Mixing
Urn problem involving two players: show that both players have the same chance to win
$begingroup$
There are two players and a urn with two kind of balls: red and blue. The first player removes two balls with reposition. If he gets the pair $(b,r)$, he wins. If he gets $(r,b)$, the second player wins. Otherwise, two balls are removed again. Prove both players have the same probability to win, which equals one half.
MY ATTEMPT
The sample space corresponding to such random experiment is given by $Omega = {(b,r),(r,b)}$, which has uniform distribution. Thus the probability to draw a blue ball and a red ball in this sequence is given by $p = 0.5$. The same applies to the other case. Hence the probability that the first player wins is given by the geometric distribution. Precisely speaking, we have
begin{align*}
textbf{P}(P_{1}) = p + p(1-p)^{2} + p(1-p)^{4} + ldots = p + p^{3} + p^{5} + ldots = frac{p}{1-p^{2}}
end{align*}
Analogously, we have
begin{align*}
textbf{P}(P_{2}) = (1-p)p + (1-p)^{3}p + (1-p)^{5}p + ldots = p^{2} + p^{4} + p^{6} + ldots = frac{p^{2}}{1-p^{2}}
end{align*}
However $textbf{P}(P_{1}) neq textbf{P}(P_{2})$. Could someone tell me what went wrong? Thanks in advance.
probability probability-theory proof-verification probability-distributions
asked Jan 25 at 2:55
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
There are two players and a urn with two kind of balls: red and blue. The first player removes two balls with reposition. If he gets the pair $(b,r)$, he wins. If he gets $(r,b)$, the second player wins. Otherwise, two balls are removed again. Prove both players have the same probability to win, which equals one half.
MY ATTEMPT
The sample space corresponding to such random experiment is given by $Omega = {(b,r),(r,b)}$, which has uniform distribution. Thus the probability to draw a blue ball and a red ball in this sequence is given by $p = 0.5$. The same applies to the other case. Hence the probability that the first player wins is given by the geometric distribution. Precisely speaking, we have
begin{align*}
textbf{P}(P_{1}) = p + p(1-p)^{2} + p(1-p)^{4} + ldots = p + p^{3} + p^{5} + ldots = frac{p}{1-p^{2}}
end{align*}
Analogously, we have
begin{align*}
textbf{P}(P_{2}) = (1-p)p + (1-p)^{3}p + (1-p)^{5}p + ldots = p^{2} + p^{4} + p^{6} + ldots = frac{p^{2}}{1-p^{2}}
end{align*}
However $textbf{P}(P_{1}) neq textbf{P}(P_{2})$. Could someone tell me what went wrong? Thanks in advance.
probability probability-theory proof-verification probability-distributions
asked Jan 25 at 2:55
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
There are two players and a urn with two kind of balls: red and blue. The first player removes two balls with reposition. If he gets the pair $(b,r)$, he wins. If he gets $(r,b)$, the second player wins. Otherwise, two balls are removed again. Prove both players have the same probability to win, which equals one half.
MY ATTEMPT
The sample space corresponding to such random experiment is given by $Omega = {(b,r),(r,b)}$, which has uniform distribution. Thus the probability to draw a blue ball and a red ball in this sequence is given by $p = 0.5$. The same applies to the other case. Hence the probability that the first player wins is given by the geometric distribution. Precisely speaking, we have
begin{align*}
textbf{P}(P_{1}) = p + p(1-p)^{2} + p(1-p)^{4} + ldots = p + p^{3} + p^{5} + ldots = frac{p}{1-p^{2}}
end{align*}
Analogously, we have
begin{align*}
textbf{P}(P_{2}) = (1-p)p + (1-p)^{3}p + (1-p)^{5}p + ldots = p^{2} + p^{4} + p^{6} + ldots = frac{p^{2}}{1-p^{2}}
end{align*}
However $textbf{P}(P_{1}) neq textbf{P}(P_{2})$. Could someone tell me what went wrong? Thanks in advance.
probability probability-theory proof-verification probability-distributions
asked Jan 25 at 2:55
user1337user1337
46110
$endgroup$
There are two players and a urn with two kind of balls: red and blue. The first player removes two balls with reposition. If he gets the pair $(b,r)$, he wins. If he gets $(r,b)$, the second player wins. Otherwise, two balls are removed again. Prove both players have the same probability to win, which equals one half.
