February 28th, 2011 at 2:58:02 PM
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I am not sure how to calculate the following probability. Any help or advice would be greatly appreciated.

When considering absolute difference, always betting on Banker and carrying the count forward from shoe to shoe as necessary. What is the probability of LOSING, or being behind by, 15 hands before ever winning, or being ahead by, 5 hands?

What about -20 vs. +5? -25 vs. +5?

When considering absolute difference, always betting on Banker and carrying the count forward from shoe to shoe as necessary. What is the probability of LOSING, or being behind by, 15 hands before ever winning, or being ahead by, 5 hands?

What about -20 vs. +5? -25 vs. +5?

February 28th, 2011 at 3:10:10 PM
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Quote:98stepsand carrying the count forward from shoe to shoe...

Memorandum

From: The Ace of Spades.

To: All Cards

Re: Meeting

There will be a brief meeting tomorrow concerning the order in which we have to line up so as to keep things proper from shoe to shoe. Please be prompt as drinks will be served afterward when the next player spills her cocktail on the felt.

February 28th, 2011 at 3:14:05 PM
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Maybe I need to clarify. By count I am refering to the hand count, not a card count. Meaning, that if the shoe is run and the hand score stays between Banker +5 and -15 hands, then when the next shoe is begun you continue counting hands rather than re-starting your count at 0-0.

February 28th, 2011 at 3:45:19 PM
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Assuming 8 deck shoe the probabilities for player and banker are:

P(Banker) = 0.458597

P(Player) = 0.446247

P(Both) = 0.904844

(For 6 deck or 1 deck shoe the probablities are slightly different, you get them from wizardofodds.com/baccarat)

Ignoring ties as they don't affect the hand count, the probabilities are:

p = P(Banker) = 0.458597/0.904844 = 0.506824

q = P(Player) = 0.446247/0.904844 = 0.493176

If you start with K units the formula to reach N units before losing K units in terms of probabilities p and q is:

Setting K = 15, N = 20 equals your question of reaching +5 count before reaching -15 count. Plugging those values into the formula gives: P = 0.7986 = 79.86%

Likewise for -20 vs. +5: K = 20, N = 25 -> P = 85.08%

Likewise for -25 vs. +5: K = 25, N = 30 -> P = 88.48%

and so on.

Note that ending up at +5 units doesn't necessarily mean you will be ahead overall, as you pay 5% commission for each banker win.

For example if it takes 100 winning banker hands before ending up at +5 result you are actually break-even at this point because of the commission.

P(Banker) = 0.458597

P(Player) = 0.446247

P(Both) = 0.904844

(For 6 deck or 1 deck shoe the probablities are slightly different, you get them from wizardofodds.com/baccarat)

Ignoring ties as they don't affect the hand count, the probabilities are:

p = P(Banker) = 0.458597/0.904844 = 0.506824

q = P(Player) = 0.446247/0.904844 = 0.493176

If you start with K units the formula to reach N units before losing K units in terms of probabilities p and q is:

1 - (q/p)^K

P = -----------

1 - (q/p)^N

Setting K = 15, N = 20 equals your question of reaching +5 count before reaching -15 count. Plugging those values into the formula gives: P = 0.7986 = 79.86%

Likewise for -20 vs. +5: K = 20, N = 25 -> P = 85.08%

Likewise for -25 vs. +5: K = 25, N = 30 -> P = 88.48%

and so on.

Note that ending up at +5 units doesn't necessarily mean you will be ahead overall, as you pay 5% commission for each banker win.

For example if it takes 100 winning banker hands before ending up at +5 result you are actually break-even at this point because of the commission.

February 28th, 2011 at 3:47:17 PM
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Well, its obvious that I still don't understand what you are talking about so I think I will bow out of the discussion because as far as I know the hand count in Baccarat can never go beyond 9.

February 28th, 2011 at 3:59:10 PM
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Quote:FleaStiffWell, its obvious that I still don't understand what you are talking about so I think I will bow out of the discussion because as far as I know the hand count in Baccarat can never go beyond 9.

If every Banker win is "+1" outcome and every Player win is "-1" outcome, he simply asked the probability of "+5" happening before "-15", which is a rather simple mathematical question.

