Expected Value with random variables for coin toss scenario
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Question:
I flip a fair coin, independently, 6 times, resulting in a sequence of heads (H) and tails (T). For each (consecutive) HTH in this sequence, you win $5.
Define the random variable X to be the "amount of dollars that you win".
For example, if the sequence is THTHTH; then X=10. What is the expected value of X?
Answer: $frac{5}{2}$
Attempt:
I am very confused on how to approach this. Like do I have to count the amount HTH possible and find the probability of it for all $64$ sequences and mulitply by 5? Writing out all the sequences will take me forever to do...
probability probability-theory random-variables expected-value
asked Dec 3 '18 at 22:51
TobyToby
1577
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add a comment |
$begingroup$
Question:
I flip a fair coin, independently, 6 times, resulting in a sequence of heads (H) and tails (T). For each (consecutive) HTH in this sequence, you win $5.
Define the random variable X to be the "amount of dollars that you win".
For example, if the sequence is THTHTH; then X=10. What is the expected value of X?
Answer: $frac{5}{2}$
Attempt:
I am very confused on how to approach this. Like do I have to count the amount HTH possible and find the probability of it for all $64$ sequences and mulitply by 5? Writing out all the sequences will take me forever to do...
probability probability-theory random-variables expected-value
asked Dec 3 '18 at 22:51
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
Question:
I flip a fair coin, independently, 6 times, resulting in a sequence of heads (H) and tails (T). For each (consecutive) HTH in this sequence, you win $5.
Define the random variable X to be the "amount of dollars that you win".
For example, if the sequence is THTHTH; then X=10. What is the expected value of X?
Answer: $frac{5}{2}$
Attempt:
I am very confused on how to approach this. Like do I have to count the amount HTH possible and find the probability of it for all $64$ sequences and mulitply by 5? Writing out all the sequences will take me forever to do...
probability probability-theory random-variables expected-value
asked Dec 3 '18 at 22:51
TobyToby
1577
$endgroup$
Question:
I flip a fair coin, independently, 6 times, resulting in a sequence of heads (H) and tails (T). For each (consecutive) HTH in this sequence, you win $5.
Define the random variable X to be the "amount of dollars that you win".
For example, if the sequence is THTHTH; then X=10. What is the expected value of X?
Answer: $frac{5}{2}$
Attempt:
I am very confused on how to approach this. Like do I have to count the amount HTH possible and find the probability of it for all $64$ sequences and mulitply by 5? Writing out all the sequences will take me forever to do...
probability probability-theory random-variables expected-value
probability probability-theory random-variables expected-value
asked Dec 3 '18 at 22:51
TobyToby
1577
asked Dec 3 '18 at 22:51
TobyToby
1577
asked Dec 3 '18 at 22:51
TobyToby
1577
asked Dec 3 '18 at 22:51
TobyToby
1577
asked Dec 3 '18 at 22:51
TobyToby
1577
1577
add a comment |
add a comment |
2 Answers
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For any three consecutive tosses, there is a $1/8$ chance of hitting HTH, so on tosses 1-3, you expect to win $5(1/8)$ dollars. Similarly for tosses 2-4, 3-5, and 4-6. By Linearity of Expectation, the expected winnings for the whole game is $4(5/8)=5/2$ dollars.
answered Dec 4 '18 at 0:21
NedNed
1,993910
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It is not that hard. Let $x$ be the random variable for the number of “HTH” in the sequence. Let $y$ be the corresponding income. Then, y=5x, so $mathbb{E}(y)=5mathbb{E}(x)$. Now, all you need to do is finding $p(x)$, which is essentially a combinatorics problem.
answered Dec 4 '18 at 0:24
D...D...
213113
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2 Answers
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2 Answers
2
active
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active
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active
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$begingroup$
For any three consecutive tosses, there is a $1/8$ chance of hitting HTH, so on tosses 1-3, you expect to win $5(1/8)$ dollars. Similarly for tosses 2-4, 3-5, and 4-6. By Linearity of Expectation, the expected winnings for the whole game is $4(5/8)=5/2$ dollars.
answered Dec 4 '18 at 0:21
NedNed
1,993910
$endgroup$
add a comment |
$begingroup$
For any three consecutive tosses, there is a $1/8$ chance of hitting HTH, so on tosses 1-3, you expect to win $5(1/8)$ dollars. Similarly for tosses 2-4, 3-5, and 4-6. By Linearity of Expectation, the expected winnings for the whole game is $4(5/8)=5/2$ dollars.
answered Dec 4 '18 at 0:21
NedNed
1,993910
$endgroup$
add a comment |
$begingroup$
For any three consecutive tosses, there is a $1/8$ chance of hitting HTH, so on tosses 1-3, you expect to win $5(1/8)$ dollars. Similarly for tosses 2-4, 3-5, and 4-6. By Linearity of Expectation, the expected winnings for the whole game is $4(5/8)=5/2$ dollars.
answered Dec 4 '18 at 0:21
NedNed
1,993910
$endgroup$
For any three consecutive tosses, there is a $1/8$ chance of hitting HTH, so on tosses 1-3, you expect to win $5(1/8)$ dollars. Similarly for tosses 2-4, 3-5, and 4-6. By Linearity of Expectation, the expected winnings for the whole game is $4(5/8)=5/2$ dollars.
answered Dec 4 '18 at 0:21
NedNed
1,993910
answered Dec 4 '18 at 0:21
NedNed
1,993910
answered Dec 4 '18 at 0:21
NedNed
1,993910
answered Dec 4 '18 at 0:21
NedNed
1,993910
1,993910
add a comment |
add a comment |
$begingroup$
It is not that hard. Let $x$ be the random variable for the number of “HTH” in the sequence. Let $y$ be the corresponding income. Then, y=5x, so $mathbb{E}(y)=5mathbb{E}(x)$. Now, all you need to do is finding $p(x)$, which is essentially a combinatorics problem.
answered Dec 4 '18 at 0:24
D...D...
213113
$endgroup$
add a comment |
$begingroup$
It is not that hard. Let $x$ be the random variable for the number of “HTH” in the sequence. Let $y$ be the corresponding income. Then, y=5x, so $mathbb{E}(y)=5mathbb{E}(x)$. Now, all you need to do is finding $p(x)$, which is essentially a combinatorics problem.
answered Dec 4 '18 at 0:24
D...D...
213113
$endgroup$
add a comment |
$begingroup$
It is not that hard. Let $x$ be the random variable for the number of “HTH” in the sequence. Let $y$ be the corresponding income. Then, y=5x, so $mathbb{E}(y)=5mathbb{E}(x)$. Now, all you need to do is finding $p(x)$, which is essentially a combinatorics problem.
answered Dec 4 '18 at 0:24
D...D...
213113
$endgroup$
It is not that hard. Let $x$ be the random variable for the number of “HTH” in the sequence. Let $y$ be the corresponding income. Then, y=5x, so $mathbb{E}(y)=5mathbb{E}(x)$. Now, all you need to do is finding $p(x)$, which is essentially a combinatorics problem.
answered Dec 4 '18 at 0:24
D...D...
213113
answered Dec 4 '18 at 0:24
D...D...
213113
answered Dec 4 '18 at 0:24
D...D...
213113
answered Dec 4 '18 at 0:24
D...D...
213113
213113
add a comment |
add a comment |
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