November 14th, 2011 at 10:01:59 AM
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Quote:guido111

With n=496 I used Excel BINOMDIST function. A spreadsheet is very handy to use for these type of calculations.

There are even online calculators available.

= 1 - BINOMDIST(281,496,0.49317517006,TRUE)

0.000456209 or 1 in 2,192

This would show the probability of there being 281 or more player wins after exactly 496 total hands have been played (66+ more player wins than banker wins) but it could be also the case that the asked event of 66+ extra player wins is achieved at some point before all those 496 hands are played and not necessarily at the very end. So would there be a solution that says, what is the chance to see 66+ surplus player wins at any point within the first 496 hands played?

So, the OP should state whether he wishes to

a) play fixed number of hands and get the probability for certain number of surplus player wins at the end of that sample (guido's solution)

or

b) Play fixed number of hands and get the probability for certain number of surplus player wins at any point within the sample (not answered)

or

c) Play indefinitely long, up until the event of certain number of surplus player wins occurs (or doesn't) and get the overall chance of it occuring (my solution)

November 14th, 2011 at 10:16:00 AM
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Quote:Jufo81

So, the OP should state whether he wishes to

a) play fixed number of hands and get the probability for certain number of surplus player wins at the end of that sample (guido's solution)

or

b) Play fixed number of hands and get the probability for certain number of surplus player wins at any point within the sample (not answered)

or

c) Play indefinitely long, up until the event of certain number of surplus player wins occurs (or doesn't) and get the overall chance of it occuring (my solution)

b) Play fixed number of hands and get the probability for certain number of surplus player wins at any point within the sample (not answered)

B) looks to be a random walk Markov Chain solution (added) or maybe a type of negative binomial distribution.

I am very rusty with those but like good math problems.

November 14th, 2011 at 3:10:33 PM
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Looks like your thread currently has 2 answers for 3 different possible questions.Quote:jms9smithI was curious to know if anyone had the maths on the ability of player results in Baccarat to exceed bank results??

Quote:jms9smithIf the outcomes must eventually reach the probabilities of the game itself,

No.

The outcomes do not have to eventually reach the probabilities.

Why?

Because, The Law of Large Numbers says as the number of trials increase the probabilities will converge towards the theoretical value. The actual numbers can get even further away from expectation but the percentages will get closer and closer to the expected probability.

Quote:jms9smithis there anyway to determine how many more than bank results the player can win before the odds of the game are realised over say a million hands?

Well if your question is about the Player being ahead after 1 million hands, answer follows.

But if it just about at any time during those 1 million hands being up X more times than the Banker, you are now getting into some interesting and challenging math areas.

So, 1,000,000 = n.

0.506824 = q (banker win)

0.493176 = p (player win)

n*q

506,824 expected number of Banker wins in 1 million trials

n*p

493,176 expected number of Player wins in 1 million trials

13,648 Banker surplus (more wins than Player)

Variance = n*p*q

249953.4

Standard Deviation = Square root of Variance

249953.4^0.5 = 499.9534309

Let's round to 500.

The Player needs to overcome 13,648 expected Banker wins.

13,648/500 = 27.3 standard deviations

To overcome 7 SDs (from wikipedia) is about 1 / 390,682,215,445

I do not even want to go past 7 SDs.

(I'm assuming that you know what variance and standard deviation are)

Jufo81 method, without a finite number of trials, is quite simple and easy to work with if that is what you are looking for.

Let's see if we can come up with a solution to his B) question.

Quote:jms9smithI know that in theory player could win 400,000 hands in a row and then bank win the rest to make odds add up but what is the maths/probability behind the difference that player and bank can achieve? Mathematically, how many hands can the player win, more than the bank??

Edit: Jufo81 pointed out a typo in his below post :)

If this were to happen, the Banker would not have to catch up with more wins in your 1 million trial example.

With more trials the Banker could in fact catch up numbers wise, by actual wins, but will get closer and closer to the expected percentages.

p | p run total | b | b run total | diff | session | total hands | p | b |
---|---|---|---|---|---|---|---|---|

400,000 | 400,000 | 0 | 0 | -400,000 | 400,000 | 400,000 | 100.0000% | 0.0000% |

197,270 | 597,270 | 202,730 | 202,730 | -394,540 | 400,000 | 800,000 | 74.6588% | 25.3413% |

394,540 | 991,810 | 405,460 | 608,190 | -383,620 | 800,000 | 1,600,000 | 61.9881% | 38.0119% |

789,080 | 1,780,890 | 810,920 | 1,419,110 | -361,780 | 1,600,000 | 3,200,000 | 55.6528% | 44.3472% |

