Probabilty of loing x hands in a row for blackjack

[Mr. Ratface]

Will repeat the title:

What is the probability that someone loses x hands in a row for blackjack assuming proper strategy? Is there any formula that someone could give me where all I have to do is replace x by a number?

Thanks

[Siegmund]

x or more hands in a row: p^x, where p is the chance of losing 1 hand.

exactly x in a row: p^x*(1-p).

I'm sure someone else knows p off the top of his head, but something in the .53 to .55 range, I believe. (You lose considerably more than half of your hands, but make up for it by winning extra after splits and blackjacks.)

[lgas]

The short answer is that the "Martingale" is not a magical unbeatable blackjack strategy.

[BruceZ]

[ QUOTE ]
I'm sure someone else knows p off the top of his head, but something in the .53 to .55 range, I believe. (You lose considerably more than half of your hands, but make up for it by winning extra after splits and blackjacks.)

[/ QUOTE ]

You don't lose over half of your hands unless you ignore pushes. You lose about 48%, win 43%, and push 9%. This means that you lose about 0.48/(0.48 + 0.43) =~ 53% of all non-pushed hands, so the probability of losing the next x or more non-pushed hands in a row is about (0.53)^x. If we don't ignore pushes, then the probability of losing the next x or more hands in a row is about (0.48)^x.

These numbers do not change much for different rules, different count systems, or even for most realistic counts that arise during a shoe.

EDIT: Changed "deck compositions" to "counts".

[Siegmund]

Yes, I should have said non-pushed hands... thanks for spelling it out in more detail Bruce

[mykey1961]

You would have to factor in the rate at which hands are dealt, and how long your average trip to the loo takes.