jms9smith
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November 13th, 2011 at 2:17:47 PM permalink
I was curious to know if anyone had the maths on the ability of player results in Baccarat to exceed bank results?? If the outcomes must eventually reach the probabilities of the game itself, is 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? I 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??
SOOPOO
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November 13th, 2011 at 2:47:44 PM permalink
jms, i think you have to ask the question a different way, and then someone here can help you..... How does this sound....
What are the odds of player winning baccarat 500,001 or more out of 1,000,000 (count a tie as 1/2 win for player and 1/2 win for banker)?
TheNightfly
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November 13th, 2011 at 3:03:37 PM permalink
Quote: jms9smith

I know that in theory player could win 400,000 hands in a row and then bank win the rest to make odds add up...

In theory, the player hand could win 400,000 hands in a row... and then win the next 600,000 in a row. In theory, anything can happen. In reality, the odds of this happening are extremely small. In reality, the odds will, over an infinite number of hands (or let's say 10 billion to be a little more realistic) end up being very close to what the math of the game suggests. The math of the game only tells you what, over the long run, you should expect to see happen. You play in reality though so you'll never be able to know with any certainty what is going to happen next based upon what has happened in the past. To answer your question a little more clearly, you could in your lifetime see a shoe where player wins every hand but it would be extremely unlikely. Using math (and keep in mind, mine is a little fuzzy) the odds of having player win every hand in an 80 hand shoe (excluding ties) are approximately 2.76596E-25.
Happiness is underrated
jms9smith
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November 13th, 2011 at 3:21:39 PM permalink
Hi guys, thanks for the replies. I know my question wasn't worded too well, so i'm sorry for that. what i really was curious on was; is there a way to work out the probability of player winning more hands than bank?
For example, if i wanted to know what the odds of player winning 50 more hands than bank, what would the math be?
Also has anyone experienced player winning say 100 more hands than bank?
thank you for the replies.
jms9smith
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November 13th, 2011 at 3:23:50 PM permalink
Quote: SOOPOO

jms, i think you have to ask the question a different way, and then someone here can help you..... How does this sound....
What are the odds of player winning baccarat 500,001 or more out of 1,000,000 (count a tie as 1/2 win for player and 1/2 win for banker)?



that's a really good question as well...... is there a way to work that out?
Jufo81
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November 13th, 2011 at 5:22:26 PM permalink
Quote: jms9smith

I was curious to know if anyone had the maths on the ability of player results in Baccarat to exceed bank results?? If the outcomes must eventually reach the probabilities of the game itself, is 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? I 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??



You should ask the question more precisely. Like for example: What is the probability that after 1,000 Baccarat hands there will be more player wins than banker wins in total? It would be easy to find an answer to questions like these.

I thought of the following that might be something that you are seeking:

If we play Baccarat indefinitely long then what is the probability that at some point there will be 10 more player wins than banker wins?

To answer this, I took the player's and banker's winning probabilities from 8 deck game:

P = P(Player wins) = 0.493176
Q = P(Banker wins) = 0.506824

Note that these probabilities have been formed by eliminating ties (non-relevant outcome) already.

The answer to the above question is obtained from a surprisingly simple risk of ruin formula: PP = (P/Q)^T, where T is number of number of surplus player wins we are looking for.

Results in my next post.

[Post edited to make equations simpler].
jms9smith
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November 14th, 2011 at 1:45:54 AM permalink
thanks for that but i see that on the wizard of odds site an example of 282 player and 214 bank (a difference of only 68) has a probability of 0.000393, or 1 in 2,544. That's less than the value you gave for the N=100 example. How come?
Are the odds difference if all you want to know is what is the chance of player winning 100 more hands than bank? Does anyone have a formula for working out what the odds of player winning 50 or 100 or N more hands than bank?
Jufo81
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November 14th, 2011 at 4:34:09 AM permalink
Quote: jms9smith

thanks for that but i see that on the wizard of odds site an example of 282 player and 214 bank (a difference of only 68) has a probability of 0.000393, or 1 in 2,544. That's less than the value you gave for the N=100 example. How come?