MY ATTEMPT
The sample space corresponding to such random experiment is given by $Omega = {(b,r),(r,b)}$, which has uniform distribution. Thus the probability to draw a blue ball and a red ball in this sequence is given by $p = 0.5$. The same applies to the other case. Hence the probability that the first player wins is given by the geometric distribution. Precisely speaking, we have
begin{align*}
textbf{P}(P_{1}) = p + p(1-p)^{2} + p(1-p)^{4} + ldots = p + p^{3} + p^{5} + ldots = frac{p}{1-p^{2}}
end{align*}
Analogously, we have
begin{align*}
textbf{P}(P_{2}) = (1-p)p + (1-p)^{3}p + (1-p)^{5}p + ldots = p^{2} + p^{4} + p^{6} + ldots = frac{p^{2}}{1-p^{2}}
end{align*}
However $textbf{P}(P_{1}) neq textbf{P}(P_{2})$. Could someone tell me what went wrong? Thanks in advance.
probability probability-theory proof-verification probability-distributions
probability probability-theory proof-verification probability-distributions
asked Jan 25 at 2:55
user1337user1337
46110
asked Jan 25 at 2:55
user1337user1337
46110
asked Jan 25 at 2:55
user1337user1337
46110
asked Jan 25 at 2:55
user1337user1337
46110
asked Jan 25 at 2:55
user1337user1337
46110
46110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A few things are very obviously wrong.
The first wrong thing is that you seem to have neglected the fact that if Player $1$ draws $(r,b)$ as the first pair of balls, Player $2$ wins instantly. We do not have to draw again to give Player $2$ a chance to win.
The second wrong thing is that you have set up the sample space so that the only two events in it are events that end the game ($(b,r)$ or $(r,b)$), but then you somehow have got the idea that you need a geometric series. What for? The reason you would need a geometric series is if there was a way for the game to continue after observing an event from your (initial) sample space.
A third (and fourth?) wrong thing is you summed each of your series incorrectly, though that's kind of irrelevant since they were the wrong series anyway.
(Also, technically, you need a sample space with an infinite number of elements in order to make a sound probability argument for an infinite geometric series, but let's leave that aside for now.)
So it's hard to explain what you should have done, because you have the beginnings of two possible approaches but have put forth things that contradict each of them.
One approach is to put the other possible two-ball combinations into the (initial) sample space so that you have at least three outcomes for the first pair of balls
and you have a non-zero probability to draw again.
Then the geometric series becomes appropriate.
But the series is actually
$$ mathbf P(P_1) = mathbf P(P_2) = p + (1-2p)p + (1-2p)^2p + (1-2p)^3p + cdots = frac12 quadtext{($p leq frac12$)}.$$
Another approach is that your sample space is simply the last pair of balls drawn.
Then indeed it is $Omega = {(b,r),(r,b)}$.
But then there is no geometric series, because this is the end of the game already.
Either Player $1$ wins or Player $2$ wins due to this pair of balls, no third alternative.
Since $Omega = {(b,r),(r,b)}$ is not a typical sample space for drawing two balls from an urn, however, I would actually take the trouble to explain why the distribution is uniform on it
(unless I had just done another problem with the same sample space recently, in which case I would refer to the earlier finding).
answered Jan 25 at 3:30
David KDavid K
55.2k344120
$endgroup$
$begingroup$
Thanks for descriptive answer, David. It helped a lot.
$endgroup$
– user1337
Jan 25 at 13:02
add a comment |
$begingroup$
After a sequence of rolling $(r,r)$ or $(b,b)$ the game terminates on the first occurrence of $(r,b)$ or $(b,r)$. If the game terminates on an odd count, player one wins if the ball is $(b,r)$ otherwise player two does. When the game terminates on an even count, then player two wins if the ball is $(b,r)$ otherwise player one does.
