge on the don't pass", except for that bet nobody really cares about the 0.04% difference. Here, the difference between counting pushes vs. not is enormous -- over 4.7%. That's a huge difference in the mind of a gambler or a casino operator. 9.43% is right in the middle of the prop bet range and is therefore very reasonable. 14.14% is the worst EV on the table except for any 7.
I'll tell you my inclination in a bit -- and why I think it informs my opinion of the don't pass bet -- but first, what are your thoughts?
I broke it into two stages of the bet:
1. Since only point numbers affect the bet, and the naturals are no action/push, then Hard point wins 5:1 (true odds), and an easy point loses - but winning the bet parlays up at 4:1 so, it's 16.667% HE - just like the any 7 bet, where 5:1 true odds pays 4:1.
2. If won, it is parlayed up the the particular hardways thrown for 5 units (the 4 won plus the original bet). On a 6 or 8 hardways, those 5 units will pay 10 for 1 (9:1) and down for 50 for 1 at .0909% HE at this stage; on a hard 4 or 10, it pays 8 for 1 (7:1) and down, for .111% HE at this stage, and should be a total of 40 for 1 on 4 & 10.
If it also pays 10 for 1 and down TOTAL on a point of 4 and 10 after the parlay to still get a 50 for 1 payout, this stage of the bet gives a huge player edge, as the player makes about 12% edge (-12% HE) - getting paid 10 for 1 instead of 8 for 1 on the 4 & 10 hard points. Combined with the original stage of the parlay, the combined bet still has a house edge at about 5%, but is a real player discount if a point of 4 or 10 is rolled, as a 5% HE on a 50 for 1 payout is a great value. Bets with payouts of 50:1 should have a 15% HE.
This because the 4 & 10 hardways bet stage pays 7:1 with HE, pays 8:1 at true odds, but is now paying 9:1 (or 10 for 1) as the second stage of the bet. This stage of the bet actually gives a players edge - but they cannot get here unless they've gone through the first parlay stage, which has a strong house edge.
If the bet's final payout is 40 for 1 on 4 & 10, and 50 for 1 on 6 & 8, it'll be 14+ %, a good percentage for very high payout prop bets. If it pays 50 for 1 on all, it'll be ~ 9%, which cuts it closer for the house, and is a good value for the players, considering the payout.
If you combine the bet as you have here, or if you separate it out by making it a "manually parlayed" bet by moving the "hit" hard points on a fresh bet to the hardways, both look like exciting bets to play.
Quote: MathExtremist
9.43% is right in the middle of the prop bet range and is therefore very reasonable. 14.14% is the worst EV on the table except for any 7.
I'll tell you my inclination in a bit -- and why I think it informs my opinion of the don't pass bet -- but first, what are your thoughts?
I feel that since this bet pays 50 for 1, then 14% is okay, and 9% might be cutting it close. The 12 bet pays 30:1 at 16.67%, or 31:1 at 13.89% - and that's less than 50 for 1.
EZ Pai Gow has a "Queen's Dragon" bet that pays 50:1 if the dealer gets a Queen-high Pai Gow hand, for a 10% edge at that payout, but the original bet was 45:1 at 18%. I admit it gets a lot more action at 50:1, which is what you want on a new game.
On the other hand, I agree that the EV is -9.43%. But the EV and HA are different figures.
Consider a lame game where you roll a die: push on 1-5 and lose $6 on 6. The EV is 0*5/6 - 6 * 1/6 = -$1.
If you calc the house edge over all bets, it would imply a HA of 1/6 = 16.7%
This figure is misleading, because you have no chance of winning... You could do better at keno, even with a worse house edge.
The HA on my lame dice game should be stated as 100%, to show you are guaranteed to lose.
If you make it a combined bet "Any hard pass" to win on a hard point being made, I assume that version can only be played on the come-out roll. It'll sit on the Hard pass box until hardways point (win), or easyways point or seven-out (lose.) Then it starts again.
If you broke it out as a "Hard pass/come" bet, and moved the parlays to the hardways, then the player can make the bet before any roll - increasing action. He can make the bet with his come bets, if he is a come bet player!
Moving it to the hardways will have the 4 & 10 bets at their regular 8 for 1, and on the 6 & 8 it'll pay 10 for 1; again, parlayed from 5 units, it's 40 for 1 and 50 for 1.