That's because in the example it's odds of player winning 68 more hands than banker in 282+214 = 496 hands played. With so low number of hands the probability is low. My formulas on the other hand gave the odds with no constraint for the total number of hands (ie. playing indefinitely long).

Quote: jms9smith


Are the odds difference if all you want to know is what is the chance of player winning 100 more hands than bank? Does anyone have a formula for working out what the odds of player winning 50 or 100 or N more hands than bank?



I gave you the formula in my previous reply. But like said that formula gives the answer for the "infinite case" ie. not restricting the total number of hands played. If you state the question as: "What is the probability for player winning 50 or 100 more hands than bank in XXX hands played" then that requires a different formula.
Jufo81
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November 14th, 2011 at 4:45:55 AM permalink
Actually in my previous post the stopping condition "Banker wins N more hands than player" is not necessary. So one can play infinitely long. The probability that in such infinite Baccarat session there will be T more player wins than banker wins is obtained simply from equation:

(P/Q)^T

Where

P = Prob(Player wins)
Q = Prob(Banker wins)

For T = 1, we get 97.3%, which shows that there is a 97.3% chance that such infinite session will have one more player wins than banker wins at some point in the future.

Values for other T are:

T= 10, PP = 0.7611
T = 100, PP = 0.065
T = 200, PP = 0.0042555
T = 300, PP = 1 / 3602
T = 500, PP = 1 / 846485

So we see that between T = 200 and T = 300 the probabilities start to fall under less than 1 in 1000 chance region. This means that it is extremely unlikely to find a sequence of Baccarat results (no matter how long they are) where player has won more than 300 hands than the banker.
guido111
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November 14th, 2011 at 8:51:53 AM permalink
Quote: jms9smith

thanks for that but i see that on the wizard of odds site an example of 282 player and 214 bank (a difference of only 68) has a probability of 0.000393, or 1 in 2,544. That's less than the value you gave for the N=100 example. How come?
Are the odds difference if all you want to know is what is the chance of player winning 100 more hands than bank? Does anyone have a formula for working out what the odds of player winning 50 or 100 or N more hands than bank?


askthewizard/baccarat
Question#4
The Wizard used the normal distribution to answer that question and it is an approximation to the binomial distribution.
Since we know the number of hands (trials) this can be calculated very easily.

Jufo81 solution, a gamblers ruin solution, assumes no exact number of trials and a nice solution I might add.

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

Here is a sample of my Binomial Dist table.
You can create your own and/or compare the results to the Wizards using his formula of expected value, standard deviation and z-score.

The AB diff column (absolute difference) is the column to find the difference between Player and Banker wins.
Expectation is 245 wins (the 5th row)

For 298 Player wins in 496 trials, (a 100 win difference) the "1 in (or more) column shows a 1 in 1,046,790.716 of Player being 100 wins or more than the Banker

For 310 Player wins in 496 trials, (a 124 win difference) the "1 in (or more) column shows a 1 in 415,584,611.3 of Player being 124 wins or more than the Banker