February 28th, 2011 at 7:51:13 PM
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Quote:Jufo81

1 - (q/p)^K

P = -----------

1 - (q/p)^N

Setting K = 15, N = 20 equals your question of reaching +5 count before reaching -15 count. Plugging those values into the formula gives: P = 0.7986 = 79.86%

Likewise for -20 vs. +5: K = 20, N = 25 -> P = 85.08%

Likewise for -25 vs. +5: K = 25, N = 30 -> P = 88.48%

and so on.

Using your formula

=(1-(q/p)^K)/(1-(q/p)^N) in a spreadsheet

K = 15, N = 20; 0.798635722

K = 20, N = 25; 0.850576291

K = 25, N = 30; 0.884674152

You did some interesting rounding on the last 2 just like I do at times.

I could not make my results match yours. Then I saw why.

March 1st, 2011 at 3:43:23 AM
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Quote:guido111

Using your formula

=(1-(q/p)^K)/(1-(q/p)^N) in a spreadsheet

K = 15, N = 20; 0.798635722

K = 20, N = 25; 0.850576291

K = 25, N = 30; 0.884674152

You did some interesting rounding on the last 2 just like I do at times.

I could not make my results match yours. Then I saw why.

Ah sorry, I was tired when posting so I got the last two digits wrong. But the formula itself is more important than numerical values.

March 1st, 2011 at 11:30:40 AM
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Quote:98stepsWhen considering absolute difference, always betting on Banker and carrying the count forward from shoe to shoe as necessary.

The Absolute difference. Scary subject.

Brings up another point in Baccarat.

Almost all Bac players, at least the many I know and play with, think that the Banker wins more hands than the Player and they are correct.

But how much more?

Most will say, 5 to 10 hands per shoe or even more.

So I did a quick study of the The Wizards 1000 Baccarat shoes. A small sample but the results match the math.

The average absolute difference per shoe is 1.028. SD is 8.539. Meaning the Banker wins 38 hands per shoe,on average, and Player wins 37.(rounded numbers).

Totals all Shoes

Banker Wins = 38384 0.458919

Player Wins = 37356 0.446628

Tie Wins = 7900 0.0944524

It is the high SD that makes Bac players "think" the Banker should always win more hands per shoe and leads many to swear at the results when the Player wins more. I always laugh at those players. My wife is one of them.

Distribution from The Wizards 1000 Bac shoes below

There were 46 shoes that ended in a tie. The absolute difference was 0.

Looking at who won the most shoes seems to show Banker wins most shoes, by absolute difference that is.

Banker shoes won 531

Player shoes won 423

The distributions below also shows the Banker winning comparing the same absolute differences.

Banker Player

0 46

1 50 -1 47

2 54 -2 34

3 45 -3 36

4 32 -4 34

5 37 -5 37

6 41 -6 36

7 43 -7 25

8 37 -8 33

9 31 -9 25

10 29 -10 29

11 27 -11 18

12 21 -12 11

13 13 -13 19

14 17 -14 8

15 14 -15 11

16 6 -16 3

17 6 -17 5

18 8 -18 3

19 3 -19 5

20 4 -20 2

21 3 -21 0

22 3 -22 0

23 3 -23 0

24 1 -24 0

25 2 -25 1

26 0 -26 1

27 0 -27 0

28 0 -28 0

29 0 -29 0

30 0 -30 0

31 0 -31 0

32 0 -32 0

33 0 -33 0

34 1 -34 0

35 0 -35 0

36 0 -36 0

37 0 -37 0

38 0 -38 0

39 0 -39 0

40 0 -40 0

In the end, the absolute difference between the two bets is just a statistic, since one can not bet that the Banker or Player will win more hands than the other per shoe. One can win more wagering on the Banker since the lower house edge will cause one to lose less.

March 1st, 2011 at 12:06:05 PM
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In addition to the formula which gives the probability to end at +5 hand units before ending up at -15, it also possible to derive a formula which gives the expected number of rounds played in the game (the game is terminated when either +5 result or -15 result is reached). The formula is:

T = K/(q-p) - N/(q-p) * [1-(q/p)^K] / [1-(q/p)^N], for p != q

T = K*(N-K), for p = q

So for the OPs question of playing until reaching either +5 Banker wins or -15 Player Wins, the expected number of total rounds played is 71.27, and this is ignoring all ties.

T = K/(q-p) - N/(q-p) * [1-(q/p)^K] / [1-(q/p)^N], for p != q

T = K*(N-K), for p = q

So for the OPs question of playing until reaching either +5 Banker wins or -15 Player Wins, the expected number of total rounds played is 71.27, and this is ignoring all ties.