Dealer Busting

How often can the dealer go without busting in BJ?

What calculation would allow me to calculate all of the following:

Or, if dealer's busts 28% of every hand deal as an average (just over 1 in 4 hands), how often does the dealer go:
5 hands without busting?
6 hands without busting?
7 hands without busting?
8 hands without busting?
9 hands without busting?
etc..
And what is the realistic most hands you can play and before the dealer busts?

Also does anyone know how to post to the site in the BJ/21 section?

If the dealer is one-on-one with a player, the answer is different then if he is dealing to a full table.

But basically, if the dealer busts with a frequency of 0.28 then he has an 0.72 chance of not busting in any given hand.

5 hands without busting is 0.72^5.

Thank you for the reply.

It got me to thinking. Is there any way to know if the dealer does not bust for 30 hands in a row, how many of those hands will the player win versus the dealer?

In other words, what % of the time does the player win, when the dealer does not bust?

Hey gordonm888 I'm just curious why would the answer be different if the player is one-on-one vs a full table?

Quote: danepeterson

Hey gordonm888 I'm just curious why would the answer be different if the player is one-on-one vs a full table?

danepeterson,

IANAgordonm888, but I can answer this one. If you're playing one spot heads-up vs. the dealer, any time you have a BJ or bust (for split hands, bust ALL of them), the dealer will (normally) not complete her hand and so will not have the opportunity to bust. For a full table, the only time the dealer will not complete her hand is if EVERY PLAYER has either a BJ or a bust.

I said "normally" above because certain sidebets force the dealer to complete her hand even if all the players' hands are BJ's or busts.

Hope this helps!

Dog Hand

Quote: NewtoTown

(snip)And what is the realistic most hands you can play and before the dealer busts?(snip)

It depends, roughly how many hands do you play per session and how many sessions per year?

Going by the probability figures given earlier in this thread, you would expect:

"30 hands , no dealer bust" to be: about 1/19,000
^^^ "41 hands , no dealer bust" to be: about 1/707,000
"50 hands , no dealer bust" to be: about 1/13,600,000
"60 hands , no dealer bust" to be: about 1/363,100,000

^^^: I read somewhere that you have about a 1/700,000 chance of being struck by lightning in any given year, so that is why I put the "41 hands ..." figure in, for comparison.

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Quote: NewtoTown

Thank you for the reply.

It got me to thinking. Is there any way to know if the dealer does not bust for 30 hands in a row, how many of those hands will the player win versus the dealer?

In other words, what % of the time does the player win, when the dealer does not bust?

I can't properly answer the above question, but if I had to guess, then somewhere between 10 and 11 hands.

Note: A better reply may be posted, but I won't do it, because it would be too time consuming for me to work out.

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Spelling not checked

Quote: DogHand

danepeterson,

IANAgordonm888, but I can answer this one. If you're playing one spot heads-up vs. the dealer, any time you have a BJ or bust (for split hands, bust ALL of them), the dealer will (normally) not complete her hand and so will not have the opportunity to bust. For a full table, the only time the dealer will not complete her hand is if EVERY PLAYER has either a BJ or a bust. . .

Yes. Thanks, Doghand.

Thanks DogHand!