Neither the player nor the dealer is dealt a blackjack

[Physics_wiz]

I am trying to find the probability of neither the player nor the dealer being dealt a blackjack (an ace with a ten, jack, queen, or king) in a blackjack game with a freshly shuffled deck. This is what I did:

1 - P(at least one gets a blackjack)

= 1 - [ 4*16/(52c2) + 3*15/(50c2)]

However, I get an answer that's about one percent higher than it should be. The right answer is .9052.

 

[robert Ihnot]

According to conditional probability, we have P(~A and ~B) =P(~A)xP(~B|~A) Which I read, the probability of not A and not B is equal to the probability of not A times the probability of not B given not A)

So in this case, we have not A = (1-(4*16/52C2) and not B given not A =1-64/50C2.

If that is correct, then we have (.9478)(.9517) = .9020.

HOWEVER, I see that this reasoning has a flaw in it, because it does not take into account that the first player may get, for example, an ace, but not have a blackjack.

 

[robert Ihnot]

It turns out this problem has been solved by a group of actuaries. http://actuary.com/actuarial-discussion-forum/showthread.php?t=5022

The solution is: P(AUB)=P(A)+P(B)-P(AB).

In this case, P(A)=P(B), because both cases are independent and they both have equal odds. However, P(AB) = 4x16/52C2 * 3x15/50C2. Thus the probability of at least one blackjack is .0947579032. And the probability of no blackjack is .905242097.

If this seems strange, a simpler example is: If two cards are dealt, what is the probability that at least one is red?

P(AUB) = P(A) + P(B) -P(AB) = 1/2 + 1/2 -(26/52)(25/51) =.75490.

This can be checked by looking at tree probability:

1/2R to 25/51 R or 26/51 B

And 1/2B to 26/51R or 25/51B.

Thus if both are black we have (1/2)(25/51) =25/(102) = .24510; and at least one is red = 1-.24510= .75490.

 

[Physics_wiz]

Yep, I posted the problem over there too...I was anxious to know the answer. I was going to post the solution here but forgot, thanks robert.

 

[robert Ihnot]

We could look at this in the case of three players. First we consider just three cards and being dealt. Consider the color red:

P(aUbUc) = P(a)+P(b)+P(c) - {P(ab)+P(ac) + P(bc)} + P(abc). (P(ab) means P(a intersection b)).

But the probabilities in each group are the same, so this becomes:

P(aUbUc) = 3P(a)-3P(ab)+P(abc)

The P(a) = 1/2; P(ab)=(1/2)(25/51)=25/102; P(abc) = (1/2)(25/51)(24/50) =2/17.
Thus P(aUbUc) = 3/2-75/102 + 2/17 = 15/17. While if all three cards are black, we independently find: (1/2)(25/51)(24/50) = 2/17.

Applying the same method to the blackjack problem, we arrive at: The probability that at least one hand of three will have a blackjack is .139521.., and thus chance of no blackjack is approximately 86%.

 

[dhanushka190]

I am trying to find the probability of neither the player nor the dealer being dealt a blackjack (an ace with a ten, jack, queen, or king) in a blackjack game with a freshly shuffled deck. This is what I did:

Let ; p(a1) be the probability that only the player gets the black jack
p(a2) be the probability that only the dealer gets the black jack
p(a1Ua2) will be the probability both gets the black jack

we know p(a1Ua2) = p(a1)+ p(a2)-p(a1a2) ------- Eq(1)
and p(a1Ua2)^c = 1- p(a1Ua2)
= 1- p(a1)+ p(a2)-p(a1a2) --- Eq(2)


p(a1) = ((4choose1)*(16choose1))/(52 choose2)
p(a2) = ((4choose1)*(16choose1))/(50 choose2)
p(a1Ua2) = {((4choose1)*(16choose1))/(52 choose2)}*{((3choose1)*(15choose1))/(50 choose2)}
rest is algebra...
gppd luck!