The NET column is from the Player wager of $10 every hand.
I hope this helps out.
n=496
winsAb difflosses(or less)exact1 in (exact)(or more)1 in (or more)winsNETEDGE
241-142550.38992330.03398299129.426485620.6440596911.552651118241-140-0.028225806
242-122540.4247674390.03484413928.699231250.61007671.63913816242-120-0.024193548
243-102530.4602079920.03544055428.216263410.5752325611.73842732243-100-0.02016129
244-82520.4959660990.03575810627.965686610.5397920081.852565406244-80-0.016129032
245-62510.5317553230.03578922527.941370910.5040339011.983993532245-60-0.012096774
246-42500.5672885160.03553319228.142700990.4682446772.135635598246-40-0.008064516
247-22490.6022846920.03499617728.57455010.4327114842.311008688247-20-0.004032258
24802480.6364756750.03419098329.247477490.3977153082.51436135424800
24922470.6696122210.03313654630.178160360.3635243252.750847553249200.004032258
25042460.7014694140.03185719331.390085430.3303877793.026746337250400.008064516
25162450.7318511230.0303817132.914540180.2985305863.349740514251600.012096774
25282440.7605933940.0287422734.791962710.2681488773.729271634252800.016129032
253102430.787566670.02697327637.073731840.2394066064.1769941732531000.02016129
254122420.8126768350.02511016539.824508780.212433334.7073592352541200.024193548
255142410.835865090.02318825643.125279460.1873231655.3383680522551400.028225806
256162400.8571067510.02124166147.077297670.164134916.0925491262561600.032258065
257182390.8764090890.01930233851.807196020.1428932496.9982312692571800.036290323
258202380.8938083750.01739928657.473622720.1235909118.0912098982582000.040322581
259222370.909366310.01555793564.275884270.1061916259.4169385112592200.044354839
260242360.9231660290.01379971972.465241250.0906336911.033424752602400.048387097
261262350.9353078850.01214185682.359732450.07683397113.015076422612600.052419355
262282340.9459051790.01059729494.363716350.06469211515.457834432622800.056451613
263302330.9550800160.009174838108.99375320.05409482118.486057922633000.060483871
264322320.9629594270.00787941126.91305370.04491998422.26180692643200.064516129
265342310.9696718470.00671242148.97756220.03704057326.99742222653400.068548387
266362300.9753440630.005672217176.29792490.03032815332.972663792663600.072580645
267382290.9800986490.004754586210.32326070.02465593740.558183262673800.076612903
268402280.9840519210.003953272252.95501740.01990135150.247845312684000.080645161
269422270.9873124090.003260488306.70256740.01594807962.70347742694200.084677419
270442260.9899798080.002667399374.89702910.01268759178.817168962704400.088709677
271462250.9921443720.002164564461.98676160.01002019299.798488392714600.092741935
272482240.993886690.001742318573.94806640.007855628127.29727572724800.096774194
273502230.9952777830.001391093718.85931740.00611331163.57751232735000.100806452
274522220.9963794590.001101676907.70825750.004722217211.76494132745200.10483871
275542210.9972448590.0008654011155.5338870.003620541276.20178462755400.108870968
276562200.9979191450.0006742861483.0513030.002755141362.95788682765600.112903226
277582190.9984402560.0005211111918.9777680.002080855480.57167312775800.116935484
278602180.9988397150.0003994592503.3830340.001559744641.13073482786000.120967742
279622170.9991434320.0003037163292.5448130.001160285861.85733522796200.125
280642160.9993724730.0002290414366.0295670.0008565681167.4490962806400.129032258
281662150.9995437910.0001713185837.0837090.0006275271593.5560532816600.133064516
282682140.9996708890.0001270977867.9800540.0004562092191.9781912826800.137096774
283702130.9997644099.35206E-0510692.828290.0003291113038.4839722837000.141129032
284722120.9998326616.82515E-0514651.699730.0002355914244.6467042847200.14516129
285742110.9998820634.94022E-0520242.012520.0001673395975.8789362857400.149193548
286762100.9999175283.54655E-0528196.418460.0001179378479.0889942867600.153225806
287782090.999942782.52515E-0539601.64538.24717E-0512125.371332877800.157258065
288802080.9999606111.78313E-0556081.046815.72202E-0517476.338132888000.161290323
289822070.9999730991.2488E-0580076.908793.93889E-0525387.87032898200.165322581
290842060.9999817738.67378E-06115289.99822.69009E-0537173.485412908400.169354839
291862050.9999877485.97484E-06167368.63781.82271E-0554863.317782918600.173387097
292882040.9999918294.08169E-06244996.42781.22523E-0581617.462722928800.177419355
293902030.9999945952.76532E-06361621.2028.17059E-06122390.21592939000.181451613
294922020.9999964531.85797E-06538222.51125.40526E-06185004.86792949200.185483871
295942010.9999976911.23797E-06807772.72553.5473E-06281904.90062959400.189516129
296962000.9999985098.1801E-071222479.3142.30932E-06433027.23442969600.193548387
297981990.9999990455.36013E-071865626.291.49131E-06670549.67622979800.197580645
2981001980.9999993933.48302E-072871074.8919.55301E-071046790.71629810000.201612903
2991021970.9999996172.24436E-074455610.3026.06999E-071647448.67329910200.205645161
3001041960.9999997611.43411E-076972987.8483.82563E-072613948.28630010400.209677419
3011061950.9999998529.08686E-0811004897.782.39152E-074181432.48530110600.213709677
3021081940.9999999095.70933E-0817515196.941.48284E-076743822.24930210800.217741935
3031101930.9999999443.55703E-0828113350.289.11906E-0810966044.5630311000.221774194
3041121920.9999999662.19743E-0845507770.445.56203E-0817979049.1130411200.225806452
3051141910.999999981.34604E-0874291794.653.3646E-0829721192.9530511400.22983871
3061161900.9999999888.1755E-09122316653.12.01856E-0849540295.2930611600.233870968
3071181890.9999999934.92349E-09203108006.71.20101E-0883263346.3830711800.237903226
3081201880.9999999962.93986E-09340151692.97.0866E-09141111438.130812000.241935484
3091221870.9999999981.74048E-09574552802.34.14673E-09241153701.130912200.245967742
3101241860.9999999991.02163E-09978828745.32.40625E-09415584611.331012400.25


