Martingale Math Question

I'm trying to prove a point to someone, and it has to do with the Martingale system. But, the math part of it, which is what I want to know, doesn't necessarily have to relate to the Martingale system.

The Wizard states that in the game of baccarat, ignoring ties, the banker win is 0.5068, which makes the player win to be 0.4932. I figure that the odds of getting 6 player wins in a row is equal to 0.4932^6, or 0.0144.

So, what I want to do is to use 6 player wins as an ending condition for some sequence of hands being dealt (=n). The question is whether or not the chance of this occurring increases as the value of n increases.

For example if the sequence occurs at the end of 7 hands, then the first hand HAD to be a banker hand, in order for the following 6 hands to be player. The first hand could not be player since that would mean the sequence would end after 6 hands instead of 7 hands. I figure that the chance on this is equal (0.5068) * (0.4932^6).

It stands to reason that the n-6 hand must be a banker hand for this reason. However, the hands before that could ALMOST be anything. In other words, in a sequence of 9 hands, the final 6 would be player hands (indicating the end of the sequence), the n-6 hand (3rd hand) would be banker. While the first 2 hands could be anything. I think I can even figure this out. (1^2)*(.5068)*(0.4932^6).

But, I get lost when it gets to larger number of n. For example, if n= 100, then I have to make sure that 6 player winning hands didn't emerge somewhere before the 94th hand.

Basically, I'm looking for a function for the value of n. Can anybody help?

If I understand correctly...

All you really need is the probability that your stopping condition, 6 wins in a row, will start on any given hand is 0.0144.

So the probability that it starts on the first hand is 0.0144. This results in an end condition of n=6.

The probability that it starts on the second hand is the probability that it didn't start on the first hand, times the probability that it starts on the second hand = (1-0.0144) x 0.0144. n = 7.

For any n, the probability is (1-0.0144)^(n-6) x 0.0144.

This results in the probability that you will hit your stopping condition on the nth hand. Ultimately, you will probably be interested in the likelihood that you will hit your stopping condition BY the nth hand. For this, you'll have to sum up these probability for k = 6 to n. There's a closed form solution for this with a fairly elegant derivation, but I'd have to look it up. Haven't used it in a few years.

Quote: PapaChubby

If I understand correctly...

All you really need is the probability that your stopping condition, 6 wins in a row, will start on any given hand is 0.0144.

So the probability that it starts on the first hand is 0.0144. This results in an end condition of n=6.

The probability that it starts on the second hand is the probability that it didn't start on the first hand, times the probability that it starts on the second hand = (1-0.0144) x 0.0144. n = 7.

For any n, the probability is (1-0.0144)^(n-6) x 0.0144.

This results in the probability that you will hit your stopping condition on the nth hand. Ultimately, you will probably be interested in the likelihood that you will hit your stopping condition BY the nth hand. For this, you'll have to sum up these probability for k = 6 to n. There's a closed form solution for this with a fairly elegant derivation, but I'd have to look it up. Haven't used it in a few years.

Maybe I'm having a braino or something, but I think your formula is wrong. Using your formula for n=7 gives 0.01419264 but I get 0.00729792 (banker followed by 6 players).

Quote: miplet

Maybe I'm having a braino or something, but I think your formula is wrong. Using your formula for n=7 gives 0.01419264 but I get 0.00729792 (banker followed by 6 players).

Interesting. I think the issue with my formulation is that the sequence not starting on the first hand, but then starting on the second hand, are not independent events. There are lots of ways that the sequence doesn't start on the first hand, but the only way that it doesn't start on the first but does start on the second is if the first hand is a banker win. Back to the drawing board.

I made a google docs spreadsheet. The last column is a cumulitive odds of failure.
6=0.014388195146279
7=0.021680492151291
8=0.028972789156304
9=0.036265086161317
10=0.043557383166329
25=0.14745962714397
50=0.29612499937927
75=0.41886621190224
94=0.49762343491503
95=0.50145932699473
100=0.52020390073376

perhaps find the probability of not winning 6 in a row in n-6 games. So for n<=12, this probability is 1.
I tried analyzing for n-6=13 so that I can come up with a generalization.
So basically since position matters, we have a total of 8192 ways to combine different wins and losses. If we look at the scenarios where we have less than a total of 6 wins for n-6, then we include all combinations of those (combin(13,8),combin(13,9)combin(13,10),combin(13,11),combin(13,12),combin(13,13)).
For 6 or more wins:
For 6 wins, we just want to avoid having 6 consecutive wins, so we look at those as one unit, then we can have 7 different combinations with 6 consecutive wins (combin(7,1)):
Loss Loss Loss Loss Loss Loss Loss 6_Wins
Loss Loss Loss Loss Loss Loss 6_Wins Loss
Loss Loss Loss Loss Loss 6_Wins Loss Loss
Loss Loss Loss Loss 6_Wins Loss Loss Loss
Loss Loss Loss 6_Wins Loss Loss Loss Loss
Loss Loss 6_Wins Loss Loss Loss Loss Loss
Loss 6_Wins Loss Loss Loss Loss Loss Loss
6_Wins Loss Loss Loss Loss Loss Loss Loss

so total combinations is combin(13,7)-7

For 7 wins we have 3 groups: single win, group of 6 consecutive wins and losses.
So total number of outcomes here is combin(13,6)-combin(8,1,1)

For 8 wins we have 3 groups: 2 wins, 6 consecutive wins and 5 losses.
So total here is combin(13,5)-combin(8,2,1)

For 9, 10 wins, same thing

So we just sum up the number of outcomes where there is no combination of 6 consecutive wins.
I don't believe there is a simple formula for this, but I'm pretty sure an algorithm will do the trick

I hope this helps

Miplet, I just finished an Excel spreadsheet to analyze Konceptum's problem, and my results agree exactly with yours. And thanks for introducing me to Google Docs!