Mixing
Law of total probability with basketball players and uncertainty
$begingroup$
Suppose you have a player from team A and a player from team B. You know that one of them has a 60% chance to make a shot and other has 40% chance to make a shot, but you are not sure which is which.
Suppose you choose 1 of the 2 players to shoot, how would you go about calculating his probability of missing.
My train of thought was that I could calculate it using law of total probability, but I got stuck along the way.
let $G$ = event that player is the better of the two
My equation was
$$ P(miss) = P(miss | G) * P(G) + (miss | overline G) * P (overline G) $$, but I was unsure of how to find $P(G)$
Alternatively, I thought maybe you could just average out the two chances and it would end up being 50/50, but that seemed off the mark.
Is there a better way of going about solving this or did I just miss something along the way?
Thanks
probability conditional-probability bayes-theorem
asked Feb 2 at 7:11
FlyromFlyrom
265
$endgroup$
add a comment |
$begingroup$
Suppose you have a player from team A and a player from team B. You know that one of them has a 60% chance to make a shot and other has 40% chance to make a shot, but you are not sure which is which.
Suppose you choose 1 of the 2 players to shoot, how would you go about calculating his probability of missing.
My train of thought was that I could calculate it using law of total probability, but I got stuck along the way.
let $G$ = event that player is the better of the two
My equation was
$$ P(miss) = P(miss | G) * P(G) + (miss | overline G) * P (overline G) $$, but I was unsure of how to find $P(G)$
Alternatively, I thought maybe you could just average out the two chances and it would end up being 50/50, but that seemed off the mark.
Is there a better way of going about solving this or did I just miss something along the way?
Thanks
probability conditional-probability bayes-theorem
asked Feb 2 at 7:11
FlyromFlyrom
265
$endgroup$
add a comment |
$begingroup$
Suppose you have a player from team A and a player from team B. You know that one of them has a 60% chance to make a shot and other has 40% chance to make a shot, but you are not sure which is which.
Suppose you choose 1 of the 2 players to shoot, how would you go about calculating his probability of missing.
My train of thought was that I could calculate it using law of total probability, but I got stuck along the way.
let $G$ = event that player is the better of the two
My equation was
$$ P(miss) = P(miss | G) * P(G) + (miss | overline G) * P (overline G) $$, but I was unsure of how to find $P(G)$
Alternatively, I thought maybe you could just average out the two chances and it would end up being 50/50, but that seemed off the mark.
Is there a better way of going about solving this or did I just miss something along the way?
Thanks
probability conditional-probability bayes-theorem
asked Feb 2 at 7:11
FlyromFlyrom
265
$endgroup$
Suppose you have a player from team A and a player from team B. You know that one of them has a 60% chance to make a shot and other has 40% chance to make a shot, but you are not sure which is which.
Suppose you choose 1 of the 2 players to shoot, how would you go about calculating his probability of missing.
My train of thought was that I could calculate it using law of total probability, but I got stuck along the way.
let $G$ = event that player is the better of the two
My equation was
$$ P(miss) = P(miss | G) * P(G) + (miss | overline G) * P (overline G) $$, but I was unsure of how to find $P(G)$
Alternatively, I thought maybe you could just average out the two chances and it would end up being 50/50, but that seemed off the mark.
Is there a better way of going about solving this or did I just miss something along the way?
Thanks
probability conditional-probability bayes-theorem
probability conditional-probability bayes-theorem
asked Feb 2 at 7:11
FlyromFlyrom
265
asked Feb 2 at 7:11
FlyromFlyrom
265
asked Feb 2 at 7:11
FlyromFlyrom
265
asked Feb 2 at 7:11
FlyromFlyrom
265
asked Feb 2 at 7:11
FlyromFlyrom
265
265
add a comment |
add a comment |
1 Answer
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$begingroup$
Your initial approach seems to be correct. If the question does not tell you anything about how you choose the player, you can assume that both players have equal probability of being chosen, i.e. $P(G)=P(overline G)=frac{1}{2}$. Now the answer should be easy to obtain.
