Mixing
If I used a fair coin flip to determine bets on binary outcomes, will I be correct half the time?
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Suppose I am betting on binary outcomes, lets say randomly drawing a red ball from an urn with only red and yellow balls That is, I say either "red" or "yellow", then a ball is drawn from the urn, and I win the bet if I am correct.
If I choose to say "red" or "yellow" on the basis of flipping a fair
coin will I be correct half the time?
Assume the coin is independent of the distribution of balls in the urn, as well as of the process of choosing a ball from the urn (because random and Its weird to have a coin flip related to this)
My attempt at this is as follows. (In what follows suppose I guess red when the coin is heads)
Suppose there are $N$ balls in the urn, $K$ of which are red. Then the probability of a red ball being drawn is $frac{K}{N}$.
Define a random variable $X$ such that $X=1$ if I win the bet (I guess correctly) and $X=0$ if I guess incorrectly. Then the question reduces to finding the expected value of $X$ (when the probability that $X=1$ is calculated according to the betting strategy).
The expected value of $X$ is $$1cdot P(X=1) = P(X=1) = \.5P(Redvert Heads) + .5P(Yellowvert Tails) = .5P(RED)+.5(P(Tails) = .5$$
So I would be correct half the time.
Could someone tell me how I could formally derive the probability of
being correct, i.e. $P(X=1)$. That is, how can I mathematically show
that $$P(X=1) = .5P(Redvert Heads) + .5P(Yellowvert Tails)$$
I guess I would need to define a random variable for the outcome of the urn, and a random variable for the outcome of the coin flip, find the joint distribution of these random variables, and then $P(Correct)=P(Red,Heads) + P(Yellow, tails)$?
By independence though I guess $P(Red,Heads) =P(RED)cdot P(Heads)) and similar for tails, which gives the formula.
probability examples-counterexamples
asked Jan 13 at 17:13
user106860user106860
390214
$endgroup$
add a comment |
$begingroup$
Suppose I am betting on binary outcomes, lets say randomly drawing a red ball from an urn with only red and yellow balls That is, I say either "red" or "yellow", then a ball is drawn from the urn, and I win the bet if I am correct.
If I choose to say "red" or "yellow" on the basis of flipping a fair
coin will I be correct half the time?
Assume the coin is independent of the distribution of balls in the urn, as well as of the process of choosing a ball from the urn (because random and Its weird to have a coin flip related to this)
My attempt at this is as follows. (In what follows suppose I guess red when the coin is heads)
Suppose there are $N$ balls in the urn, $K$ of which are red. Then the probability of a red ball being drawn is $frac{K}{N}$.
Define a random variable $X$ such that $X=1$ if I win the bet (I guess correctly) and $X=0$ if I guess incorrectly. Then the question reduces to finding the expected value of $X$ (when the probability that $X=1$ is calculated according to the betting strategy).
The expected value of $X$ is $$1cdot P(X=1) = P(X=1) = \.5P(Redvert Heads) + .5P(Yellowvert Tails) = .5P(RED)+.5(P(Tails) = .5$$
So I would be correct half the time.
Could someone tell me how I could formally derive the probability of
being correct, i.e. $P(X=1)$. That is, how can I mathematically show
that $$P(X=1) = .5P(Redvert Heads) + .5P(Yellowvert Tails)$$
I guess I would need to define a random variable for the outcome of the urn, and a random variable for the outcome of the coin flip, find the joint distribution of these random variables, and then $P(Correct)=P(Red,Heads) + P(Yellow, tails)$?
By independence though I guess $P(Red,Heads) =P(RED)cdot P(Heads)) and similar for tails, which gives the formula.
probability examples-counterexamples
asked Jan 13 at 17:13
user106860user106860
390214
$endgroup$
$begingroup$
Not all binary events have equal probability. My lottery ticket will either win or lose today....but not with equal probability.
$endgroup$
– lulu
Jan 13 at 17:14
$begingroup$
The binary even doesn't need to have equal probability for betting on a coin flip to be make me correct half the time? At least, I didn't assume in the question that the binary event has equal probability, but perhaps I made a mistake somewhere?
$endgroup$
– user106860
Jan 13 at 17:18
$begingroup$
No, sorry. What you said is correct. sticking with my lottery ticket, you'll be right half the time regardless of whether I win or lose.
$endgroup$
– lulu
Jan 13 at 17:22
add a comment |
$begingroup$
Suppose I am betting on binary outcomes, lets say randomly drawing a red ball from an urn with only red and yellow balls That is, I say either "red" or "yellow", then a ball is drawn from the urn, and I win the bet if I am correct.
If I choose to say "red" or "yellow" on the basis of flipping a fair
coin will I be correct half the time?
Assume the coin is independent of the distribution of balls in the urn, as well as of the process of choosing a ball from the urn (because random and Its weird to have a coin flip related to this)
My attempt at this is as follows. (In what follows suppose I guess red when the coin is heads)
Suppose there are $N$ balls in the urn, $K$ of which are red. Then the probability of a red ball being drawn is $frac{K}{N}$.
