What is the expected value from these two different coin tossing games?
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Consider these two games:
Game $1$: Toss $4$ coins. If coins $1$ and $2$ are heads, you win $$5$. If coins $3$ and $4$ are heads, you win an additional $$5$.
Game $2$: Toss $3$ coins. If coins $1$ and $2$ are heads, you win $$5$. If coins $2$ and $3$ are heads, you win an additional $$5$.
My question is, which game will have the highest expected value for winnings? The case of game $1$ is simple. Receiving $$5$ from tossing coins $1$ and $2$ and receiving $$5$ from tossing coins $3$ and $4$ are independent events. Therefore, if the coins lands heads with probability $p$, the expected value of the winnings is$$
E(text{Game} ; 1) = 5 p^2 + 5 p^2 = 10p^2.
$$
In game $2$, we still have the probability $p^2$ of winning the first $$5$. However, the probability of receiving the additional $$5$from coin $3$ is dependent upon having a heads in coin $2$. I figured so far that
$$E(text{Game};2) = 5p^2 + x$$
I am not sure what $x$ is. Which result links the expected value of an event when conditional probabilities are involved?
EDIT: My solution approach is wrong, please see my comment below.
probability conditional-expectation conditional-probability expected-value
asked Jan 9 at 4:39
Lucas AlanisLucas Alanis
5343926
$endgroup$
add a comment |
$begingroup$
Consider these two games:
Game $1$: Toss $4$ coins. If coins $1$ and $2$ are heads, you win $$5$. If coins $3$ and $4$ are heads, you win an additional $$5$.
Game $2$: Toss $3$ coins. If coins $1$ and $2$ are heads, you win $$5$. If coins $2$ and $3$ are heads, you win an additional $$5$.
My question is, which game will have the highest expected value for winnings? The case of game $1$ is simple. Receiving $$5$ from tossing coins $1$ and $2$ and receiving $$5$ from tossing coins $3$ and $4$ are independent events. Therefore, if the coins lands heads with probability $p$, the expected value of the winnings is$$
E(text{Game} ; 1) = 5 p^2 + 5 p^2 = 10p^2.
$$
In game $2$, we still have the probability $p^2$ of winning the first $$5$. However, the probability of receiving the additional $$5$from coin $3$ is dependent upon having a heads in coin $2$. I figured so far that
$$E(text{Game};2) = 5p^2 + x$$
I am not sure what $x$ is. Which result links the expected value of an event when conditional probabilities are involved?
EDIT: My solution approach is wrong, please see my comment below.
probability conditional-expectation conditional-probability expected-value
asked Jan 9 at 4:39
Lucas AlanisLucas Alanis
5343926
$endgroup$
2
$begingroup$
If in Game 1 you don't get two heads from tossing Coins 1 and 2, does it matter what you get from tossing Coins 3 and 4? Your solution approach implies that it matters, but the problem statement seems to suggest otherwise.
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– Fabio Somenzi
Jan 9 at 5:36
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@FabioSomenzi It matters. One can win $$$5 if coins 3 and 4 are heads regardless of what coins 1 and 2 are. My solution approach is wrong.
$endgroup$
– Lucas Alanis
Jan 9 at 7:11
add a comment |
$begingroup$
Consider these two games:
Game $1$: Toss $4$ coins. If coins $1$ and $2$ are heads, you win $$5$. If coins $3$ and $4$ are heads, you win an additional $$5$.
Game $2$: Toss $3$ coins. If coins $1$ and $2$ are heads, you win $$5$. If coins $2$ and $3$ are heads, you win an additional $$5$.
My question is, which game will have the highest expected value for winnings? The case of game $1$ is simple. Receiving $$5$ from tossing coins $1$ and $2$ and receiving $$5$ from tossing coins $3$ and $4$ are independent events. Therefore, if the coins lands heads with probability $p$, the expected value of the winnings is$$
E(text{Game} ; 1) = 5 p^2 + 5 p^2 = 10p^2.
$$
In game $2$, we still have the probability $p^2$ of winning the first $$5$. However, the probability of receiving the additional $$5$from coin $3$ is dependent upon having a heads in coin $2$. I figured so far that
$$E(text{Game};2) = 5p^2 + x$$
I am not sure what $x$ is. Which result links the expected value of an event when conditional probabilities are involved?