On any draw, let $P$ be the probability for drawing $(b,r)$. It is also the probability for drawing $(r,b)$. The probability for drawing same coloured balls is $(1-2P)$.
Player one wins if (i) after an even number of draws with same coloured balls $(b,r)$ is drawn, or (ii) after an odd number of draws with same colours, $(r,b)$ is finally drawn.$$begin{split}& Psum_{k=0}^infty (1-2P)^{2k}+Psum_{k=0}^infty(1-2P)^{2k+1}\[1ex] =~& P(2-2P)sum_{k=0}^infty (1-2P)^{2k}\[1ex]=~&dfrac{2P(1-P)}{1-(1-2P)^2}\vdots\[2ex]=~&dfrac 12end{split}$$
answered Jan 25 at 4:36
Graham KempGraham Kemp
86.8k43579
$endgroup$
$begingroup$
Thanks for the contribution, Graham. I see how to proceed now :)
$endgroup$
– user1337
Jan 25 at 13:04
add a comment |
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2 Answers
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2 Answers
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$begingroup$
A few things are very obviously wrong.
The first wrong thing is that you seem to have neglected the fact that if Player $1$ draws $(r,b)$ as the first pair of balls, Player $2$ wins instantly. We do not have to draw again to give Player $2$ a chance to win.
The second wrong thing is that you have set up the sample space so that the only two events in it are events that end the game ($(b,r)$ or $(r,b)$), but then you somehow have got the idea that you need a geometric series. What for? The reason you would need a geometric series is if there was a way for the game to continue after observing an event from your (initial) sample space.
A third (and fourth?) wrong thing is you summed each of your series incorrectly, though that's kind of irrelevant since they were the wrong series anyway.
(Also, technically, you need a sample space with an infinite number of elements in order to make a sound probability argument for an infinite geometric series, but let's leave that aside for now.)
So it's hard to explain what you should have done, because you have the beginnings of two possible approaches but have put forth things that contradict each of them.
One approach is to put the other possible two-ball combinations into the (initial) sample space so that you have at least three outcomes for the first pair of balls
and you have a non-zero probability to draw again.
Then the geometric series becomes appropriate.
But the series is actually
$$ mathbf P(P_1) = mathbf P(P_2) = p + (1-2p)p + (1-2p)^2p + (1-2p)^3p + cdots = frac12 quadtext{($p leq frac12$)}.$$
Another approach is that your sample space is simply the last pair of balls drawn.
Then indeed it is $Omega = {(b,r),(r,b)}$.
But then there is no geometric series, because this is the end of the game already.
Either Player $1$ wins or Player $2$ wins due to this pair of balls, no third alternative.
Since $Omega = {(b,r),(r,b)}$ is not a typical sample space for drawing two balls from an urn, however, I would actually take the trouble to explain why the distribution is uniform on it
(unless I had just done another problem with the same sample space recently, in which case I would refer to the earlier finding).
answered Jan 25 at 3:30
David KDavid K
55.2k344120
$endgroup$
$begingroup$
Thanks for descriptive answer, David. It helped a lot.
$endgroup$
– user1337
Jan 25 at 13:02
add a comment |
$begingroup$
A few things are very obviously wrong.
The first wrong thing is that you seem to have neglected the fact that if Player $1$ draws $(r,b)$ as the first pair of balls, Player $2$ wins instantly. We do not have to draw again to give Player $2$ a chance to win.
The second wrong thing is that you have set up the sample space so that the only two events in it are events that end the game ($(b,r)$ or $(r,b)$), but then you somehow have got the idea that you need a geometric series. What for? The reason you would need a geometric series is if there was a way for the game to continue after observing an event from your (initial) sample space.
A third (and fourth?) wrong thing is you summed each of your series incorrectly, though that's kind of irrelevant since they were the wrong series anyway.
