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Four coins with reflip problem?
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I came across the following problem today:
Flip four coins. For every head, you get $1. You may reflip one coin after the four flips. Calculate the expected returns.
I know that the expected value without the extra flip is $2. However, I am unsure of how to condition on the extra flips. I am tempted to claim that having the reflip simply adds $frac12$ to each case with tails since the only thing which affects the reflip is whether there are tails or not, but my gut tells me this is wrong. I am also told the correct returns is $frac7932$ and I have no idea where this comes from. Could someone help me out please? Thanks!
probability conditional-expectation conditional-probability expected-value
asked 1 hour ago
user107224
1526
add a comment |Â
up vote
4
down vote
favorite
I came across the following problem today:
Flip four coins. For every head, you get $1. You may reflip one coin after the four flips. Calculate the expected returns.
I know that the expected value without the extra flip is $2. However, I am unsure of how to condition on the extra flips. I am tempted to claim that having the reflip simply adds $frac12$ to each case with tails since the only thing which affects the reflip is whether there are tails or not, but my gut tells me this is wrong. I am also told the correct returns is $frac7932$ and I have no idea where this comes from. Could someone help me out please? Thanks!
probability conditional-expectation conditional-probability expected-value
asked 1 hour ago
user107224
1526
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I came across the following problem today:
Flip four coins. For every head, you get $1. You may reflip one coin after the four flips. Calculate the expected returns.
I know that the expected value without the extra flip is $2. However, I am unsure of how to condition on the extra flips. I am tempted to claim that having the reflip simply adds $frac12$ to each case with tails since the only thing which affects the reflip is whether there are tails or not, but my gut tells me this is wrong. I am also told the correct returns is $frac7932$ and I have no idea where this comes from. Could someone help me out please? Thanks!
probability conditional-expectation conditional-probability expected-value
asked 1 hour ago
user107224
1526
I came across the following problem today:
Flip four coins. For every head, you get $1. You may reflip one coin after the four flips. Calculate the expected returns.
I know that the expected value without the extra flip is $2. However, I am unsure of how to condition on the extra flips. I am tempted to claim that having the reflip simply adds $frac12$ to each case with tails since the only thing which affects the reflip is whether there are tails or not, but my gut tells me this is wrong. I am also told the correct returns is $frac7932$ and I have no idea where this comes from. Could someone help me out please? Thanks!
probability conditional-expectation conditional-probability expected-value
probability conditional-expectation conditional-probability expected-value
asked 1 hour ago
user107224
1526
asked 1 hour ago
user107224
1526
asked 1 hour ago
user107224
1526
asked 1 hour ago
user107224
1526
asked 1 hour ago
user107224
1526
1526
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
Let $X_i=1$ if $i$-th toss is head and $0$ otherwise.
The reward is $$sum_i=1^4X_i + X_5left( 1-prod_i=1^4X_iright)=sum_i=1^5X_i-prod_i=1^5X_i$$
Hence
$$mathbbEleft(sum_i=1^5X_i-prod_i=1^5X_iright)=left(sum_i=1^5mathbbE[X_i]-prod_i=1^5mathbbE[X_i]right)=frac52-frac132=frac7932$$
answered 1 hour ago
Siong Thye Goh
82.4k1456104
Very clear, thank you!
– user107224
29 mins ago
add a comment |Â
up vote
1
down vote
Your temptation is right and your gut is wrong. You do get an extra $frac12$ if you got tails at least once. The probability that you don't have a tail to reflip is $frac116$, so you get an extra $frac12left(1-frac116right)=frac1532$.
answered 1 hour ago
joriki
168k10180335
add a comment |Â
up vote
1
down vote
Expectation of first 4 flips is $2.
Expectation of the fifth flip, condition on there is a tail in the first 4 flips, is 0.5.
Expectation of the fifth flip, condition on there is no tail, is 0.
Probability there is a tail in the first 4 flips is 15/16.
Total expectation = 2 + 0.5*(15/16) + 0*(1/16) = 79/32
answered 1 hour ago
William Wong
1214
add a comment |Â
up vote
0
down vote
Your answer is $frac 8032$. The $frac1 32$ is from where all four flips are heads.
answered 1 hour ago
abc...
2,164529
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $X_i=1$ if $i$-th toss is head and $0$ otherwise.
The reward is $$sum_i=1^4X_i + X_5left( 1-prod_i=1^4X_iright)=sum_i=1^5X_i-prod_i=1^5X_i$$
Hence
$$mathbbEleft(sum_i=1^5X_i-prod_i=1^5X_iright)=left(sum_i=1^5mathbbE[X_i]-prod_i=1^5mathbbE[X_i]right)=frac52-frac132=frac7932$$
answered 1 hour ago
Siong Thye Goh
82.4k1456104
Very clear, thank you!
– user107224
29 mins ago
add a comment |Â
up vote
1
down vote
accepted
Let $X_i=1$ if $i$-th toss is head and $0$ otherwise.
The reward is $$sum_i=1^4X_i + X_5left( 1-prod_i=1^4X_iright)=sum_i=1^5X_i-prod_i=1^5X_i$$
Hence
$$mathbbEleft(sum_i=1^5X_i-prod_i=1^5X_iright)=left(sum_i=1^5mathbbE[X_i]-prod_i=1^5mathbbE[X_i]right)=frac52-frac132=frac7932$$
answered 1 hour ago
Siong Thye Goh
82.4k1456104
Very clear, thank you!
– user107224
29 mins ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $X_i=1$ if $i$-th toss is head and $0$ otherwise.
