Probability that Both Player and Dealer are not Dealt Blackjack
$begingroup$
Let us say we have a rather simple blackjack game between one dealer and one player with a single 52 card deck. We want to compute the probability that neither the dealer nor the player is dealt blackjack (initially). Let $A$ be the event that the player is dealt blackjack, $B$ the same, except for the dealer. We will compute:
$$
1 - mathbb{P}(A cup B) = 1 - (mathbb{P}(A) + mathbb{P}(B) - mathbb{P}(A cap B))
$$
We see that:
$$
mathbb{P}(A) = frac{{{16}choose{1}} {{4}choose{1}}}{52 choose 2}
$$
As we have that out of the total number of hands we can be dealt, there are 64 possible hands that give us blackjack. Now, we compute the probability for the dealer, $mathbb{P}(B)$. It seems that by symmetry, we may say this is the same as the probability for the player. However, my question here is, why is that so? Namely, would the order of dealing not matter? I.e if the dealer deals my cards first (which is usually the case, I think) would I not have to condition on the event that I am dealt blackjack first and use the fact that the dealer is working with a smaller deck?
probability combinatorics
asked Dec 24 '18 at 0:04
rubikscube09rubikscube09
1,277719
$endgroup$
add a comment |
$begingroup$
Let us say we have a rather simple blackjack game between one dealer and one player with a single 52 card deck. We want to compute the probability that neither the dealer nor the player is dealt blackjack (initially). Let $A$ be the event that the player is dealt blackjack, $B$ the same, except for the dealer. We will compute:
$$
1 - mathbb{P}(A cup B) = 1 - (mathbb{P}(A) + mathbb{P}(B) - mathbb{P}(A cap B))
$$
We see that:
$$
mathbb{P}(A) = frac{{{16}choose{1}} {{4}choose{1}}}{52 choose 2}
$$
As we have that out of the total number of hands we can be dealt, there are 64 possible hands that give us blackjack. Now, we compute the probability for the dealer, $mathbb{P}(B)$. It seems that by symmetry, we may say this is the same as the probability for the player. However, my question here is, why is that so? Namely, would the order of dealing not matter? I.e if the dealer deals my cards first (which is usually the case, I think) would I not have to condition on the event that I am dealt blackjack first and use the fact that the dealer is working with a smaller deck?
probability combinatorics
asked Dec 24 '18 at 0:04
rubikscube09rubikscube09
1,277719
$endgroup$
add a comment |
$begingroup$
Let us say we have a rather simple blackjack game between one dealer and one player with a single 52 card deck. We want to compute the probability that neither the dealer nor the player is dealt blackjack (initially). Let $A$ be the event that the player is dealt blackjack, $B$ the same, except for the dealer. We will compute:
$$
1 - mathbb{P}(A cup B) = 1 - (mathbb{P}(A) + mathbb{P}(B) - mathbb{P}(A cap B))
$$
We see that:
$$
mathbb{P}(A) = frac{{{16}choose{1}} {{4}choose{1}}}{52 choose 2}
$$
As we have that out of the total number of hands we can be dealt, there are 64 possible hands that give us blackjack. Now, we compute the probability for the dealer, $mathbb{P}(B)$. It seems that by symmetry, we may say this is the same as the probability for the player. However, my question here is, why is that so? Namely, would the order of dealing not matter? I.e if the dealer deals my cards first (which is usually the case, I think) would I not have to condition on the event that I am dealt blackjack first and use the fact that the dealer is working with a smaller deck?
probability combinatorics
asked Dec 24 '18 at 0:04
rubikscube09rubikscube09
1,277719
$endgroup$
Let us say we have a rather simple blackjack game between one dealer and one player with a single 52 card deck. We want to compute the probability that neither the dealer nor the player is dealt blackjack (initially). Let $A$ be the event that the player is dealt blackjack, $B$ the same, except for the dealer. We will compute:
$$
1 - mathbb{P}(A cup B) = 1 - (mathbb{P}(A) + mathbb{P}(B) - mathbb{P}(A cap B))
$$
We see that:
$$
mathbb{P}(A) = frac{{{16}choose{1}} {{4}choose{1}}}{52 choose 2}
$$
As we have that out of the total number of hands we can be dealt, there are 64 possible hands that give us blackjack. Now, we compute the probability for the dealer, $mathbb{P}(B)$. It seems that by symmetry, we may say this is the same as the probability for the player. However, my question here is, why is that so? Namely, would the order of dealing not matter? I.e if the dealer deals my cards first (which is usually the case, I think) would I not have to condition on the event that I am dealt blackjack first and use the fact that the dealer is working with a smaller deck?
probability combinatorics
probability combinatorics
asked Dec 24 '18 at 0:04
rubikscube09rubikscube09
1,277719
asked Dec 24 '18 at 0:04
rubikscube09rubikscube09
1,277719
asked Dec 24 '18 at 0:04
rubikscube09rubikscube09
1,277719
asked Dec 24 '18 at 0:04
rubikscube09rubikscube09
1,277719
asked Dec 24 '18 at 0:04
rubikscube09rubikscube09
1,277719
1,277719
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The probabilities $P(B)$ and $P(A)$ are indeed equal, but $A$ and $B$ are not independent. After all, there are only four aces in the deck, and we can see two of them if both the player and the dealer have blackjack. We should expect $P(Acap B)$ to be strictly less than $P(A)cdot P(B)$.