Note: There are limits to using Excel Binomial distribution function. Large values and small probabilities can easily max out the results and return no values at all.
That is anther good reason to understand how to use the normal distribution solution that the Wizard used.
Jufo81
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November 14th, 2011 at 10:01:59 AM permalink
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)
guido111
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November 14th, 2011 at 10:16:00 AM permalink
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.
7craps
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November 14th, 2011 at 3:10:33 PM permalink
Quote: jms9smith

I was curious to know if anyone had the maths on the ability of player results in Baccarat to exceed bank results??

Looks like your thread currently has 2 answers for 3 different possible questions.
Quote: jms9smith

If 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: jms9smith

is 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: jms9smith

I 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.



pp run totalbb run totaldiffsessiontotal handspb
400,000400,00000-400,000400,000400,000100.0000%0.0000%
197,270597,270202,730202,730-394,540400,000800,00074.6588%25.3413%
394,540991,810405,460608,190-383,620800,0001,600,00061.9881%38.0119%
789,0801,780,890810,9201,419,110-361,7801,600,0003,200,00055.6528%44.3472%
1,578,1603,359,0501,621,8403,040,950-318,1003,200,0006,400,00052.4852%47.5148%
3,156,3206,515,3703,243,6806,284,630-230,7406,400,00012,800,00050.9013%49.0987%
6,312,64012,828,0106,487,36012,771,990-56,02012,800,00025,600,00050.1094%49.8906%
2,023,99014,852,0002,080,01014,852,00004,104,00029,704,00050.0000%50.0000%
197,27015,049,270202,73015,054,7305,460400,00030,104,00049.9909%50.0091%
394,54015,443,810405,46015,460,19016,380800,00030,904,00049.9735%50.0265%
789,08016,232,890810,92016,271,11038,2201,600,00032,504,00049.9412%50.0588%
1,578,16017,811,0501,621,84017,892,95081,9003,200,00035,704,00049.8853%50.1147%
3,156,32020,967,3703,243,68021,136,630169,2606,400,00042,104,00049.7990%50.2010%
6,312,64027,280,0106,487,36027,623,990343,98012,800,00054,904,00049.6867%50.3133%
12,625,28039,905,29012,974,72040,598,710693,42025,600,00080,504,00049.5693%50.4307%
      expected49.3175%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)
Jufo81
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November 15th, 2011 at 2:00:36 AM permalink
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.
jms9smith
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November 18th, 2011 at 2:46:03 PM permalink
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
jms9smith
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November 18th, 2011 at 2:53:09 PM permalink
Quote: Jufo81

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)




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 :)
Jufo81
Jufo81
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November 21st, 2011 at 2:54:58 PM permalink
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.
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