Gambling Probablility Question. Math Wiz Please Help!

[stocktrader23]

Hi, I've been having great success with a betting strategy I've been using for blackjack and baccarat. However, I have a couple of things I can't figure out for the life of me.

Assuming a 48% chance of winning any given hand, what are the odds of NOT winning 2 in a row for 10, 20, etc hands? If someone could tell me how to figure this out it would be much appreciated.

If you need any other info PLMK. Thanks.

[XChamp]

Odds of not winning are 52%, so odds of not winning X amounts of hands in a row is (0.52^X)


Does that answer your question?

[stocktrader23]

[ QUOTE ]
Odds of not winning are 52%, so odds of not winning X amounts of hands in a row is (0.52^X)


Does that answer your question?

[/ QUOTE ]

Actually I'm looking for the odds of not getting 2 in a row anywhere in X hands. For example, odds of 20 hands without once winning twice in a row.

[HesseJam]

That took me a while. It's about 0,47% for 1:1 chances on 20 hands. Obviously, it is higher in your case, but not by that much.

Edt HIGHER, of course

[HesseJam]

You'll have to build the formula yourself.

I did the following:

Step 1: Nothing happens, because only one event. But (in your case) 48% have won and thus are endangered to be eliminated in the next step.

Step 2: 48% of 48% (0.48 * 0.48) have won two times in a row, leaving you with 1 - (0.48)^2. Of these a further 48% become endangered for the next step.

Step 3: (1 - (0.48)^3) * 0.48 log the double win and have to be deducted from the remainder of Step 2.

and so on...

[elitegimp]

I'm not sure if this is correct (someone please verify / disprove [img]/images/graemlins/smile.gif[/img], but I thought of this in 3 stages:

1) You have not won your last game
2) You won your last game, but lost the one before that
3) You won two in a row

I then set up a Markov chain with the following probabilities:
P(1->1) = 0.52 (you lose, you stay at stage 1)
P(1->2) = 0.48 (you win, you move to stage 2)
P(1->3) = 0 (you only play one hand, you can't win it twice)
P(2->1) = 0.52 (you lose, you move back to stage 1)
P(2->2) = 0 (you either win or lose)
P(2->3) = 0.48 (you win, you move to stage 3)

The tricky part is here:
P(3->1) = 0
P(3->2) = 0
P(3->3) = 1
Once you've won two games in a row, we call it a success and stay Stage 3.

So we throw this all into a matrix we'll call A (apologies for appearance, but code tags didn't work):
| 0.52 0.48 0 |
| 0.52 0 0.48 |
| 0 0 1 |

And the (i,j) component of A^y gives us the probability of going from stage (i) to stage (j) after y attempts.

In this case, we are interested in going from stage 1 to stage 3 in 20 attempts, and the probability of this is 0.9764 (the top right corner of the matrix A^20).

Translation: There is a 97.64% chance you win at least 2 in a row at some point over the course of 20 hands, implying a 2.36% chance you don't win at least 2 in a row.

[pzhon]

[ QUOTE ]
Hi, I've been having great success with a betting strategy I've been using for blackjack and baccarat.

[/ QUOTE ]
Uh oh. As long as you don't expect to add negative numbers together to get a positive number, I suppose a betting system can be entertaining. It can help you to disguise your card counting in blackjack. It can't help you to beat the game on average if you are behind on every hand.

[ QUOTE ]
Assuming a 48% chance of winning any given hand, what are the odds of NOT winning 2 in a row for 10, 20, etc hands? If someone could tell me how to figure this out it would be much appreciated.

[/ QUOTE ]

State 0: You did not win the last hand.

State 1: You won the last hand.

You start with a 100% chance to be in state 0. At each step, if you are in state 0, you have a 52% chance to stay in state 0, and a 48% chance to move to state 1. If you are in state 1, you have a 52% chance to move to state 0. You can use a program such as Excel to compute the probability of being in each state after n trials. Another method you can use is to build a transfer matrix.

[0.52 0.52]^n
[0.48 0 ]

is a transfer matrix for playing n hands. The upper left hand corner tells you the probability to go from state 0->0. The lower left tells you the probability to go from state 0->1. You start in state 0, so what you want is the sum of the entries in the left column of the matrix raised to the nth power.

After 10 hands, the probability of no streak of two wins is 16.535%. After 20 hands, 2.363%. After 50 hands, 0.006898%.

The probability that there were no two wins in a row in n hands is roughly 1.157 (0.8232)^n. This can be seen by diagonalizing the transfer matrix or by analyzing its eigenvalues and eigenvectors. In general, the probability that you do not observe a pattern like this falls exponentially.

[alThor]

[ QUOTE ]
Hi, I've been having great success with a betting strategy I've been using for blackjack and baccarat. ...
Assuming a 48% chance of winning any given hand, what are the odds of NOT winning 2 in a row for 10, 20, etc hands?

[/ QUOTE ]

I know you didn't ask, but progression betting strategies do not give you an edge. You have a high chance of winning a small amount, and a small chance of losing a huge amount. But on average, the house keeps the edge.

alThor

[pzhon]

<ul type="square">
Re: Gambling Probablility Question. Math Wiz Please Help! elitegimp 02/04/05 12:<font color="red">16</font> PM
Re: Gambling Probablility Question. Math Wiz Please Help! pzhon 02/04/05 12:<font color="red">17</font> PM [/list]Kicker!

[stocktrader23]

[ QUOTE ]
[ QUOTE ]
Hi, I've been having great success with a betting strategy I've been using for blackjack and baccarat. ...
Assuming a 48% chance of winning any given hand, what are the odds of NOT winning 2 in a row for 10, 20, etc hands?

[/ QUOTE ]

I know you didn't ask, but progression betting strategies do not give you an edge. You have a high chance of winning a small amount, and a small chance of losing a huge amount. But on average, the house keeps the edge.

alThor

[/ QUOTE ]

Yes I know. I have been a winning blackjack player for years. By strategy I mean playing in a way where if I win I win big but if I lose it's a limited loss. I've done ok with it be it a very long good run or whatever. It's fun, don't have to think as much as poker. [img]/images/graemlins/wink.gif[/img]