What is minimum and maximum probability
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What is meant by minimum and maximum probability of an event. I came across a questions that asks minimum and maximum probability of three heads in three coins are flipped. We need to find minimum and maximum probability in two cases
- When all three coins are not independent
- All pairs of coins are mutually independent
The probability of head and tail on each individual coin is 0.5. I am more concerned on how to approach this problem rather than its solution. So far, I am unable to find any material on minimum and maximum probability.
probability probability-theory
asked Sep 16 '16 at 20:47
Muhammad Adeel Zahid
1627
add a comment |Â
up vote
2
down vote
favorite
What is meant by minimum and maximum probability of an event. I came across a questions that asks minimum and maximum probability of three heads in three coins are flipped. We need to find minimum and maximum probability in two cases
- When all three coins are not independent
- All pairs of coins are mutually independent
The probability of head and tail on each individual coin is 0.5. I am more concerned on how to approach this problem rather than its solution. So far, I am unable to find any material on minimum and maximum probability.
probability probability-theory
asked Sep 16 '16 at 20:47
Muhammad Adeel Zahid
1627
For the first one, you can alter the probability by altering the dependence. For example, suppose the second toss can't be the same as the first. Then the probability of getting three $H's$ in a row is $0$. Can't get lower than that, but you still have to find the max.
– lulu
Sep 16 '16 at 20:56
Can you elaborate a bit on this please?
– Muhammad Adeel Zahid
Sep 17 '16 at 9:01
Sure. If the tosses are independent, the answer is $frac 18$. However, if they are dependent the answer will depend on the dependence. I gave one form of the dependence which made the answer $0$. On the other side, if we assume that all tosses are the same then the answer is $frac 12$. That's the maximum.
– lulu
Sep 17 '16 at 10:03
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
What is meant by minimum and maximum probability of an event. I came across a questions that asks minimum and maximum probability of three heads in three coins are flipped. We need to find minimum and maximum probability in two cases
- When all three coins are not independent
- All pairs of coins are mutually independent
The probability of head and tail on each individual coin is 0.5. I am more concerned on how to approach this problem rather than its solution. So far, I am unable to find any material on minimum and maximum probability.
probability probability-theory
asked Sep 16 '16 at 20:47
Muhammad Adeel Zahid
1627
What is meant by minimum and maximum probability of an event. I came across a questions that asks minimum and maximum probability of three heads in three coins are flipped. We need to find minimum and maximum probability in two cases
- When all three coins are not independent
- All pairs of coins are mutually independent
The probability of head and tail on each individual coin is 0.5. I am more concerned on how to approach this problem rather than its solution. So far, I am unable to find any material on minimum and maximum probability.
probability probability-theory
probability probability-theory
asked Sep 16 '16 at 20:47
Muhammad Adeel Zahid
1627
asked Sep 16 '16 at 20:47
Muhammad Adeel Zahid
1627
asked Sep 16 '16 at 20:47
Muhammad Adeel Zahid
1627
asked Sep 16 '16 at 20:47
Muhammad Adeel Zahid
1627
asked Sep 16 '16 at 20:47
Muhammad Adeel Zahid
1627
1627
For the first one, you can alter the probability by altering the dependence. For example, suppose the second toss can't be the same as the first. Then the probability of getting three $H's$ in a row is $0$. Can't get lower than that, but you still have to find the max.
– lulu
Sep 16 '16 at 20:56
Can you elaborate a bit on this please?
– Muhammad Adeel Zahid
Sep 17 '16 at 9:01
Sure. If the tosses are independent, the answer is $frac 18$. However, if they are dependent the answer will depend on the dependence. I gave one form of the dependence which made the answer $0$. On the other side, if we assume that all tosses are the same then the answer is $frac 12$. That's the maximum.
– lulu
Sep 17 '16 at 10:03
add a comment |Â
For the first one, you can alter the probability by altering the dependence. For example, suppose the second toss can't be the same as the first. Then the probability of getting three $H's$ in a row is $0$. Can't get lower than that, but you still have to find the max.
– lulu
Sep 16 '16 at 20:56
Can you elaborate a bit on this please?
– Muhammad Adeel Zahid
Sep 17 '16 at 9:01
Sure. If the tosses are independent, the answer is $frac 18$. However, if they are dependent the answer will depend on the dependence. I gave one form of the dependence which made the answer $0$. On the other side, if we assume that all tosses are the same then the answer is $frac 12$. That's the maximum.
– lulu
Sep 17 '16 at 10:03
For the first one, you can alter the probability by altering the dependence. For example, suppose the second toss can't be the same as the first. Then the probability of getting three $H's$ in a row is $0$. Can't get lower than that, but you still have to find the max.