Define a random variable $X$ such that $X=1$ if I win the bet (I guess correctly) and $X=0$ if I guess incorrectly. Then the question reduces to finding the expected value of $X$ (when the probability that $X=1$ is calculated according to the betting strategy).
The expected value of $X$ is $$1cdot P(X=1) = P(X=1) = \.5P(Redvert Heads) + .5P(Yellowvert Tails) = .5P(RED)+.5(P(Tails) = .5$$
So I would be correct half the time.
Could someone tell me how I could formally derive the probability of
being correct, i.e. $P(X=1)$. That is, how can I mathematically show
that $$P(X=1) = .5P(Redvert Heads) + .5P(Yellowvert Tails)$$
I guess I would need to define a random variable for the outcome of the urn, and a random variable for the outcome of the coin flip, find the joint distribution of these random variables, and then $P(Correct)=P(Red,Heads) + P(Yellow, tails)$?
By independence though I guess $P(Red,Heads) =P(RED)cdot P(Heads)) and similar for tails, which gives the formula.
probability examples-counterexamples
asked Jan 13 at 17:13
user106860user106860
390214
$endgroup$
Suppose I am betting on binary outcomes, lets say randomly drawing a red ball from an urn with only red and yellow balls That is, I say either "red" or "yellow", then a ball is drawn from the urn, and I win the bet if I am correct.
If I choose to say "red" or "yellow" on the basis of flipping a fair
coin will I be correct half the time?
Assume the coin is independent of the distribution of balls in the urn, as well as of the process of choosing a ball from the urn (because random and Its weird to have a coin flip related to this)
My attempt at this is as follows. (In what follows suppose I guess red when the coin is heads)
Suppose there are $N$ balls in the urn, $K$ of which are red. Then the probability of a red ball being drawn is $frac{K}{N}$.
Define a random variable $X$ such that $X=1$ if I win the bet (I guess correctly) and $X=0$ if I guess incorrectly. Then the question reduces to finding the expected value of $X$ (when the probability that $X=1$ is calculated according to the betting strategy).
The expected value of $X$ is $$1cdot P(X=1) = P(X=1) = \.5P(Redvert Heads) + .5P(Yellowvert Tails) = .5P(RED)+.5(P(Tails) = .5$$
So I would be correct half the time.
Could someone tell me how I could formally derive the probability of
being correct, i.e. $P(X=1)$. That is, how can I mathematically show
that $$P(X=1) = .5P(Redvert Heads) + .5P(Yellowvert Tails)$$
I guess I would need to define a random variable for the outcome of the urn, and a random variable for the outcome of the coin flip, find the joint distribution of these random variables, and then $P(Correct)=P(Red,Heads) + P(Yellow, tails)$?
By independence though I guess $P(Red,Heads) =P(RED)cdot P(Heads)) and similar for tails, which gives the formula.
probability examples-counterexamples
probability examples-counterexamples
asked Jan 13 at 17:13
user106860user106860
390214
asked Jan 13 at 17:13
user106860user106860
390214
edited Jan 13 at 17:24
user106860
asked Jan 13 at 17:13
user106860user106860
390214
asked Jan 13 at 17:13
user106860user106860
390214
asked Jan 13 at 17:13
user106860user106860
390214
390214
$begingroup$
Not all binary events have equal probability. My lottery ticket will either win or lose today....but not with equal probability.
$endgroup$
– lulu
Jan 13 at 17:14
$begingroup$
The binary even doesn't need to have equal probability for betting on a coin flip to be make me correct half the time? At least, I didn't assume in the question that the binary event has equal probability, but perhaps I made a mistake somewhere?
$endgroup$
– user106860
Jan 13 at 17:18
$begingroup$
No, sorry. What you said is correct. sticking with my lottery ticket, you'll be right half the time regardless of whether I win or lose.
$endgroup$
– lulu
Jan 13 at 17:22
add a comment |
$begingroup$
Not all binary events have equal probability. My lottery ticket will either win or lose today....but not with equal probability.
$endgroup$
– lulu
Jan 13 at 17:14
$begingroup$
The binary even doesn't need to have equal probability for betting on a coin flip to be make me correct half the time? At least, I didn't assume in the question that the binary event has equal probability, but perhaps I made a mistake somewhere?
$endgroup$
– user106860
Jan 13 at 17:18
$begingroup$
No, sorry. What you said is correct. sticking with my lottery ticket, you'll be right half the time regardless of whether I win or lose.
$endgroup$
– lulu
Jan 13 at 17:22
$begingroup$
Not all binary events have equal probability. My lottery ticket will either win or lose today....but not with equal probability.
$endgroup$
– lulu
Jan 13 at 17:14
$begingroup$
Not all binary events have equal probability. My lottery ticket will either win or lose today....but not with equal probability.
$endgroup$
– lulu
Jan 13 at 17:14
$begingroup$
The binary even doesn't need to have equal probability for betting on a coin flip to be make me correct half the time? At least, I didn't assume in the question that the binary event has equal probability, but perhaps I made a mistake somewhere?
