Blackjack Statistics Question [on hold]
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Suppose you start a hand of Blackjack with a fresh deck of cards. Your first 2 cards total 16 without using an ace, and you are reasonably sure the dealer's two cards total 20, also, with no ace. What is the probability you will win this hand by hitting and obtaining one or more cards worth a total value of 5?
statistics
asked 15 hours ago
P.M.
4
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put on hold as off-topic by Rebellos, GNUSupporter 8964民主女神 地下教會, ArsenBerk, Chris Custer, user10354138 5 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rebellos, GNUSupporter 8964民主女神 地下教會, ArsenBerk, Chris Custer, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
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Suppose you start a hand of Blackjack with a fresh deck of cards. Your first 2 cards total 16 without using an ace, and you are reasonably sure the dealer's two cards total 20, also, with no ace. What is the probability you will win this hand by hitting and obtaining one or more cards worth a total value of 5?
statistics
asked 15 hours ago
P.M.
4
New contributor
P.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Rebellos, GNUSupporter 8964民主女神 地下教會, ArsenBerk, Chris Custer, user10354138 5 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rebellos, GNUSupporter 8964民主女神 地下教會, ArsenBerk, Chris Custer, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
1
To the asker: please show your working on the question, then we may help you with this. To the down voters, you should be saying this to the OP, not me!
– Rhys Hughes
15 hours ago
you need to work out what the further combinations of cards you can get are. The fact that you know that the 4 aces are in the deck, but other cards could be already drawn, is a factor
– Cato
14 hours ago
P(Ace & 4)+P(3 & 2)=0.0142
– P.M.
14 hours ago
(4/48)(4/47) + (4/48 + 4/47) = 0.0142
– P.M.
14 hours ago
1
A. You have left off a lot of cases. You could just draw a $5$,say. Or two $A's$ and a $3$. Or three $A's$ and a $2$. B. Your computation of the probabilities is wrong.
– lulu
13 hours ago
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show 3 more comments
up vote
-2
down vote
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up vote
-2
down vote
favorite
Suppose you start a hand of Blackjack with a fresh deck of cards. Your first 2 cards total 16 without using an ace, and you are reasonably sure the dealer's two cards total 20, also, with no ace. What is the probability you will win this hand by hitting and obtaining one or more cards worth a total value of 5?
statistics
asked 15 hours ago
P.M.
4
New contributor
P.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Suppose you start a hand of Blackjack with a fresh deck of cards. Your first 2 cards total 16 without using an ace, and you are reasonably sure the dealer's two cards total 20, also, with no ace. What is the probability you will win this hand by hitting and obtaining one or more cards worth a total value of 5?
statistics
statistics
asked 15 hours ago
P.M.
4
New contributor
P.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 hours ago
P.M.
4
New contributor
P.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 hours ago
P.M.
4
New contributor
P.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 hours ago
P.M.
4
asked 15 hours ago
P.M.
4
4
New contributor
P.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
P.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
P.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Rebellos, GNUSupporter 8964民主女神 地下教會, ArsenBerk, Chris Custer, user10354138 5 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rebellos, GNUSupporter 8964民主女神 地下教會, ArsenBerk, Chris Custer, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Rebellos, GNUSupporter 8964民主女神 地下教會, ArsenBerk, Chris Custer, user10354138 5 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rebellos, GNUSupporter 8964民主女神 地下教會, ArsenBerk, Chris Custer, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
1
To the asker: please show your working on the question, then we may help you with this. To the down voters, you should be saying this to the OP, not me!
– Rhys Hughes
15 hours ago
you need to work out what the further combinations of cards you can get are. The fact that you know that the 4 aces are in the deck, but other cards could be already drawn, is a factor
– Cato
14 hours ago
P(Ace & 4)+P(3 & 2)=0.0142
– P.M.
14 hours ago
(4/48)(4/47) + (4/48 + 4/47) = 0.0142
– P.M.
14 hours ago
1
A. You have left off a lot of cases. You could just draw a $5$,say. Or two $A's$ and a $3$. Or three $A's$ and a $2$. B. Your computation of the probabilities is wrong.
– lulu
13 hours ago
|
show 3 more comments
1
To the asker: please show your working on the question, then we may help you with this. To the down voters, you should be saying this to the OP, not me!
– Rhys Hughes
15 hours ago
you need to work out what the further combinations of cards you can get are. The fact that you know that the 4 aces are in the deck, but other cards could be already drawn, is a factor
– Cato
14 hours ago
P(Ace & 4)+P(3 & 2)=0.0142
– P.M.
14 hours ago
(4/48)(4/47) + (4/48 + 4/47) = 0.0142
– P.M.
14 hours ago
1
A. You have left off a lot of cases. You could just draw a $5$,say. Or two $A's$ and a $3$. Or three $A's$ and a $2$. B. Your computation of the probabilities is wrong.
– lulu
13 hours ago
1
1
To the asker: please show your working on the question, then we may help you with this. To the down voters, you should be saying this to the OP, not me!
– Rhys Hughes
15 hours ago
To the asker: please show your working on the question, then we may help you with this. To the down voters, you should be saying this to the OP, not me!
– Rhys Hughes
15 hours ago
you need to work out what the further combinations of cards you can get are. The fact that you know that the 4 aces are in the deck, but other cards could be already drawn, is a factor
– Cato
14 hours ago
you need to work out what the further combinations of cards you can get are. The fact that you know that the 4 aces are in the deck, but other cards could be already drawn, is a factor
– Cato
14 hours ago
P(Ace & 4)+P(3 & 2)=0.0142
– P.M.
14 hours ago
P(Ace & 4)+P(3 & 2)=0.0142
– P.M.
14 hours ago
(4/48)(4/47) + (4/48 + 4/47) = 0.0142
– P.M.
14 hours ago
(4/48)(4/47) + (4/48 + 4/47) = 0.0142
– P.M.
14 hours ago
1
1
A. You have left off a lot of cases. You could just draw a $5$,say. Or two $A's$ and a $3$. Or three $A's$ and a $2$. B. Your computation of the probabilities is wrong.
– lulu
13 hours ago
A. You have left off a lot of cases. You could just draw a $5$,say. Or two $A's$ and a $3$. Or three $A's$ and a $2$. B. Your computation of the probabilities is wrong.
– lulu
13 hours ago
|
show 3 more comments
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To the asker: please show your working on the question, then we may help you with this. To the down voters, you should be saying this to the OP, not me!
– Rhys Hughes
15 hours ago
you need to work out what the further combinations of cards you can get are. The fact that you know that the 4 aces are in the deck, but other cards could be already drawn, is a factor
– Cato
14 hours ago
P(Ace & 4)+P(3 & 2)=0.0142
– P.M.
14 hours ago
(4/48)(4/47) + (4/48 + 4/47) = 0.0142
– P.M.
14 hours ago
1
A. You have left off a lot of cases. You could just draw a $5$,say. Or two $A's$ and a $3$. Or three $A's$ and a $2$. B. Your computation of the probabilities is wrong.
– lulu
13 hours ago