– lulu
Sep 16 '16 at 20:56
For the first one, you can alter the probability by altering the dependence. For example, suppose the second toss can't be the same as the first. Then the probability of getting three $H's$ in a row is $0$. Can't get lower than that, but you still have to find the max.
– lulu
Sep 16 '16 at 20:56
Can you elaborate a bit on this please?
– Muhammad Adeel Zahid
Sep 17 '16 at 9:01
Can you elaborate a bit on this please?
– Muhammad Adeel Zahid
Sep 17 '16 at 9:01
Sure. If the tosses are independent, the answer is $frac 18$. However, if they are dependent the answer will depend on the dependence. I gave one form of the dependence which made the answer $0$. On the other side, if we assume that all tosses are the same then the answer is $frac 12$. That's the maximum.
– lulu
Sep 17 '16 at 10:03
Sure. If the tosses are independent, the answer is $frac 18$. However, if they are dependent the answer will depend on the dependence. I gave one form of the dependence which made the answer $0$. On the other side, if we assume that all tosses are the same then the answer is $frac 12$. That's the maximum.
– lulu
Sep 17 '16 at 10:03
add a comment |Â
2 Answers
2
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up vote
1
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Assign probability $1/4$ to each of the outcomes TTT, HHT, THH, THH. (That is, spread the mass uniformly on the subset where there is an even number of heads. One can do this by flipping the first two coins independently, and then using the parity of the first two to determine the output of the 3d flip. At any rate, in this case, $P(HHH)=0$. This strange probability law evidently minimizes $P(HHH)$.
Now assign probability $1/2$ to the outcomes HHH and TTT. Now $P(HHH)=1/2$, which is maximal. (If $P(HHH)>1/2$ we'd make the first coin unfair, giving it probability $>1/2$.)
That's the answer. A general principle here is that the joint probability law is constrained by having certain marginals. The constraints are linear (these probabilities must add up to that) and so the OP's problem boils down to a linear programming problem.
answered Jul 13 '17 at 20:00
kimchi lover
8,91031128
add a comment |Â
up vote
-1
down vote
The probability of three heads with three FAIR coins is $(1over 2)^3$. The min/max may relate to conditional probabilities where the chance of getting heads on the seconds coin depends on what happened with the first coin etc. I'm only guessing that it may mean this as I have never encountered min/max in probability questions such as this.
answered Sep 16 '16 at 21:14
Susan
525
"The probability of three heads with three FAIR coins is $(1over 2)^3$" If they are independent, yes. Otherwise, not always.
– Did
Mar 20 at 15:32
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Assign probability $1/4$ to each of the outcomes TTT, HHT, THH, THH. (That is, spread the mass uniformly on the subset where there is an even number of heads. One can do this by flipping the first two coins independently, and then using the parity of the first two to determine the output of the 3d flip. At any rate, in this case, $P(HHH)=0$. This strange probability law evidently minimizes $P(HHH)$.
Now assign probability $1/2$ to the outcomes HHH and TTT. Now $P(HHH)=1/2$, which is maximal. (If $P(HHH)>1/2$ we'd make the first coin unfair, giving it probability $>1/2$.)
That's the answer. A general principle here is that the joint probability law is constrained by having certain marginals. The constraints are linear (these probabilities must add up to that) and so the OP's problem boils down to a linear programming problem.
answered Jul 13 '17 at 20:00
kimchi lover
8,91031128
add a comment |Â
up vote
1
down vote
Assign probability $1/4$ to each of the outcomes TTT, HHT, THH, THH. (That is, spread the mass uniformly on the subset where there is an even number of heads. One can do this by flipping the first two coins independently, and then using the parity of the first two to determine the output of the 3d flip. At any rate, in this case, $P(HHH)=0$. This strange probability law evidently minimizes $P(HHH)$.
Now assign probability $1/2$ to the outcomes HHH and TTT. Now $P(HHH)=1/2$, which is maximal. (If $P(HHH)>1/2$ we'd make the first coin unfair, giving it probability $>1/2$.)
That's the answer. A general principle here is that the joint probability law is constrained by having certain marginals. The constraints are linear (these probabilities must add up to that) and so the OP's problem boils down to a linear programming problem.
answered Jul 13 '17 at 20:00
kimchi lover
8,91031128
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assign probability $1/4$ to each of the outcomes TTT, HHT, THH, THH. (That is, spread the mass uniformly on the subset where there is an even number of heads. One can do this by flipping the first two coins independently, and then using the parity of the first two to determine the output of the 3d flip. At any rate, in this case, $P(HHH)=0$. This strange probability law evidently minimizes $P(HHH)$.