For each roll, Dealers will pay or take the regular hardways first, THEN move the new hard pass/come bets to the hardways.
I was dealing craps the other night, and on a dead game bounced this idea by them. Their reaction?
"Good bet - players might love it - but Ug! - more work."
I added it might make a good tip bet. "Nickel Hard pass for the dealers" hitting would be $250 in the box. Sure beats the crap out of a dollar hard six!
Quote: dwheatleyYou should measure the house edge per RESOLVED bet. I wouldn't count pushes. Then I get the same result as boyminbo, 14.14%.
On the other hand, I agree that the EV is -9.43%. But the EV and HA are different figures.
Consider a lame game where you roll a die: push on 1-5 and lose $6 on 6. The EV is 0*5/6 - 6 * 1/6 = -$1.
If you calc the house edge over all bets, it would imply a HA of 1/6 = 16.7%
This figure is misleading, because you have no chance of winning... You could do better at keno, even with a worse house edge.
The HA on my lame dice game should be stated as 100%, to show you are guaranteed to lose.
Interesting theory. If you put $1 on each of the 38 spots on the roulette wheel, what would you say your HA is? I think we agree the EV is -5.26% or -$2 actual.
Roulette gets resolved on every spin. Craps does not. The 14.14 percent was based on resolved bet because I assumed that the player would not take the bet down after a push.
All I know is that if you successfully market this game at 50-1 per resolved bet and you push on the 7-11 or craps, I want a piece of the action.
You can talk about combined house edges on sequential bets (in 3 card poker where you have to raise), but these 38 bets are simultaneous and dependent. How much money is actually at risk? I have argued only $2 is at risk, that is, only $2 worth of bets are being resolved.
Then, the HA on the roulette cover is 2/2 = 100%.
Quote: boymimboRoulette gets resolved on every spin. Craps does not. The 14.14 percent was based on resolved bet because I assumed that the player would not take the bet down after a push.
Right, but blackjack gets resolved every bet too, and there are a lot of pushes. What's the HA on blackjack when you factor out the pushes? If the EV of blackjack is -0.5%, and the push percentage is 8% (roughly) then would you say the HA is -0.5% / 92% = -.54%?
Quote: boymimboAll I know is that if you successfully market this game at 50-1 per resolved bet and you push on the 7-11 or craps, I want a piece of the action.
Stay tuned for the answer to the "successfully" question. In the meanwhile, if you want another piece of the action, see my latest blog post. I'm going to Vegas next week...
Quote: NowTheSerpentThe question, "Is the house edge on the 'Don't' -1.40% or -1.36%?" is answered by saying that the house edge can only be one single value, either -1.40% or -1.36%. It can't be both. The two formulas
[(Total of # Of Wins)(Payout Per Win) - (Total of # Losses)(Payout Per Loss) + (Total of # Of Ties)(Payout per Tie)]/(Total Bet On ALL Contingent Ways)
and
(Total Return Over All Contingent Ways - Total Bet Over All Contingent Ways)/(Total Bet Over All Contingent Ways)
must both produce the same number. The popular practice in many books of dividing -27 by 1,925 in the first formula to get -1.4026% is based on the fallacy of assuming that because the come-out '12' result doesn't get paid, those 55 ways don't count as betted ways. When you place a bet on the "Don't", the outcome of that bet is still future, and all contingencies are bet on - the bet "covers" all possible wins, losses, and draws. It's not true that results that count as wins but don't get paid don't count as covered by the bet - yes they do! Dividing by 1,925 instead of by 1,980 has never been correct - just propagated by many parrots, so don't be misled.
I think you meant to say "it's not true that results that count as *draws* but don't get paid don't count as covered by the bet" -- I'm unaware of any bet where a "win" doesn't get paid. But help me understand your position. You computed the house edge on the don't as:
[949(1) - 976(1) + 55(0)]/[949 + 55 + 976] = (-27)/(1980) = -1.36%
Therefore, is the house edge on the Place 6 bet properly computed as follows:
[5(7/6) - 6(1) + 25(0)]/[5+6+25] = (-1/6)/(36) = -0.463%
?
In other words, do you dispute the standard figure of -1.52%?