(Also, technically, you need a sample space with an infinite number of elements in order to make a sound probability argument for an infinite geometric series, but let's leave that aside for now.)
So it's hard to explain what you should have done, because you have the beginnings of two possible approaches but have put forth things that contradict each of them.
One approach is to put the other possible two-ball combinations into the (initial) sample space so that you have at least three outcomes for the first pair of balls
and you have a non-zero probability to draw again.
Then the geometric series becomes appropriate.
But the series is actually
$$ mathbf P(P_1) = mathbf P(P_2) = p + (1-2p)p + (1-2p)^2p + (1-2p)^3p + cdots = frac12 quadtext{($p leq frac12$)}.$$
Another approach is that your sample space is simply the last pair of balls drawn.
Then indeed it is $Omega = {(b,r),(r,b)}$.
But then there is no geometric series, because this is the end of the game already.
Either Player $1$ wins or Player $2$ wins due to this pair of balls, no third alternative.
Since $Omega = {(b,r),(r,b)}$ is not a typical sample space for drawing two balls from an urn, however, I would actually take the trouble to explain why the distribution is uniform on it
(unless I had just done another problem with the same sample space recently, in which case I would refer to the earlier finding).
answered Jan 25 at 3:30
David KDavid K
55.2k344120
$endgroup$
$begingroup$
Thanks for descriptive answer, David. It helped a lot.
$endgroup$
– user1337
Jan 25 at 13:02
add a comment |
$begingroup$
A few things are very obviously wrong.
The first wrong thing is that you seem to have neglected the fact that if Player $1$ draws $(r,b)$ as the first pair of balls, Player $2$ wins instantly. We do not have to draw again to give Player $2$ a chance to win.
The second wrong thing is that you have set up the sample space so that the only two events in it are events that end the game ($(b,r)$ or $(r,b)$), but then you somehow have got the idea that you need a geometric series. What for? The reason you would need a geometric series is if there was a way for the game to continue after observing an event from your (initial) sample space.
A third (and fourth?) wrong thing is you summed each of your series incorrectly, though that's kind of irrelevant since they were the wrong series anyway.
(Also, technically, you need a sample space with an infinite number of elements in order to make a sound probability argument for an infinite geometric series, but let's leave that aside for now.)
So it's hard to explain what you should have done, because you have the beginnings of two possible approaches but have put forth things that contradict each of them.
One approach is to put the other possible two-ball combinations into the (initial) sample space so that you have at least three outcomes for the first pair of balls
and you have a non-zero probability to draw again.
Then the geometric series becomes appropriate.
But the series is actually
$$ mathbf P(P_1) = mathbf P(P_2) = p + (1-2p)p + (1-2p)^2p + (1-2p)^3p + cdots = frac12 quadtext{($p leq frac12$)}.$$
Another approach is that your sample space is simply the last pair of balls drawn.
Then indeed it is $Omega = {(b,r),(r,b)}$.
But then there is no geometric series, because this is the end of the game already.
Either Player $1$ wins or Player $2$ wins due to this pair of balls, no third alternative.
Since $Omega = {(b,r),(r,b)}$ is not a typical sample space for drawing two balls from an urn, however, I would actually take the trouble to explain why the distribution is uniform on it
(unless I had just done another problem with the same sample space recently, in which case I would refer to the earlier finding).
answered Jan 25 at 3:30
David KDavid K
55.2k344120
$endgroup$
A few things are very obviously wrong.
The first wrong thing is that you seem to have neglected the fact that if Player $1$ draws $(r,b)$ as the first pair of balls, Player $2$ wins instantly. We do not have to draw again to give Player $2$ a chance to win.
The second wrong thing is that you have set up the sample space so that the only two events in it are events that end the game ($(b,r)$ or $(r,b)$), but then you somehow have got the idea that you need a geometric series. What for? The reason you would need a geometric series is if there was a way for the game to continue after observing an event from your (initial) sample space.
A third (and fourth?) wrong thing is you summed each of your series incorrectly, though that's kind of irrelevant since they were the wrong series anyway.
