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How does the house make a 1.4% edge on the pass line if it is winning so many times more frequently than the Don't Pass side?

Maybe I have just seen a very, very long Pass line win streak, or does the Pass line always win more often than the Don't Pass side and the house makes it odds some other way?

Quote:NewtoTownMaybe I have just seen a very, very long Pass line win streak, or does the Pass line always win more often than the Don't Pass side and the house makes it odds some other way?

You saw a long Pass Line win streak.

The Pass Line, and the Don’t Pass, each win 49.3% of the winning hands. The DP has a slightly lower House Edge due to the Push on the 12.

P W: 49.293% P L: 50.707%

DP W: 47.929% DP T : 2.778% DP L: 49.293%

If you ignored the standoffs DP wins 49.299% and loses 50.701%, so has a slightly lower House Edge.

you did not define 'far more often'Quote:NewtoTownI have used a craps simulator and the pass line wins far more often than the don't pass side.

I have also used many craps simulators but

using the "'Perfect 1980" one could see over 1,980,000 comeout rolls (pass line decisions) the pass line is expected to win

976,000 times (976/1980 - win probability)

while the Dpass wins 949,000 times (949/1980 - win probability)

for a difference of 27,000 times in favor of the Pass line.

that could be seen as 'far more often' (some call that the frequency)

as to your question on the house edge, below are a few tables to use for the calculations

Pass line house edge = 28/1980 (7/495) about: 0.014141414

Dpass line house edge = 27/1980 (3/320) about: 0.013636364

Pass Line

Event | ways | pct |
---|---|---|

comeout win | 440 | 22.22% |

comeout loss | 220 | 11.11% |

win on 6 | 125 | 6.31% |

loss on 6 | 150 | 7.58% |

win on 8 | 125 | 6.31% |

loss on 8 | 150 | 7.58% |

win on 5 | 88 | 4.44% |

loss on 5 | 132 | 6.67% |

win on 9 | 88 | 4.44% |

loss on 9 | 132 | 6.67% |

win on 4 | 55 | 2.78% |

loss on 4 | 110 | 5.56% |

win on 10 | 55 | 2.78% |

loss on 10 | 110 | 5.56% |

total | 1980 | 100.00% |

DPass

Event | ways | pct |
---|---|---|

comeout win | 165 | 8.33% |

comeout loss | 440 | 22.22% |

push | 55 | 2.78% |

win on 6 | 150 | 7.58% |

loss on 6 | 125 | 6.31% |

win on 8 | 150 | 7.58% |

loss on 8 | 125 | 6.31% |

win on 5 | 132 | 6.67% |

loss on 5 | 88 | 4.44% |

win on 9 | 132 | 6.67% |

loss on 9 | 88 | 4.44% |

win on 4 | 110 | 5.56% |

loss on 4 | 55 | 2.78% |

win on 10 | 110 | 5.56% |

loss on 10 | 55 | 2.78% |

total | 1980 | 100.00% |

as pointed out the DPass Standoff (PUSH) is the difference and it does happen often enough for the house to offer the bet (been around since the early 1900s)

If I'm alone at the table, maybe I'll set the dice for a 5 or a 9 on the come-out, and set for a 7 after the point is established.

I was thinking there must be an upper limit at some point.

In other words, after how many total decisions should it be before I start to see a 1.41% edge for the house (or in the case of the current simulator, at 500 decisions and climbing I am still way in favor of the pass line.

If you play for several hours a day, try to quit ahead well before you get to 5,000 decisions. It only gets worse the longer you go on. Like start each day with a simple buy-in and have win goals like double or triple your money unless you're on a hot streak. For tax reasons, my gambling diary works better if I quit the day ahead. Losses may not be tax deductible, so I try to limit those.

I've seen graphs where simple PL & DP results just show a negligible downward slope that adds up over time, but when you add odds bets you can seriously affect your balance, be it positive or negative. Some people like to play a couple come bets with odds and a long roll can make up for a couple days of losses.

My scenario is after almost 2,200 decisions I am at a Pass ratio of 170 to DP ratio of 143 (170 wins per 143 losses, had I bet only the P side). This seems like a lot of wins on the pass side for a very long time period... although I get it, in the scheme of things, it could/should reverse back to where the Pass side is at 49.28 wins (P) to 50.72 losses (DP).

I was wondering if this number of winning pass decisions (as a ratio of wins to losses) is atypical for a string of 2,200 decisions??