The reward is $$sum_i=1^4X_i + X_5left( 1-prod_i=1^4X_iright)=sum_i=1^5X_i-prod_i=1^5X_i$$
Hence
$$mathbbEleft(sum_i=1^5X_i-prod_i=1^5X_iright)=left(sum_i=1^5mathbbE[X_i]-prod_i=1^5mathbbE[X_i]right)=frac52-frac132=frac7932$$
answered 1 hour ago
Siong Thye Goh
82.4k1456104
Let $X_i=1$ if $i$-th toss is head and $0$ otherwise.
The reward is $$sum_i=1^4X_i + X_5left( 1-prod_i=1^4X_iright)=sum_i=1^5X_i-prod_i=1^5X_i$$
Hence
$$mathbbEleft(sum_i=1^5X_i-prod_i=1^5X_iright)=left(sum_i=1^5mathbbE[X_i]-prod_i=1^5mathbbE[X_i]right)=frac52-frac132=frac7932$$
answered 1 hour ago
Siong Thye Goh
82.4k1456104
answered 1 hour ago
Siong Thye Goh
82.4k1456104
answered 1 hour ago
Siong Thye Goh
82.4k1456104
answered 1 hour ago
Siong Thye Goh
82.4k1456104
82.4k1456104
Very clear, thank you!
– user107224
29 mins ago
add a comment |Â
Very clear, thank you!
– user107224
29 mins ago
Very clear, thank you!
– user107224
29 mins ago
Very clear, thank you!
– user107224
29 mins ago
add a comment |Â
up vote
1
down vote
Your temptation is right and your gut is wrong. You do get an extra $frac12$ if you got tails at least once. The probability that you don't have a tail to reflip is $frac116$, so you get an extra $frac12left(1-frac116right)=frac1532$.
answered 1 hour ago
joriki
168k10180335
add a comment |Â
up vote
1
down vote
Your temptation is right and your gut is wrong. You do get an extra $frac12$ if you got tails at least once. The probability that you don't have a tail to reflip is $frac116$, so you get an extra $frac12left(1-frac116right)=frac1532$.
answered 1 hour ago
joriki
168k10180335
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your temptation is right and your gut is wrong. You do get an extra $frac12$ if you got tails at least once. The probability that you don't have a tail to reflip is $frac116$, so you get an extra $frac12left(1-frac116right)=frac1532$.
answered 1 hour ago
joriki
168k10180335
Your temptation is right and your gut is wrong. You do get an extra $frac12$ if you got tails at least once. The probability that you don't have a tail to reflip is $frac116$, so you get an extra $frac12left(1-frac116right)=frac1532$.
answered 1 hour ago
joriki
168k10180335
answered 1 hour ago
joriki
168k10180335
answered 1 hour ago
joriki
168k10180335
answered 1 hour ago
joriki
168k10180335
168k10180335
add a comment |Â
add a comment |Â
up vote
1
down vote
Expectation of first 4 flips is $2.
Expectation of the fifth flip, condition on there is a tail in the first 4 flips, is 0.5.
Expectation of the fifth flip, condition on there is no tail, is 0.
Probability there is a tail in the first 4 flips is 15/16.
Total expectation = 2 + 0.5*(15/16) + 0*(1/16) = 79/32
answered 1 hour ago
William Wong
1214
add a comment |Â
up vote
1
down vote
Expectation of first 4 flips is $2.
Expectation of the fifth flip, condition on there is a tail in the first 4 flips, is 0.5.
Expectation of the fifth flip, condition on there is no tail, is 0.
Probability there is a tail in the first 4 flips is 15/16.
Total expectation = 2 + 0.5*(15/16) + 0*(1/16) = 79/32
answered 1 hour ago
William Wong
1214
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Expectation of first 4 flips is $2.
Expectation of the fifth flip, condition on there is a tail in the first 4 flips, is 0.5.
Expectation of the fifth flip, condition on there is no tail, is 0.
Probability there is a tail in the first 4 flips is 15/16.
Total expectation = 2 + 0.5*(15/16) + 0*(1/16) = 79/32
answered 1 hour ago
William Wong
1214
Expectation of first 4 flips is $2.
Expectation of the fifth flip, condition on there is a tail in the first 4 flips, is 0.5.
Expectation of the fifth flip, condition on there is no tail, is 0.
Probability there is a tail in the first 4 flips is 15/16.
Total expectation = 2 + 0.5*(15/16) + 0*(1/16) = 79/32
answered 1 hour ago
William Wong
1214
answered 1 hour ago
William Wong
1214
answered 1 hour ago
William Wong
1214
answered 1 hour ago
William Wong
1214
1214
add a comment |Â
add a comment |Â
up vote
0
down vote
Your answer is $frac 8032$. The $frac1 32$ is from where all four flips are heads.
answered 1 hour ago
abc...
2,164529
add a comment |Â
up vote
0
down vote
Your answer is $frac 8032$. The $frac1 32$ is from where all four flips are heads.
answered 1 hour ago
abc...
2,164529
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your answer is $frac 8032$. The $frac1 32$ is from where all four flips are heads.
answered 1 hour ago
abc...
2,164529
Your answer is $frac 8032$. The $frac1 32$ is from where all four flips are heads.
answered 1 hour ago
abc...
2,164529
answered 1 hour ago
abc...
2,164529
answered 1 hour ago
abc...
2,164529
answered 1 hour ago
abc...
2,164529
2,164529
add a comment |Â
add a comment |Â
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