Probability of Blackjack

bdj

Apr 2005
2
0
What are the chances of getting 21?
 
Apr 2005
171
0
This is a complicated question but I'll simplify the answer.

If you are talking about getting 21 on the inital deal of two cards then that is a fairly easy calculation. You need to first find the chances of getting a 10 card and then find the chances of getting an 11 card.

10 card

out of 52 possible cards there are 4 each of: 10s, Js, Qs, Ks which means you have 16 ways of getting 10 out of 52 cards.

11 card

out of 52 possible cards there are only 4 kinds of Ace so 4 out of 52.

Then you multiply the probability of getting each card and you have your answer: (16/52) * (4/52) = 64/2704 = 2.37%

The answer can get more complicated when you start to think about how there is only 51 cards left after you take the first card etc.
 
Apr 2005
171
0
Blackjack-mania

Also dont forget about the possibility of getting 21 with 3 cards or 4 cards or 5 etc.
 
Nov 2005
3
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Also, don't forget about the fact that most casinos use multiple decks which makes the odds even worse for you.
 
Jul 2006
2
0
Craiova, RO
blackjack formulas

There is a website with theoretical and applied probability tutorials with a consistent blackjack section, holding some interesting formulas for the probability of being hit favorable cards in the various ensembles of conditions: http://probability.infarom.ro
 
In his text, "Playing Blackjack as a Business", Lawrence Revere stated that in a single deck game one would get a BLACKJACK once every 20.7 hands
and that for a FOUR-DECK SHOE that would drop to once every 21 hands.

Revere never explained the discrepancy... I believe Peter Griffin in his book, "Theory of Blackjack" explained the 0.3 discrepancy...

..but I don't recall it if he did.
 
Dec 2006
26
2
This is a complicated question but I'll simplify the answer.

If you are talking about getting 21 on the inital deal of two cards then that is a fairly easy calculation. You need to first find the chances of getting a 10 card and then find the chances of getting an 11 card.

10 card

out of 52 possible cards there are 4 each of: 10s, Js, Qs, Ks which means you have 16 ways of getting 10 out of 52 cards.

11 card

out of 52 possible cards there are only 4 kinds of Ace so 4 out of 52.

Then you multiply the probability of getting each card and you have your answer: (16/52) * (4/52) = 64/2704 = 2.37%

The answer can get more complicated when you start to think about how there is only 51 cards left after you take the first card etc.
It is true that you have 16 cards representing "10" and 4 representing "11" in a single deck of cards.
To calculate the probability to get "Black Jack" or 21 on the initial deal, we need to consider both events

A: first card "10" and second "11"
B: first card "11" and second "10"

We have

\(\displaystyle P("Black Jack") = P(A) + P(B) =\frac{16}{52}\cdot\frac{4}{51}+\frac{4}{52}\cdot\frac{16}{51}=\frac{32}{663}\approx 0.04827\)

where we take into account the fact that only 51 cards remain in the deck after first card is drawn. This is the probability of getting blackjack when using one deck of cards and the deck is "perfectly" shuffled between subsequent deals.
It corresponds fairly to 1/20.7 as proposed by "Playing Blackjack as a Business" accordning to kahlmy_ishmael_xxiii's post.

Also, don't forget about the fact that most casinos use multiple decks which makes the odds even worse for you.
Suppose we have n decks perfectly shuffled between subsequent deals in a one player Black Jack game. Let the events be A and B as denoted above.
Now

\(\displaystyle P(A)=P(B)=\frac{16n}{52n}\cdot\frac{4n}{(52n-1)}=\frac{16}{13}\cdot\frac{n}{(52n-1)}\)

\(\displaystyle P("Black Jack") = P(A) + P(B) =2P(A)=\frac{32}{13}\cdot\frac{n}{(52n-1)} \)

With n = 4 we get a probability of 128/2691 or about 0.04757. This is close to the proposed value of 1/21 by "Playing Blackjack as a Business" accordning to kahlmy_ishmael_xxiii's post.

The situations get more complicated as many hands are dealt after another without shuffling. Also... to find the probability to get 21 with many cards, knowing what cards are dealt, maybe knowing the number of decks, maybe knowing the decks aren't tampered with etc. ...is quite a complicated task...
 
Last edited:

CaptainBlack

MHF Hall of Fame
Nov 2005
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5,271
someplace
In his text, "Playing Blackjack as a Business", Lawrence Revere stated that in a single deck game one would get a BLACKJACK once every 20.7 hands
and that for a FOUR-DECK SHOE that would drop to once every 21 hands.

Revere never explained the discrepancy... I believe Peter Griffin in his book, "Theory of Blackjack" explained the 0.3 discrepancy...

..but I don't recall it if he did.
with a 52 card deck the probability that a random hand is balckjack is:

2*16*4/52/51 ~= 1/20.7

If you have an infinite number of decks in the shoe the 51 becomes a 52
as we are now effectivly doing the dealing with replacement, and the
probability becomes:

2*16*4/52/52 ~= 1/21.125.

Now the 4 deck case is getting close to the infinite case. Doing the
calculation exactly it is:

2*(16*4)*(4*4)/(4*52)/(4*52-1) =1/21.023..

RonL
 
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Jun 2007
3
0
Since this is a Black Jack thread, I might as well ask
What does it mean to count cards?
 

ThePerfectHacker

MHF Hall of Fame
Nov 2005
10,616
3,268
New York City
What does it mean to count cards?
It means to keep track of some of cards that leave the deck by memory. And then being able to compute the necessarily probabilities. Casinos hate it because people who train themselves to do this do have a small advantage, enough to win money.