EDIT: My solution approach is wrong, please see my comment below.
probability conditional-expectation conditional-probability expected-value
asked Jan 9 at 4:39
Lucas AlanisLucas Alanis
5343926
$endgroup$
Consider these two games:
Game $1$: Toss $4$ coins. If coins $1$ and $2$ are heads, you win $$5$. If coins $3$ and $4$ are heads, you win an additional $$5$.
Game $2$: Toss $3$ coins. If coins $1$ and $2$ are heads, you win $$5$. If coins $2$ and $3$ are heads, you win an additional $$5$.
My question is, which game will have the highest expected value for winnings? The case of game $1$ is simple. Receiving $$5$ from tossing coins $1$ and $2$ and receiving $$5$ from tossing coins $3$ and $4$ are independent events. Therefore, if the coins lands heads with probability $p$, the expected value of the winnings is$$
E(text{Game} ; 1) = 5 p^2 + 5 p^2 = 10p^2.
$$
In game $2$, we still have the probability $p^2$ of winning the first $$5$. However, the probability of receiving the additional $$5$from coin $3$ is dependent upon having a heads in coin $2$. I figured so far that
$$E(text{Game};2) = 5p^2 + x$$
I am not sure what $x$ is. Which result links the expected value of an event when conditional probabilities are involved?
EDIT: My solution approach is wrong, please see my comment below.
probability conditional-expectation conditional-probability expected-value
probability conditional-expectation conditional-probability expected-value
asked Jan 9 at 4:39
Lucas AlanisLucas Alanis
5343926
asked Jan 9 at 4:39
Lucas AlanisLucas Alanis
5343926
edited Jan 9 at 7:14
Lucas Alanis
asked Jan 9 at 4:39
Lucas AlanisLucas Alanis
5343926
asked Jan 9 at 4:39
Lucas AlanisLucas Alanis
5343926
asked Jan 9 at 4:39
Lucas AlanisLucas Alanis
5343926
5343926
2
$begingroup$
If in Game 1 you don't get two heads from tossing Coins 1 and 2, does it matter what you get from tossing Coins 3 and 4? Your solution approach implies that it matters, but the problem statement seems to suggest otherwise.
$endgroup$
– Fabio Somenzi
Jan 9 at 5:36
$begingroup$
@FabioSomenzi It matters. One can win $$$5 if coins 3 and 4 are heads regardless of what coins 1 and 2 are. My solution approach is wrong.
$endgroup$
– Lucas Alanis
Jan 9 at 7:11
add a comment |
2
$begingroup$
If in Game 1 you don't get two heads from tossing Coins 1 and 2, does it matter what you get from tossing Coins 3 and 4? Your solution approach implies that it matters, but the problem statement seems to suggest otherwise.
$endgroup$
– Fabio Somenzi
Jan 9 at 5:36
$begingroup$
@FabioSomenzi It matters. One can win $$$5 if coins 3 and 4 are heads regardless of what coins 1 and 2 are. My solution approach is wrong.
$endgroup$
– Lucas Alanis
Jan 9 at 7:11
2
2
$begingroup$
If in Game 1 you don't get two heads from tossing Coins 1 and 2, does it matter what you get from tossing Coins 3 and 4? Your solution approach implies that it matters, but the problem statement seems to suggest otherwise.
$endgroup$
– Fabio Somenzi
Jan 9 at 5:36
$begingroup$
If in Game 1 you don't get two heads from tossing Coins 1 and 2, does it matter what you get from tossing Coins 3 and 4? Your solution approach implies that it matters, but the problem statement seems to suggest otherwise.
$endgroup$
– Fabio Somenzi
Jan 9 at 5:36
$begingroup$
@FabioSomenzi It matters. One can win $$$5 if coins 3 and 4 are heads regardless of what coins 1 and 2 are. My solution approach is wrong.
