Using Bayes Theorem in Blackjack

Hello forum, I was wondering if you could help me with a part of an article I read. The article is on using Markov Chains in the game of Blackjack/21.

It is assumed that the chance of drawing any card from the deck is 1/13, which is a fair assumption when many decks are involved. The paper is giving an example to derive a basic strategy which is a strategy of mathematically correct plays by which the player should follow to keep his expectancy favorable.

One of the assumptions used is that the strategy would only apply given neither player nor dealer have Blackjack, which again makes sense as the player will only have to make his play given that he does not hold Blackjack (which is being dealt an Ace and a card valued 10 in the first 2 cards dealt to the player, for those who are unaware of the game rules).

In the paper, a calculation is used to determine what the probability is that the face down card of the dealer is valued between 2 through 9 or an Ace given the dealers other card is an Ace. It uses Bayes theorem. They get an answer of 1/9, but I do not see how they arrived at this answer. What I have tried:

http://upload.wikimedia.org/math/d/9...a3690cbd31.png is Baye's theorem.

So P(the face down card is 2-9 or an A|the dealers face up card is an A)=P( the dealers face up card is an A|the face down card is 2-9 or an A)*P(the dealers face up card is an A)/the face down card is 2-9 or an A

I calculated an answer of 8/13 but as i stated before, the answer is 1/9. Could someone perhaps shed some light on this?

Thank you.

Re: Using Bayes Theorem in Blackjack

I think you need to clarify the problem being solved. I believe they get 1/9 because they are determining the probability that the face down card is an ace, given that the face up card is an ace. This automatically implies that the face-down is not a 10, J, Q, or K, or the dealer would have already turned it over. The face-down card must be one of only 9 possible choices: either an ace or 2-9. So the probability of it being an ace is 1/9. The problem as you stated it - "that the face down card of the dealer is valued between 2 through 9 or an Ace given the dealers other card is an Ace" has a probability of 1 - it can't be anything but A or 2-9 or else the dealer would have ended the game.

Re: Using Bayes Theorem in Blackjack

Here is the extract in question-

"Finally, in calculating the player's strategy we ignore the possibility of either

the player or the dealer having \blackjack," meaning being dealt an ace and a

ten or face card in the initial hand. If this happens the game ends and the party

with the blackjack wins automatically before the player having the chance to

take any action. Therefore, given that the player is even in a position to make a

strategy choice, it must be that neither party has blackjack. The eect of this is

to change the probabilities of what the dealer's down card may be. E.g., if the

dealer's up card is an ace, we may assume that her down card is not a ten, so

by Bayes' rule the probability of the down card being any card in two through

nine or an ace is now 1=9 rather than 1=13"

Re: Using Bayes Theorem in Blackjack

This part is defintely in error:

Quote:

Originally Posted by

**kiranizzle** ...the probability of the down card being any card in two through

nine or an ace is now 1=9 rather than 1=13"

I asume you meant to type "1/9 rather than 1/13." Clearly this makes no sense - no way can the probability of card being A or 2-9 be anything near as low as 1/9 or 1/13. I think they meant to say "the probability of the down card being an ace is now 1/9 rather than 1/13."