Probability Question

[Serge B]

Hey Fellas,

I was just wondering how to figure out how many hands, on average, I would be able to play before I lost 5,6,7 and 8 consecutive hands.

Thats assuming the following rules,6 Deck, 1.5 - 2.0 Penetration, H17, DOA, DAS, One card on Aces. Also assuming I only play basic strategy.

I realise Matingale systems dont work but I just want to see how high the numbers are..

> Thanks,

[BJStanko]

...not really sure but I think is something like 1 in 500 for 8 straight hands.

[aussiecounter]

Its lower than that I think. I read somewhere about 1-170 odds for losing.
That seems pretty low. I assume winning would be slightly worse odds, like 1-180 or something.

Look it up for yourself is your best bet. Try searches and look at BJMATH, BJSTATS etc. Don't people know how to search for stuff anymore?

[Mikeaber]

Originally Posted by Serge B

Hey Fellas,

I was just wondering how to figure out how many hands, on average, I would be able to play before I lost 5,6,7 and 8 consecutive hands.

Thats assuming the following rules,6 Deck, 1.5 - 2.0 Penetration, H17, DOA, DAS, One card on Aces. Also assuming I only play basic strategy.

I realise Matingale systems dont work but I just want to see how high the numbers are..

> Thanks,

Here is what I go by on streaks. I can't remember where I found it. I think I have a foumula laying around somewhere for it.

The odds on winning any given hand are 47.5% and loosing it are 52.5% (note that I believe my percentages above are off a little because around 8% of your hands are pushes):

Here are odds on any streak occurring:
2 ------- .2304 ------ 1 vs 3
3 ------- .1106 ------ 1 vs 8
4 ------- .0531 ------ 1 vs 18
5 ------- .0255 ------ 1 vs 38
6 ------- .0122 ------ 1 vs 81
7 ------- .0059 ------ 1 vs 168
8 ------- .0028 ------ 1 vs 356
9 ------- .0014 ------ 1 vs 713
10 ------ .0006 ------ 1 vs 1666
11 ------ .0003 ------ 1 vs 3332
12 ------ .0001 ------ 1 vs 9999

[Midnite]

Hey Mike, I know where you found it.
I sent it to you.
It is based on Bernoulli's formula.
They say the memory is the first to go. lol

Midnite

[newyorkbear]

Why is it that the odds roughly double 2 thru 11 but triple on 12?

[aussiecounter]

Because he used only 4 decimal places. i.e. 0.0001 isnt really that, it might be 0.00014, so it would be close to half 0.0003, which itself might really be 0.00029, or something like that. And he has used these innacurate numbers to calculate odds, rather than the real, longer numbers

[Mikeaber]

Originally Posted by Midnite

Hey Mike, I know where you found it.
I sent it to you.
It is based on Bernoulli's formula.
They say the memory is the first to go. lol

Midnite

I should've know that Midnite! I've used this table as a rule of thumb for quite some time now. Close enough for government work.