1,578,160 | 3,359,050 | 1,621,840 | 3,040,950 | -318,100 | 3,200,000 | 6,400,000 | 52.4852% | 47.5148% |

3,156,320 | 6,515,370 | 3,243,680 | 6,284,630 | -230,740 | 6,400,000 | 12,800,000 | 50.9013% | 49.0987% |

6,312,640 | 12,828,010 | 6,487,360 | 12,771,990 | -56,020 | 12,800,000 | 25,600,000 | 50.1094% | 49.8906% |

2,023,990 | 14,852,000 | 2,080,010 | 14,852,000 | 0 | 4,104,000 | 29,704,000 | 50.0000% | 50.0000% |

197,270 | 15,049,270 | 202,730 | 15,054,730 | 5,460 | 400,000 | 30,104,000 | 49.9909% | 50.0091% |

394,540 | 15,443,810 | 405,460 | 15,460,190 | 16,380 | 800,000 | 30,904,000 | 49.9735% | 50.0265% |

789,080 | 16,232,890 | 810,920 | 16,271,110 | 38,220 | 1,600,000 | 32,504,000 | 49.9412% | 50.0588% |

1,578,160 | 17,811,050 | 1,621,840 | 17,892,950 | 81,900 | 3,200,000 | 35,704,000 | 49.8853% | 50.1147% |

3,156,320 | 20,967,370 | 3,243,680 | 21,136,630 | 169,260 | 6,400,000 | 42,104,000 | 49.7990% | 50.2010% |

6,312,640 | 27,280,010 | 6,487,360 | 27,623,990 | 343,980 | 12,800,000 | 54,904,000 | 49.6867% | 50.3133% |

12,625,280 | 39,905,290 | 12,974,720 | 40,598,710 | 693,420 | 25,600,000 | 80,504,000 | 49.5693% | 50.4307% |

expected | 49.3175% | 50.6825% |

It would take ~29.7 million Player/Banker hands for the Banker to equal the Player wins assuming expectation holds true. I do not think any one player could witness that in their lifetime :)

winsome johnny (not Win some johnny)

November 15th, 2011 at 2:00:36 AM
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Quote:7craps

If this were to happen, the Banker would not have to catch up with more wins. With more trials the Banker could in fact never catch up numbers wise, by actual wins, but will get closer and closer to the expected percentages.

I am not sure if I understood you correctly, but banker wins will exceed player wins, no matter what, if you play long enough. Banker will always catch up numbers wise eventually even if you started at a position where player has 400,000 more wins than banker.

November 18th, 2011 at 2:46:03 PM
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thank you for doing all that.. i was curious to know if there is a formula for working out the probability of player being ahead of bank by x number of wins at any given time over any given number of hands? a formula which had no restrictions on how many hands were played but instead could be used to calculate the probability that after x number of hands the player was ahead n amount of wins. further, it could be used to calculate how many hands the player could win more than the bank.. for example what is the formula for calculating the probability that player will win 1 more hand than bank? do we need to know how many hands this is out of? what are the odds that player will win 100 more hands than bank at any time? im really sorry if the answer has already been given but i can't really follow that well. thanks guys

November 18th, 2011 at 2:53:09 PM
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Quote:Jufo81This would show the probability of there being 281 or more player wins after exactly 496 total hands have been played (66+ more player wins than banker wins) but it could be also the case that the asked event of 66+ extra player wins is achieved at some point before all those 496 hands are played and not necessarily at the very end. So would there be a solution that says, what is the chance to see 66+ surplus player wins at any point within the first 496 hands played?

So, the OP should state whether he wishes to

a) play fixed number of hands and get the probability for certain number of surplus player wins at the end of that sample (guido's solution)

or

b) Play fixed number of hands and get the probability for certain number of surplus player wins at any point within the sample (not answered)

or

c) Play indefinitely long, up until the event of certain number of surplus player wins occurs (or doesn't) and get the overall chance of it occuring (my solution)

yes i think i seek (c) i want to know what the chance of seeing say 100 (or n) surplus player hands is at any point over infinite trials (hands). thank you for clearly stating that for me :)

November 21st, 2011 at 2:54:58 PM
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Quote:jms9smith

yes i think i seek (c) i want to know what the chance of seeing say 100 (or n) surplus player hands is at any point over infinite trials (hands). thank you for clearly stating that for me :)

I already gave you the answer for that in my previous post in this thread. If you play indefinitely long starting from 0 banker wins, 0 player wins, the chance that at some point in the future there are 100 surplus Player wins is 6.5%. The general formula is (P/Q)^T where T is the number of surplus wins.

Other persons already answered the different question of the probability there being X or more surplus player wins after N total hands played.