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?






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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.









  • 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

















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







share|cite|improve this question







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















up vote
-2
down vote

favorite









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







share|cite|improve this question







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






share|cite|improve this question







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.











share|cite|improve this question







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.









share|cite|improve this question




share|cite|improve this question






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




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
















  • 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

















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