$endgroup$
– user106860
Jan 13 at 17:18
$begingroup$
The binary even doesn't need to have equal probability for betting on a coin flip to be make me correct half the time? At least, I didn't assume in the question that the binary event has equal probability, but perhaps I made a mistake somewhere?
$endgroup$
– user106860
Jan 13 at 17:18
$begingroup$
No, sorry. What you said is correct. sticking with my lottery ticket, you'll be right half the time regardless of whether I win or lose.
$endgroup$
– lulu
Jan 13 at 17:22
$begingroup$
No, sorry. What you said is correct. sticking with my lottery ticket, you'll be right half the time regardless of whether I win or lose.
$endgroup$
– lulu
Jan 13 at 17:22
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You will win with probability $dfrac KN$ half of the time and win with probability $dfrac{N-K}N$ the other half. On average, $dfrac12$.
Said differently, whatever the outcome, you will choose this outcome half of the time.
answered Jan 13 at 17:19
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
Thank you. Any chance you could include how I could formally derive the the probability of winning?
$endgroup$
– user106860
Jan 13 at 17:20
$begingroup$
@user106860: $1/2 K/N+1/2 (N-K)/N=1/2$.
$endgroup$
– Yves Daoust
Jan 13 at 17:21
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
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$begingroup$
You will win with probability $dfrac KN$ half of the time and win with probability $dfrac{N-K}N$ the other half. On average, $dfrac12$.
Said differently, whatever the outcome, you will choose this outcome half of the time.
answered Jan 13 at 17:19
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
Thank you. Any chance you could include how I could formally derive the the probability of winning?
$endgroup$
– user106860
Jan 13 at 17:20
$begingroup$
@user106860: $1/2 K/N+1/2 (N-K)/N=1/2$.
$endgroup$
– Yves Daoust
Jan 13 at 17:21
add a comment |
$begingroup$
You will win with probability $dfrac KN$ half of the time and win with probability $dfrac{N-K}N$ the other half. On average, $dfrac12$.
Said differently, whatever the outcome, you will choose this outcome half of the time.
answered Jan 13 at 17:19
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
Thank you. Any chance you could include how I could formally derive the the probability of winning?
$endgroup$
– user106860
Jan 13 at 17:20
$begingroup$
@user106860: $1/2 K/N+1/2 (N-K)/N=1/2$.
$endgroup$
– Yves Daoust
Jan 13 at 17:21
add a comment |
$begingroup$
You will win with probability $dfrac KN$ half of the time and win with probability $dfrac{N-K}N$ the other half. On average, $dfrac12$.
Said differently, whatever the outcome, you will choose this outcome half of the time.
answered Jan 13 at 17:19
Yves DaoustYves Daoust
127k673226
$endgroup$
You will win with probability $dfrac KN$ half of the time and win with probability $dfrac{N-K}N$ the other half. On average, $dfrac12$.
Said differently, whatever the outcome, you will choose this outcome half of the time.
answered Jan 13 at 17:19
Yves DaoustYves Daoust
127k673226
edited Jan 13 at 17:21
answered Jan 13 at 17:19
Yves DaoustYves Daoust
127k673226
answered Jan 13 at 17:19
Yves DaoustYves Daoust
127k673226
answered Jan 13 at 17:19
Yves DaoustYves Daoust
127k673226
127k673226
$begingroup$
Thank you. Any chance you could include how I could formally derive the the probability of winning?
$endgroup$
– user106860
Jan 13 at 17:20
$begingroup$
@user106860: $1/2 K/N+1/2 (N-K)/N=1/2$.
$endgroup$
– Yves Daoust
Jan 13 at 17:21
add a comment |
$begingroup$
Thank you. Any chance you could include how I could formally derive the the probability of winning?
$endgroup$
– user106860
Jan 13 at 17:20
$begingroup$
@user106860: $1/2 K/N+1/2 (N-K)/N=1/2$.
$endgroup$
– Yves Daoust
Jan 13 at 17:21
$begingroup$
Thank you. Any chance you could include how I could formally derive the the probability of winning?
$endgroup$
– user106860
Jan 13 at 17:20
$begingroup$
Thank you. Any chance you could include how I could formally derive the the probability of winning?
$endgroup$
– user106860
Jan 13 at 17:20
$begingroup$
@user106860: $1/2 K/N+1/2 (N-K)/N=1/2$.
$endgroup$
– Yves Daoust
Jan 13 at 17:21
$begingroup$
@user106860: $1/2 K/N+1/2 (N-K)/N=1/2$.
$endgroup$
– Yves Daoust
Jan 13 at 17:21
add a comment |
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$begingroup$
Not all binary events have equal probability. My lottery ticket will either win or lose today....but not with equal probability.
$endgroup$
– lulu
Jan 13 at 17:14
$begingroup$
The binary even doesn't need to have equal probability for betting on a coin flip to be make me correct half the time? At least, I didn't assume in the question that the binary event has equal probability, but perhaps I made a mistake somewhere?
$endgroup$
– user106860
Jan 13 at 17:18
$begingroup$
No, sorry. What you said is correct. sticking with my lottery ticket, you'll be right half the time regardless of whether I win or lose.
$endgroup$
– lulu
Jan 13 at 17:22