Expected number of tosses for n heads in a row. [on hold]
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The probability of getting a head after a coin toss is $p$. Let $t$ be the number of tosses made until we get $n$ heads in a row. What’s the expected value of $t$?
probability
put on hold as off-topic by Y. Forman, Mark, Rebellos, jgon, Shailesh 20 hours ago
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The probability of getting a head after a coin toss is $p$. Let $t$ be the number of tosses made until we get $n$ heads in a row. What’s the expected value of $t$?
probability
put on hold as off-topic by Y. Forman, Mark, Rebellos, jgon, Shailesh 20 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." – Y. Forman, Mark, Rebellos, jgon, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
2
What is the source of this question? This is the second time it has appeared today.
– lulu
yesterday
See here: math.stackexchange.com/questions/364038/… e=2(2^n-1)
– Ben W
yesterday
@lulu I read it in the book Mathematics in Games, Sports and Gambling, of Ronald J. Gould.
– Sergio Enrique Yarza Acuña
yesterday
The answer is $(1-p^n)/((p^n)(1-p))$ which can be derived in a number of ways. Here's one source: www2.bc.edu/ned-rosen/public/CoinFlips.pdf
– Ned
yesterday
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up vote
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up vote
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down vote
favorite
The probability of getting a head after a coin toss is $p$. Let $t$ be the number of tosses made until we get $n$ heads in a row. What’s the expected value of $t$?
probability
The probability of getting a head after a coin toss is $p$. Let $t$ be the number of tosses made until we get $n$ heads in a row. What’s the expected value of $t$?
probability
probability
asked yesterday
Sergio Enrique Yarza Acuña
693314
693314
put on hold as off-topic by Y. Forman, Mark, Rebellos, jgon, Shailesh 20 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." – Y. Forman, Mark, Rebellos, jgon, Shailesh
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 Y. Forman, Mark, Rebellos, jgon, Shailesh 20 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." – Y. Forman, Mark, Rebellos, jgon, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
2
What is the source of this question? This is the second time it has appeared today.
– lulu
yesterday
See here: math.stackexchange.com/questions/364038/… e=2(2^n-1)
– Ben W
yesterday
@lulu I read it in the book Mathematics in Games, Sports and Gambling, of Ronald J. Gould.
– Sergio Enrique Yarza Acuña
yesterday
The answer is $(1-p^n)/((p^n)(1-p))$ which can be derived in a number of ways. Here's one source: www2.bc.edu/ned-rosen/public/CoinFlips.pdf
– Ned
yesterday
add a comment |
2
What is the source of this question? This is the second time it has appeared today.
– lulu
yesterday
See here: math.stackexchange.com/questions/364038/… e=2(2^n-1)
– Ben W
yesterday
@lulu I read it in the book Mathematics in Games, Sports and Gambling, of Ronald J. Gould.
– Sergio Enrique Yarza Acuña
yesterday
The answer is $(1-p^n)/((p^n)(1-p))$ which can be derived in a number of ways. Here's one source: www2.bc.edu/ned-rosen/public/CoinFlips.pdf
– Ned
yesterday
2
2
What is the source of this question? This is the second time it has appeared today.
– lulu
yesterday
What is the source of this question? This is the second time it has appeared today.
– lulu
yesterday
See here: math.stackexchange.com/questions/364038/… e=2(2^n-1)
– Ben W
yesterday
See here: math.stackexchange.com/questions/364038/… e=2(2^n-1)
– Ben W
yesterday
@lulu I read it in the book Mathematics in Games, Sports and Gambling, of Ronald J. Gould.
– Sergio Enrique Yarza Acuña
yesterday
@lulu I read it in the book Mathematics in Games, Sports and Gambling, of Ronald J. Gould.
– Sergio Enrique Yarza Acuña
yesterday
The answer is $(1-p^n)/((p^n)(1-p))$ which can be derived in a number of ways. Here's one source: www2.bc.edu/ned-rosen/public/CoinFlips.pdf
– Ned
yesterday
The answer is $(1-p^n)/((p^n)(1-p))$ which can be derived in a number of ways. Here's one source: www2.bc.edu/ned-rosen/public/CoinFlips.pdf
– Ned
yesterday
add a comment |
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2
What is the source of this question? This is the second time it has appeared today.
– lulu
yesterday
See here: math.stackexchange.com/questions/364038/… e=2(2^n-1)
– Ben W
yesterday
@lulu I read it in the book Mathematics in Games, Sports and Gambling, of Ronald J. Gould.
– Sergio Enrique Yarza Acuña
yesterday
The answer is $(1-p^n)/((p^n)(1-p))$ which can be derived in a number of ways. Here's one source: www2.bc.edu/ned-rosen/public/CoinFlips.pdf
– Ned
yesterday