As for the order - we just dealt out four cards face up, two to the player and two to the dealer. Do we really care what order we laid the cards on the table? No; what matters is which two cards the player got, which two cards the dealer got, and that the cards are all different. Either way you condition it to account for the smaller deck, the probability of double blackjack will be the same.
answered Dec 24 '18 at 2:21
jmerryjmerry
9,7331225
$endgroup$
add a comment |
$begingroup$
$B$ is the event in which the dealer makes black jack, so the probability of $B$ is the same of $A$, but in the problem clearly you have to consider the whole situation, and indeed you have to calculate $mathbb P (A cap B)$
answered Dec 24 '18 at 0:28
Claudio DelfinoClaudio Delfino
43
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add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
The probabilities $P(B)$ and $P(A)$ are indeed equal, but $A$ and $B$ are not independent. After all, there are only four aces in the deck, and we can see two of them if both the player and the dealer have blackjack. We should expect $P(Acap B)$ to be strictly less than $P(A)cdot P(B)$.
As for the order - we just dealt out four cards face up, two to the player and two to the dealer. Do we really care what order we laid the cards on the table? No; what matters is which two cards the player got, which two cards the dealer got, and that the cards are all different. Either way you condition it to account for the smaller deck, the probability of double blackjack will be the same.
answered Dec 24 '18 at 2:21
jmerryjmerry
9,7331225
$endgroup$
add a comment |
$begingroup$
The probabilities $P(B)$ and $P(A)$ are indeed equal, but $A$ and $B$ are not independent. After all, there are only four aces in the deck, and we can see two of them if both the player and the dealer have blackjack. We should expect $P(Acap B)$ to be strictly less than $P(A)cdot P(B)$.
As for the order - we just dealt out four cards face up, two to the player and two to the dealer. Do we really care what order we laid the cards on the table? No; what matters is which two cards the player got, which two cards the dealer got, and that the cards are all different. Either way you condition it to account for the smaller deck, the probability of double blackjack will be the same.
answered Dec 24 '18 at 2:21
jmerryjmerry
9,7331225
$endgroup$
add a comment |
$begingroup$
The probabilities $P(B)$ and $P(A)$ are indeed equal, but $A$ and $B$ are not independent. After all, there are only four aces in the deck, and we can see two of them if both the player and the dealer have blackjack. We should expect $P(Acap B)$ to be strictly less than $P(A)cdot P(B)$.
As for the order - we just dealt out four cards face up, two to the player and two to the dealer. Do we really care what order we laid the cards on the table? No; what matters is which two cards the player got, which two cards the dealer got, and that the cards are all different. Either way you condition it to account for the smaller deck, the probability of double blackjack will be the same.
answered Dec 24 '18 at 2:21
jmerryjmerry
9,7331225
$endgroup$
The probabilities $P(B)$ and $P(A)$ are indeed equal, but $A$ and $B$ are not independent. After all, there are only four aces in the deck, and we can see two of them if both the player and the dealer have blackjack. We should expect $P(Acap B)$ to be strictly less than $P(A)cdot P(B)$.
As for the order - we just dealt out four cards face up, two to the player and two to the dealer. Do we really care what order we laid the cards on the table? No; what matters is which two cards the player got, which two cards the dealer got, and that the cards are all different. Either way you condition it to account for the smaller deck, the probability of double blackjack will be the same.
answered Dec 24 '18 at 2:21
jmerryjmerry
9,7331225
answered Dec 24 '18 at 2:21
jmerryjmerry
9,7331225
answered Dec 24 '18 at 2:21
jmerryjmerry
9,7331225
answered Dec 24 '18 at 2:21
jmerryjmerry
9,7331225
9,7331225
add a comment |
add a comment |
$begingroup$
$B$ is the event in which the dealer makes black jack, so the probability of $B$ is the same of $A$, but in the problem clearly you have to consider the whole situation, and indeed you have to calculate $mathbb P (A cap B)$
answered Dec 24 '18 at 0:28
Claudio DelfinoClaudio Delfino
43
$endgroup$
add a comment |
$begingroup$
$B$ is the event in which the dealer makes black jack, so the probability of $B$ is the same of $A$, but in the problem clearly you have to consider the whole situation, and indeed you have to calculate $mathbb P (A cap B)$
answered Dec 24 '18 at 0:28
Claudio DelfinoClaudio Delfino
43
$endgroup$
add a comment |
$begingroup$
$B$ is the event in which the dealer makes black jack, so the probability of $B$ is the same of $A$, but in the problem clearly you have to consider the whole situation, and indeed you have to calculate $mathbb P (A cap B)$
answered Dec 24 '18 at 0:28
Claudio DelfinoClaudio Delfino
43
$endgroup$
$B$ is the event in which the dealer makes black jack, so the probability of $B$ is the same of $A$, but in the problem clearly you have to consider the whole situation, and indeed you have to calculate $mathbb P (A cap B)$
answered Dec 24 '18 at 0:28
Claudio DelfinoClaudio Delfino
43
answered Dec 24 '18 at 0:28
Claudio DelfinoClaudio Delfino
43
answered Dec 24 '18 at 0:28
Claudio DelfinoClaudio Delfino
43
answered Dec 24 '18 at 0:28
Claudio DelfinoClaudio Delfino
43
43
add a comment |
add a comment |
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