(Also, technically, you need a sample space with an infinite number of elements in order to make a sound probability argument for an infinite geometric series, but let's leave that aside for now.)
So it's hard to explain what you should have done, because you have the beginnings of two possible approaches but have put forth things that contradict each of them.
One approach is to put the other possible two-ball combinations into the (initial) sample space so that you have at least three outcomes for the first pair of balls
and you have a non-zero probability to draw again.
Then the geometric series becomes appropriate.
But the series is actually
$$ mathbf P(P_1) = mathbf P(P_2) = p + (1-2p)p + (1-2p)^2p + (1-2p)^3p + cdots = frac12 quadtext{($p leq frac12$)}.$$
Another approach is that your sample space is simply the last pair of balls drawn.
Then indeed it is $Omega = {(b,r),(r,b)}$.
But then there is no geometric series, because this is the end of the game already.
Either Player $1$ wins or Player $2$ wins due to this pair of balls, no third alternative.
Since $Omega = {(b,r),(r,b)}$ is not a typical sample space for drawing two balls from an urn, however, I would actually take the trouble to explain why the distribution is uniform on it
(unless I had just done another problem with the same sample space recently, in which case I would refer to the earlier finding).
answered Jan 25 at 3:30
David KDavid K
55.2k344120
edited Jan 25 at 3:42
answered Jan 25 at 3:30
David KDavid K
55.2k344120
answered Jan 25 at 3:30
David KDavid K
55.2k344120
answered Jan 25 at 3:30
David KDavid K
55.2k344120
55.2k344120
$begingroup$
Thanks for descriptive answer, David. It helped a lot.
$endgroup$
– user1337
Jan 25 at 13:02
add a comment |
$begingroup$
Thanks for descriptive answer, David. It helped a lot.
$endgroup$
– user1337
Jan 25 at 13:02
$begingroup$
Thanks for descriptive answer, David. It helped a lot.
$endgroup$
– user1337
Jan 25 at 13:02
$begingroup$
Thanks for descriptive answer, David. It helped a lot.
$endgroup$
– user1337
Jan 25 at 13:02
add a comment |
$begingroup$
After a sequence of rolling $(r,r)$ or $(b,b)$ the game terminates on the first occurrence of $(r,b)$ or $(b,r)$. If the game terminates on an odd count, player one wins if the ball is $(b,r)$ otherwise player two does. When the game terminates on an even count, then player two wins if the ball is $(b,r)$ otherwise player one does.
On any draw, let $P$ be the probability for drawing $(b,r)$. It is also the probability for drawing $(r,b)$. The probability for drawing same coloured balls is $(1-2P)$.
Player one wins if (i) after an even number of draws with same coloured balls $(b,r)$ is drawn, or (ii) after an odd number of draws with same colours, $(r,b)$ is finally drawn.$$begin{split}& Psum_{k=0}^infty (1-2P)^{2k}+Psum_{k=0}^infty(1-2P)^{2k+1}\[1ex] =~& P(2-2P)sum_{k=0}^infty (1-2P)^{2k}\[1ex]=~&dfrac{2P(1-P)}{1-(1-2P)^2}\vdots\[2ex]=~&dfrac 12end{split}$$
answered Jan 25 at 4:36
Graham KempGraham Kemp
86.8k43579
$endgroup$
$begingroup$
Thanks for the contribution, Graham. I see how to proceed now :)
$endgroup$
– user1337
Jan 25 at 13:04
add a comment |
$begingroup$
After a sequence of rolling $(r,r)$ or $(b,b)$ the game terminates on the first occurrence of $(r,b)$ or $(b,r)$. If the game terminates on an odd count, player one wins if the ball is $(b,r)$ otherwise player two does. When the game terminates on an even count, then player two wins if the ball is $(b,r)$ otherwise player one does.
On any draw, let $P$ be the probability for drawing $(b,r)$. It is also the probability for drawing $(r,b)$. The probability for drawing same coloured balls is $(1-2P)$.
