## Poll

 9.43% 1 vote (33.33%) 14.14% 2 votes (66.66%)

3 members have voted

NowTheSerpent
Joined: Sep 30, 2011
• Posts: 417
October 8th, 2011 at 12:56:28 AM permalink
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NowTheSerpent
Joined: Sep 30, 2011
• Posts: 417
October 8th, 2011 at 1:42:49 AM permalink
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MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
October 8th, 2011 at 2:32:31 PM permalink
Quote: NowTheSerpent

The 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%?
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
NowTheSerpent
Joined: Sep 30, 2011
• Posts: 417
October 8th, 2011 at 3:14:01 PM permalink
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MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
October 8th, 2011 at 3:53:26 PM permalink
Quote: NowTheSerpent

When 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?
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
NowTheSerpent
Joined: Sep 30, 2011
• Posts: 417
October 8th, 2011 at 4:20:43 PM permalink
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MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
October 8th, 2011 at 5:00:00 PM permalink
Quote: NowTheSerpent

When 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?
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
NowTheSerpent
Joined: Sep 30, 2011
• Posts: 417
October 8th, 2011 at 5:25:14 PM permalink
Quote: MathExtremist

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?

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.
NowTheSerpent
Joined: Sep 30, 2011
• Posts: 417
October 9th, 2011 at 9:34:13 AM permalink
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MathExtremist
Joined: Aug 31, 2010