Hoping you/someone has been playing long enough to have seen this or know the math behind if this is possible, or maybe if the simulator is skewed toward Pass Line decisions...

goatcabin

https://wizardofvegas.com/forum/off-topic/general/972-the-hoax-that-is-the-1-41-house-advantage-on-pass-line-bets/4/

1980, of which 976 are winners, 1004 losers, 784 are seven-outs, so the probability of sevening out on any given bet is 784/1980 = .396

Down 28 units after 1980 decisions. YMMV

These are decisions on your bet, not the number of times the dice are thrown.

Let's use your number of 5000 outcomes (rather than not rolls) betting $1, and use (i) Average = Np (ii) STdDev = SQRT(Npq).

Your average number of wins is 5000*(976/1980) = 2464.6465, which means you lose 2535.3535 making a loss of $70.7071.

This is weird as one Standard Deviation is SQRT ( 5000 * 976/1980 * 1004/1980 ) = SQRT ( 1249.75 ) = 35.3518.

This means that you have to be almost exactly more than 2 SDs lucky, i.e. about 2.5% chance of being ahead after 5000 outcomes ( https://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule ).

Quote:ChumpChangeFebruary 14th, 2010 at 12:26:06 AM

goatcabin

https://wizardofvegas.com/forum/off-topic/general/972-the-hoax-that-is-the-1-41-house-advantage-on-pass-line-bets/4/

1980, of which 976 are winners, 1004 losers, 784 are seven-outs, so the probability of sevening out on any given bet is 784/1980 = .396

Down 28 units after 1980 decisions. YMMV

These are decisions on your bet, not the number of times the dice are thrown.

Thanks for bringing us back to a classic WoV thread. Met the OP in Biloxi some time later, he was serious, not a troll. Of course he was seriously wrong.

I think you mixed up some dataQuote:charliepatrickIt's perfectly possible that after 5000 Pass line bets you might be ahead, in fact about 1 time in 40.

This means that you have to be almost exactly more than 2 SDs lucky, i.e. about 2.5% chance of being ahead after 5000 outcomes ( https://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule ).

I did not look to see what exactly you calculated

for 5k resolved same bet pass line bets there is a 15.5% probability of showing a profit after 5k wagers.

easily done by many over a lifetime of bets

this is actually very easy to calculate using a Markov chain

here is some data

5k bets resolved (same bet)

loss:0.83791069

even:0.00684363

ahead:0.15524568

10k bets resolved (same bet)

loss:0.91987749

even:0.00293518

ahead:0.07718733

20k bets resolved (same bet)

loss:0.97686668

even:0.00076354

ahead:0.02236978

100k bets resolved (same bet)

loss:0.99999607

even:0.00000011

ahead:0.0000037 (about 1 in 270,271)

Sorry my maths above was out by a factor of two (not enough beer)!Quote:charliepatrick...Let's use your number of 5000 outcomes (rather than not rolls) betting $1, and use (i) Average = Np (ii) STdDev = SQRT(Npq).

Your average number of wins is 5000*(976/1980) = 2464.6465, which means...

This is weird as one Standard Deviation is SQRT ( 5000 * 976/1980 * 1004/1980 ) = SQRT ( 1249.75 ) = 35.3518....

I ran a simulation (100000 runs of 5000) and agreed with your numbers:-

Ahead 15.548%, Same 0.645% Behind 83.807%.

My mistake was to double count the difference, the average number of wins for 5000 outcomes is 2464.65. It only has to get to 2500 or better to show a profit. This is only ONE Standard Deviation out. So ignoring the chances of being spot on one end of the tail is (100% - 68.27%) / 2 = 15.86% which closely matches Ahead and half the Same.

So it might well be true that a Craps player who doesn't play very much is still ahead.

and for those that do play a lot, they can still show a profit from just one bet. depends on how many bets they actually make and resolve.Quote:charliepatrickSo it might well be true that a Craps player who doesn't play very much is still ahead.

I doubt one can find a craps player that makes just one bet, the same all the time.

Take a Place 6 bettor only.

the probability of showing a net loss

after 5k resolved wagers is 0.83833106

a net profit probability of 0.16166894

Most will make a Place 6 and Place 8

the probability of showing a net loss

after 5k resolved wagers is 0.82875728

a net profit probability of 0.17124272

some may form a conclusion on this while not seeing the whole picture (or understanding the picture).

it can take many resolved wagers until the expectation plays out

(ev dominates standard deviation)