Difficult Coin Problem
$begingroup$
Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?
Official answer: 2/3, but I cannot arrive at it.
Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.
When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.
However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.
probability
asked 14 hours ago
Victor S.Victor S.
897
$endgroup$
add a comment |
$begingroup$
Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?
Official answer: 2/3, but I cannot arrive at it.
Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.
When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.
However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.
probability
asked 14 hours ago
Victor S.Victor S.
897
$endgroup$
$begingroup$
What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
$endgroup$
– Pakk
9 hours ago
$begingroup$
I have understood the flaw in my method by understanding the correct answer, but thank you anyways
$endgroup$
– Victor S.
8 hours ago
add a comment |
$begingroup$
Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?
Official answer: 2/3, but I cannot arrive at it.
Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.
When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.
However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.
probability
asked 14 hours ago
Victor S.Victor S.
897
$endgroup$
Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?
Official answer: 2/3, but I cannot arrive at it.
Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.
When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.
However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.
probability
probability
asked 14 hours ago
Victor S.Victor S.
897
asked 14 hours ago
Victor S.Victor S.
897
asked 14 hours ago
Victor S.Victor S.
897
asked 14 hours ago
Victor S.Victor S.
897
asked 14 hours ago
Victor S.Victor S.
897
897
$begingroup$
What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
$endgroup$
– Pakk
9 hours ago
$begingroup$
I have understood the flaw in my method by understanding the correct answer, but thank you anyways
$endgroup$
– Victor S.
8 hours ago
add a comment |
$begingroup$
What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
$endgroup$
– Pakk
9 hours ago
$begingroup$
I have understood the flaw in my method by understanding the correct answer, but thank you anyways
$endgroup$
– Victor S.
8 hours ago
$begingroup$
What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
$endgroup$
– Pakk
9 hours ago
$begingroup$
What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
$endgroup$
– Pakk
9 hours ago
$begingroup$
I have understood the flaw in my method by understanding the correct answer, but thank you anyways
$endgroup$
– Victor S.
8 hours ago
$begingroup$
I have understood the flaw in my method by understanding the correct answer, but thank you anyways
$endgroup$
– Victor S.
8 hours ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
$$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$
where $p$ is probability that head comes in one toss and $q=1-p$.
answered 14 hours ago
greedoidgreedoid
41.8k1152105
$endgroup$
add a comment |
$begingroup$
Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$
answered 14 hours ago
saulspatzsaulspatz
15k21330
$endgroup$
add a comment |
$begingroup$
Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
$$p = .5+ .25p$$
which yields $p=2/3$.
answered 14 hours ago
pwerthpwerth
3,233417
$endgroup$
add a comment |
$begingroup$
Here's another approach.
Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.
In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).
But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.
This gives A a $frac23$ chance of winning to $frac13$ for B.
answered 13 hours ago
paw88789paw88789
29.1k12349
$endgroup$
add a comment |
$begingroup$
P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)
A loses first round, B loses second round, A wins the third round (0.5)^3
And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2
P(A winning) = 0.5/(1-0.25)
answered 14 hours ago
Shreyas Shreyas
812
New contributor
Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
$$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$
where $p$ is probability that head comes in one toss and $q=1-p$.
answered 14 hours ago
greedoidgreedoid
41.8k1152105
$endgroup$
add a comment |
$begingroup$
So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
$$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$
where $p$ is probability that head comes in one toss and $q=1-p$.
answered 14 hours ago
greedoidgreedoid
41.8k1152105
$endgroup$
add a comment |
$begingroup$
So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
$$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$
where $p$ is probability that head comes in one toss and $q=1-p$.
answered 14 hours ago
greedoidgreedoid
41.8k1152105
$endgroup$
So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
$$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$
where $p$ is probability that head comes in one toss and $q=1-p$.