Player one wins if (i) after an even number of draws with same coloured balls $(b,r)$ is drawn, or (ii) after an odd number of draws with same colours, $(r,b)$ is finally drawn.$$begin{split}& Psum_{k=0}^infty (1-2P)^{2k}+Psum_{k=0}^infty(1-2P)^{2k+1}\[1ex] =~& P(2-2P)sum_{k=0}^infty (1-2P)^{2k}\[1ex]=~&dfrac{2P(1-P)}{1-(1-2P)^2}\vdots\[2ex]=~&dfrac 12end{split}$$
answered Jan 25 at 4:36
Graham KempGraham Kemp
86.8k43579
$endgroup$
$begingroup$
Thanks for the contribution, Graham. I see how to proceed now :)
$endgroup$
– user1337
Jan 25 at 13:04
add a comment |
$begingroup$
After a sequence of rolling $(r,r)$ or $(b,b)$ the game terminates on the first occurrence of $(r,b)$ or $(b,r)$. If the game terminates on an odd count, player one wins if the ball is $(b,r)$ otherwise player two does. When the game terminates on an even count, then player two wins if the ball is $(b,r)$ otherwise player one does.
On any draw, let $P$ be the probability for drawing $(b,r)$. It is also the probability for drawing $(r,b)$. The probability for drawing same coloured balls is $(1-2P)$.
Player one wins if (i) after an even number of draws with same coloured balls $(b,r)$ is drawn, or (ii) after an odd number of draws with same colours, $(r,b)$ is finally drawn.$$begin{split}& Psum_{k=0}^infty (1-2P)^{2k}+Psum_{k=0}^infty(1-2P)^{2k+1}\[1ex] =~& P(2-2P)sum_{k=0}^infty (1-2P)^{2k}\[1ex]=~&dfrac{2P(1-P)}{1-(1-2P)^2}\vdots\[2ex]=~&dfrac 12end{split}$$
answered Jan 25 at 4:36
Graham KempGraham Kemp
86.8k43579
$endgroup$
After a sequence of rolling $(r,r)$ or $(b,b)$ the game terminates on the first occurrence of $(r,b)$ or $(b,r)$. If the game terminates on an odd count, player one wins if the ball is $(b,r)$ otherwise player two does. When the game terminates on an even count, then player two wins if the ball is $(b,r)$ otherwise player one does.
On any draw, let $P$ be the probability for drawing $(b,r)$. It is also the probability for drawing $(r,b)$. The probability for drawing same coloured balls is $(1-2P)$.
Player one wins if (i) after an even number of draws with same coloured balls $(b,r)$ is drawn, or (ii) after an odd number of draws with same colours, $(r,b)$ is finally drawn.$$begin{split}& Psum_{k=0}^infty (1-2P)^{2k}+Psum_{k=0}^infty(1-2P)^{2k+1}\[1ex] =~& P(2-2P)sum_{k=0}^infty (1-2P)^{2k}\[1ex]=~&dfrac{2P(1-P)}{1-(1-2P)^2}\vdots\[2ex]=~&dfrac 12end{split}$$
answered Jan 25 at 4:36
Graham KempGraham Kemp
86.8k43579
answered Jan 25 at 4:36
Graham KempGraham Kemp
86.8k43579
answered Jan 25 at 4:36
Graham KempGraham Kemp
86.8k43579
answered Jan 25 at 4:36
Graham KempGraham Kemp
86.8k43579
86.8k43579
$begingroup$
Thanks for the contribution, Graham. I see how to proceed now :)
$endgroup$
– user1337
Jan 25 at 13:04
add a comment |
$begingroup$
Thanks for the contribution, Graham. I see how to proceed now :)
$endgroup$
– user1337
Jan 25 at 13:04
$begingroup$
Thanks for the contribution, Graham. I see how to proceed now :)
$endgroup$
– user1337
Jan 25 at 13:04
$begingroup$
Thanks for the contribution, Graham. I see how to proceed now :)
$endgroup$
– user1337
Jan 25 at 13:04
add a comment |
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