Now assign probability $1/2$ to the outcomes HHH and TTT. Now $P(HHH)=1/2$, which is maximal. (If $P(HHH)>1/2$ we'd make the first coin unfair, giving it probability $>1/2$.)
That's the answer. A general principle here is that the joint probability law is constrained by having certain marginals. The constraints are linear (these probabilities must add up to that) and so the OP's problem boils down to a linear programming problem.
answered Jul 13 '17 at 20:00
kimchi lover
8,91031128
Assign probability $1/4$ to each of the outcomes TTT, HHT, THH, THH. (That is, spread the mass uniformly on the subset where there is an even number of heads. One can do this by flipping the first two coins independently, and then using the parity of the first two to determine the output of the 3d flip. At any rate, in this case, $P(HHH)=0$. This strange probability law evidently minimizes $P(HHH)$.
Now assign probability $1/2$ to the outcomes HHH and TTT. Now $P(HHH)=1/2$, which is maximal. (If $P(HHH)>1/2$ we'd make the first coin unfair, giving it probability $>1/2$.)
That's the answer. A general principle here is that the joint probability law is constrained by having certain marginals. The constraints are linear (these probabilities must add up to that) and so the OP's problem boils down to a linear programming problem.
answered Jul 13 '17 at 20:00
kimchi lover
8,91031128
answered Jul 13 '17 at 20:00
kimchi lover
8,91031128
answered Jul 13 '17 at 20:00
kimchi lover
8,91031128
answered Jul 13 '17 at 20:00
kimchi lover
8,91031128
8,91031128
add a comment |Â
add a comment |Â
up vote
-1
down vote
The probability of three heads with three FAIR coins is $(1over 2)^3$. The min/max may relate to conditional probabilities where the chance of getting heads on the seconds coin depends on what happened with the first coin etc. I'm only guessing that it may mean this as I have never encountered min/max in probability questions such as this.
answered Sep 16 '16 at 21:14
Susan
525
"The probability of three heads with three FAIR coins is $(1over 2)^3$" If they are independent, yes. Otherwise, not always.
– Did
Mar 20 at 15:32
add a comment |Â
up vote
-1
down vote
The probability of three heads with three FAIR coins is $(1over 2)^3$. The min/max may relate to conditional probabilities where the chance of getting heads on the seconds coin depends on what happened with the first coin etc. I'm only guessing that it may mean this as I have never encountered min/max in probability questions such as this.
answered Sep 16 '16 at 21:14
Susan
525
"The probability of three heads with three FAIR coins is $(1over 2)^3$" If they are independent, yes. Otherwise, not always.
– Did
Mar 20 at 15:32
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
The probability of three heads with three FAIR coins is $(1over 2)^3$. The min/max may relate to conditional probabilities where the chance of getting heads on the seconds coin depends on what happened with the first coin etc. I'm only guessing that it may mean this as I have never encountered min/max in probability questions such as this.
answered Sep 16 '16 at 21:14
Susan
525
The probability of three heads with three FAIR coins is $(1over 2)^3$. The min/max may relate to conditional probabilities where the chance of getting heads on the seconds coin depends on what happened with the first coin etc. I'm only guessing that it may mean this as I have never encountered min/max in probability questions such as this.
answered Sep 16 '16 at 21:14
Susan
525
answered Sep 16 '16 at 21:14
Susan
525
answered Sep 16 '16 at 21:14
Susan
525
answered Sep 16 '16 at 21:14
Susan
525
525
"The probability of three heads with three FAIR coins is $(1over 2)^3$" If they are independent, yes. Otherwise, not always.
– Did
Mar 20 at 15:32
add a comment |Â
"The probability of three heads with three FAIR coins is $(1over 2)^3$" If they are independent, yes. Otherwise, not always.
– Did
Mar 20 at 15:32
"The probability of three heads with three FAIR coins is $(1over 2)^3$" If they are independent, yes. Otherwise, not always.
– Did
Mar 20 at 15:32
"The probability of three heads with three FAIR coins is $(1over 2)^3$" If they are independent, yes. Otherwise, not always.
– Did
Mar 20 at 15:32
add a comment |Â
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For the first one, you can alter the probability by altering the dependence. For example, suppose the second toss can't be the same as the first. Then the probability of getting three $H's$ in a row is $0$. Can't get lower than that, but you still have to find the max.
– lulu
Sep 16 '16 at 20:56
Can you elaborate a bit on this please?
– Muhammad Adeel Zahid
Sep 17 '16 at 9:01
Sure. If the tosses are independent, the answer is $frac 18$. However, if they are dependent the answer will depend on the dependence. I gave one form of the dependence which made the answer $0$. On the other side, if we assume that all tosses are the same then the answer is $frac 12$. That's the maximum.
– lulu
Sep 17 '16 at 10:03