answered 14 hours ago
greedoidgreedoid
41.8k1152105
answered 14 hours ago
greedoidgreedoid
41.8k1152105
answered 14 hours ago
greedoidgreedoid
41.8k1152105
answered 14 hours ago
greedoidgreedoid
41.8k1152105
41.8k1152105
add a comment |
add a comment |
$begingroup$
Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$
answered 14 hours ago
saulspatzsaulspatz
15k21330
$endgroup$
add a comment |
$begingroup$
Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$
answered 14 hours ago
saulspatzsaulspatz
15k21330
$endgroup$
add a comment |
$begingroup$
Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$
answered 14 hours ago
saulspatzsaulspatz
15k21330
$endgroup$
Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$
answered 14 hours ago
saulspatzsaulspatz
15k21330
answered 14 hours ago
saulspatzsaulspatz
15k21330
answered 14 hours ago
saulspatzsaulspatz
15k21330
answered 14 hours ago
saulspatzsaulspatz
15k21330
15k21330
add a comment |
add a comment |
$begingroup$
Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
$$p = .5+ .25p$$
which yields $p=2/3$.
answered 14 hours ago
pwerthpwerth
3,233417
$endgroup$
add a comment |
$begingroup$
Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
$$p = .5+ .25p$$
which yields $p=2/3$.
answered 14 hours ago
pwerthpwerth
3,233417
$endgroup$
add a comment |
$begingroup$
Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
$$p = .5+ .25p$$
which yields $p=2/3$.
answered 14 hours ago
pwerthpwerth
3,233417
$endgroup$
Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
$$p = .5+ .25p$$
which yields $p=2/3$.
answered 14 hours ago
pwerthpwerth
3,233417
answered 14 hours ago
pwerthpwerth
3,233417
answered 14 hours ago
pwerthpwerth
3,233417
answered 14 hours ago
pwerthpwerth
3,233417
3,233417
add a comment |
add a comment |
$begingroup$
Here's another approach.
Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.
In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).
But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.
This gives A a $frac23$ chance of winning to $frac13$ for B.
answered 13 hours ago
paw88789paw88789
29.1k12349
$endgroup$
add a comment |
$begingroup$
Here's another approach.
Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.
In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).
But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.
This gives A a $frac23$ chance of winning to $frac13$ for B.
answered 13 hours ago
paw88789paw88789
29.1k12349
$endgroup$
add a comment |
$begingroup$
Here's another approach.
Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.
In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).
But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.
This gives A a $frac23$ chance of winning to $frac13$ for B.
answered 13 hours ago
paw88789paw88789
29.1k12349
$endgroup$
Here's another approach.
Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.
In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).
But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.
This gives A a $frac23$ chance of winning to $frac13$ for B.
answered 13 hours ago
paw88789paw88789
29.1k12349
answered 13 hours ago
paw88789paw88789
29.1k12349
answered 13 hours ago
paw88789paw88789
29.1k12349
answered 13 hours ago
paw88789paw88789
29.1k12349
29.1k12349
add a comment |
add a comment |
$begingroup$
P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)
A loses first round, B loses second round, A wins the third round (0.5)^3
And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2
P(A winning) = 0.5/(1-0.25)
answered 14 hours ago
Shreyas Shreyas
812
New contributor
Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)
A loses first round, B loses second round, A wins the third round (0.5)^3
And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2
P(A winning) = 0.5/(1-0.25)
answered 14 hours ago
Shreyas Shreyas
812
New contributor
Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)
A loses first round, B loses second round, A wins the third round (0.5)^3
And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2
P(A winning) = 0.5/(1-0.25)
answered 14 hours ago
Shreyas Shreyas
812
New contributor
Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)
A loses first round, B loses second round, A wins the third round (0.5)^3
And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2
P(A winning) = 0.5/(1-0.25)
answered 14 hours ago
Shreyas Shreyas
812
New contributor
Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 14 hours ago
Shreyas Shreyas
812
New contributor
Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 14 hours ago
Shreyas Shreyas
812
answered 14 hours ago
Shreyas Shreyas
812
812
New contributor
Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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$begingroup$
What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
$endgroup$
– Pakk
9 hours ago
$begingroup$
I have understood the flaw in my method by understanding the correct answer, but thank you anyways
$endgroup$
– Victor S.
8 hours ago