Interestingly, if you make this assumption then the answer is the same as averaging their respective probabilities of missing. However, that follows trivially from your first approach along with the aforementioned assumption. It will not be true if the probability of choosing one player is different from the probability of choosing another.
answered Feb 2 at 7:32
s0ulr3aper07s0ulr3aper07
695112
$endgroup$
1
$begingroup$
We choose which player based on flipping a fair coin, so it should be 50/50 in that case
$endgroup$
– Flyrom
Feb 2 at 16:34
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Your initial approach seems to be correct. If the question does not tell you anything about how you choose the player, you can assume that both players have equal probability of being chosen, i.e. $P(G)=P(overline G)=frac{1}{2}$. Now the answer should be easy to obtain.
Interestingly, if you make this assumption then the answer is the same as averaging their respective probabilities of missing. However, that follows trivially from your first approach along with the aforementioned assumption. It will not be true if the probability of choosing one player is different from the probability of choosing another.
answered Feb 2 at 7:32
s0ulr3aper07s0ulr3aper07
695112
$endgroup$
1
$begingroup$
We choose which player based on flipping a fair coin, so it should be 50/50 in that case
$endgroup$
– Flyrom
Feb 2 at 16:34
add a comment |
$begingroup$
Your initial approach seems to be correct. If the question does not tell you anything about how you choose the player, you can assume that both players have equal probability of being chosen, i.e. $P(G)=P(overline G)=frac{1}{2}$. Now the answer should be easy to obtain.
Interestingly, if you make this assumption then the answer is the same as averaging their respective probabilities of missing. However, that follows trivially from your first approach along with the aforementioned assumption. It will not be true if the probability of choosing one player is different from the probability of choosing another.
answered Feb 2 at 7:32
s0ulr3aper07s0ulr3aper07
695112
$endgroup$
1
$begingroup$
We choose which player based on flipping a fair coin, so it should be 50/50 in that case
$endgroup$
– Flyrom
Feb 2 at 16:34
add a comment |
$begingroup$
Your initial approach seems to be correct. If the question does not tell you anything about how you choose the player, you can assume that both players have equal probability of being chosen, i.e. $P(G)=P(overline G)=frac{1}{2}$. Now the answer should be easy to obtain.
Interestingly, if you make this assumption then the answer is the same as averaging their respective probabilities of missing. However, that follows trivially from your first approach along with the aforementioned assumption. It will not be true if the probability of choosing one player is different from the probability of choosing another.
answered Feb 2 at 7:32
s0ulr3aper07s0ulr3aper07
695112
$endgroup$
Your initial approach seems to be correct. If the question does not tell you anything about how you choose the player, you can assume that both players have equal probability of being chosen, i.e. $P(G)=P(overline G)=frac{1}{2}$. Now the answer should be easy to obtain.
Interestingly, if you make this assumption then the answer is the same as averaging their respective probabilities of missing. However, that follows trivially from your first approach along with the aforementioned assumption. It will not be true if the probability of choosing one player is different from the probability of choosing another.
answered Feb 2 at 7:32
s0ulr3aper07s0ulr3aper07
695112
edited Feb 2 at 7:40
answered Feb 2 at 7:32
s0ulr3aper07s0ulr3aper07
695112
answered Feb 2 at 7:32
s0ulr3aper07s0ulr3aper07
695112
answered Feb 2 at 7:32
s0ulr3aper07s0ulr3aper07
695112
695112
1
$begingroup$
We choose which player based on flipping a fair coin, so it should be 50/50 in that case
$endgroup$
– Flyrom
Feb 2 at 16:34
add a comment |
1
$begingroup$
We choose which player based on flipping a fair coin, so it should be 50/50 in that case
$endgroup$
– Flyrom
Feb 2 at 16:34
1
1
$begingroup$
We choose which player based on flipping a fair coin, so it should be 50/50 in that case
$endgroup$
– Flyrom
Feb 2 at 16:34
$begingroup$
We choose which player based on flipping a fair coin, so it should be 50/50 in that case
$endgroup$
– Flyrom
Feb 2 at 16:34
add a comment |
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