$endgroup$
– Lucas Alanis
Jan 9 at 7:11
$begingroup$
@FabioSomenzi It matters. One can win $$$5 if coins 3 and 4 are heads regardless of what coins 1 and 2 are. My solution approach is wrong.
$endgroup$
– Lucas Alanis
Jan 9 at 7:11
add a comment |
1 Answer
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In game $2$ you win $10$ if all the first three come up heads, which is probability $p^3$. You win $5$ if the first three come up $HHT$ or $THH$, which is probability $2p^2(1-p)$. The total expectation is $10p^3+10p^2(1-p)=10p^3+10p^2-10p^3=10p^2$. The expectation is the same.
This is an example of the linearity of expectation. The expectation from each pair is $5p^2$. Each game has two pairs that matter. The linearity of expectation does not require independence.
answered Jan 9 at 6:06
Ross MillikanRoss Millikan
297k23198371
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add a comment |
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$begingroup$
In game $2$ you win $10$ if all the first three come up heads, which is probability $p^3$. You win $5$ if the first three come up $HHT$ or $THH$, which is probability $2p^2(1-p)$. The total expectation is $10p^3+10p^2(1-p)=10p^3+10p^2-10p^3=10p^2$. The expectation is the same.
This is an example of the linearity of expectation. The expectation from each pair is $5p^2$. Each game has two pairs that matter. The linearity of expectation does not require independence.
answered Jan 9 at 6:06
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
In game $2$ you win $10$ if all the first three come up heads, which is probability $p^3$. You win $5$ if the first three come up $HHT$ or $THH$, which is probability $2p^2(1-p)$. The total expectation is $10p^3+10p^2(1-p)=10p^3+10p^2-10p^3=10p^2$. The expectation is the same.
This is an example of the linearity of expectation. The expectation from each pair is $5p^2$. Each game has two pairs that matter. The linearity of expectation does not require independence.
answered Jan 9 at 6:06
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
In game $2$ you win $10$ if all the first three come up heads, which is probability $p^3$. You win $5$ if the first three come up $HHT$ or $THH$, which is probability $2p^2(1-p)$. The total expectation is $10p^3+10p^2(1-p)=10p^3+10p^2-10p^3=10p^2$. The expectation is the same.
This is an example of the linearity of expectation. The expectation from each pair is $5p^2$. Each game has two pairs that matter. The linearity of expectation does not require independence.
answered Jan 9 at 6:06
Ross MillikanRoss Millikan
297k23198371
$endgroup$
In game $2$ you win $10$ if all the first three come up heads, which is probability $p^3$. You win $5$ if the first three come up $HHT$ or $THH$, which is probability $2p^2(1-p)$. The total expectation is $10p^3+10p^2(1-p)=10p^3+10p^2-10p^3=10p^2$. The expectation is the same.
This is an example of the linearity of expectation. The expectation from each pair is $5p^2$. Each game has two pairs that matter. The linearity of expectation does not require independence.
answered Jan 9 at 6:06
Ross MillikanRoss Millikan
297k23198371
edited Jan 9 at 14:49
answered Jan 9 at 6:06
Ross MillikanRoss Millikan
297k23198371
answered Jan 9 at 6:06
Ross MillikanRoss Millikan
297k23198371
answered Jan 9 at 6:06
Ross MillikanRoss Millikan
297k23198371
297k23198371
add a comment |
add a comment |
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$begingroup$
If in Game 1 you don't get two heads from tossing Coins 1 and 2, does it matter what you get from tossing Coins 3 and 4? Your solution approach implies that it matters, but the problem statement seems to suggest otherwise.
$endgroup$
– Fabio Somenzi
Jan 9 at 5:36
$begingroup$
@FabioSomenzi It matters. One can win $$$5 if coins 3 and 4 are heads regardless of what coins 1 and 2 are. My solution approach is wrong.
$endgroup$
– Lucas Alanis
Jan 9 at 7:11