Quote: NowTheSerpentWhen you compute the familiar 1.52% edge figure you aren't counting as rolls any results other than "6" or "7", i. e. you don't even consider any non-6 or non-7 rolls which might happen in-between as occurring. If you do count "12" or "4" or "9" rolls in between, then you are actually applying the per-roll figure you presented above. The subtlety is not one that gets pointed out very often. When using the 1.52%, or 4%, or 6.67% figures, imagine, for each one, the dice as able to roll ONLY the place number or "7", like an 11-, 10-, or 9-sided "coin" with 6 "7" faces and 5 "6", 4 "5", or 3 "4" faces, respectively. It's mathematically, then, the same thing. The 1.52% "Place" figure applies to the bet that the number will occur before 7. The 0.463% figure you provided would apply to the bet that any non-7 will roll next. In the former case only "6" and "7" are contingencies; in the latter case, all numbers, "2" thru "12" are contingencies.
Thanks for the question.
Okay then -- basically, you're saying that the math for the place 6 bet should not count any 12 rolls because 12 doesn't resolve the wager. But 12 doesn't resolve the don't pass bet either, yet you *are* counting the 12 rolls in the house edge. If you weren't, you'd end up with 1.40%.
Why is it okay to count the 12 for the don't pass bet EV but not for the place 6 bet EV?
Quote: NowTheSerpentWhen you make a "Don't-Pass" or "Don't-Come" bet, your proposition is: "that a either '2', '3', or '12' will roll immediately
How do you figure that you're betting on a 12 when the rules explicitly state "12 does not win"?
How does 12 not winning on a don't pass bet differ than 12 not winning on a place 6 bet?
Quote: MathExtremistHow do you figure that you're betting on a 12 when the rules explicitly state "12 does not win"?
How does 12 not winning on a don't pass bet differ than 12 not winning on a place 6 bet?
To put it simply, the Don't Pass is a bet that the Pass Line wager will lose. Therefore, your Don't Pass wager covers the Craps Twelve, in the sense that the proposition implies a decision on the come-out Twelve ("Craps 12"). That decision is that the Don't wins, since the Pass Line is declared a loser, which is precisely what you proposed would happen in your Don't Pass wager. However, the casino has to cover all bets by guaranteeing a negative player expectation (NPE) on all viable bets. They do this by returning the Don't on a come-out Twelve at 1-for-1, or a "push" instead of at 2-for-1, or "even money" (This is really no different from BJ returning 11-for-5 instead of 5-for-2 in order to merely increase the NPE). With the Place-6, the only numbers you cover to begin with are the Six and the Seven because, according to the proposition, they are the only numbers which can determine the bet either way. The Place-6 that wins returns 13-for-6; losers return 0-for-1. No biggie.
Quote: NowTheSerpentYep. Sure. You're right! My bad. I just don't know my ass from a hole in the ground. Fuck me!
You know, we were merely having an innocuous conversation about the definition of house edge. If you decided you no longer wanted to participate, it would have been easier to simply not respond than to lose your cool in public and delete all your posts like the other poster did. Unless you are that poster, in which case I would expect nothing less from someone who thinks they could win the Nobel in economics for evaluating some roulette odds.
Quote: MathExtremistPeople who never question what they're told are doomed to a life of gullible subservience.
Those who always question what they are told tend to be doomed to such a life also, since large governments and corporations will give them an ultimatum no less quickly than they will to the aforementioned gullible subservients.
But in response to your original post question it appears that the proper HA figure on your bet is 14.14%
probability of setting a Point (4, 6, 8, or 10) via hardways: 4 out of 16 or 0.25.
probability of winning on a Point of 6 or 8: 1 out of 11 "6" or "8" establishments, or 0.090909090909......
probability of winning on Point of 4 or 10: 1 out of 9 "4" or "10" establishments, or 0.1111111111111.....
So, probability of winning = (prob. of setting "4" via 2&2) x (prob. of winning as 2&2) + (prob. of setting "6" via 3&3) x (prob. of winning as 3&3)
+(prob. of setting "8" via 4&4) x (prob. of winning as 4&4) + (prob. of setting "10" as 5&5) x (prob. of winning as 5&5)
= (1/16)(1/9) + (1/16)(1/11) + (1/16)(1/11) + (1/16)(1/9) = 5/198 or 2.52525....%
I believe (though I don't still remember for sure) you suggested a payout of 33 to 1 on this bet? If so, then the return is 34 x 0.025252 = 0.858585, which leaves 14.1414% to the House, which, for a payout at this level, is reasonable.
