Mixing
Probability of extracting a ball after two balls were swapped
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We have $2$ boxes, the first one contains $10$ white balls and $11$ black balls. The second box contain $12$ white balls and $13$ black balls.
We swap two balls between the boxes then we extract a ball from the first box. What is the probability that the ball is white?
A swap consists in taking a ball from the first box and put it in the second one, then taking a ball from the second box and put it in the first one.
$$P(text{white ball from box1})=frac{10}{21}$$
$$P(text{black ball from box1})=frac{11}{21}$$
$$P(text{white ball from box2})= frac{12+1}{26}frac{10}{21}+frac{12}{26}frac{11}{21}=frac{13cdot10+12cdot11}{26cdot 21}$$
$$P(text{black ball from box2})= frac{13}{26}frac{10}{21}+frac{13+1}{26}frac{11}{21}=frac{13cdot 10+14cdot 11}{26cdot 21}$$
$$P(text{white ball from box1 after swap})=frac{10+1}{22}cdotfrac{13cdot10+12cdot11}{26cdot 21}+frac{10}{22}cdotfrac{13cdot 10+14cdot 11}{26cdot 21}$$
$$=frac{2882}{12012}+frac{2840}{12012}=frac{5722}{12012}=0.4763$$
Have I done it correctly? I don't know how to verify myself with this kind of problems.
Also, is there a trick to find the probability if there were two swaps?
probability proof-verification
$endgroup$
|
show 2 more comments
$begingroup$
We have $2$ boxes, the first one contains $10$ white balls and $11$ black balls. The second box contain $12$ white balls and $13$ black balls.
We swap two balls between the boxes then we extract a ball from the first box. What is the probability that the ball is white?
A swap consists in taking a ball from the first box and put it in the second one, then taking a ball from the second box and put it in the first one.
$$P(text{white ball from box1})=frac{10}{21}$$
$$P(text{black ball from box1})=frac{11}{21}$$
$$P(text{white ball from box2})= frac{12+1}{26}frac{10}{21}+frac{12}{26}frac{11}{21}=frac{13cdot10+12cdot11}{26cdot 21}$$
$$P(text{black ball from box2})= frac{13}{26}frac{10}{21}+frac{13+1}{26}frac{11}{21}=frac{13cdot 10+14cdot 11}{26cdot 21}$$
$$P(text{white ball from box1 after swap})=frac{10+1}{22}cdotfrac{13cdot10+12cdot11}{26cdot 21}+frac{10}{22}cdotfrac{13cdot 10+14cdot 11}{26cdot 21}$$
$$=frac{2882}{12012}+frac{2840}{12012}=frac{5722}{12012}=0.4763$$
Have I done it correctly? I don't know how to verify myself with this kind of problems.
Also, is there a trick to find the probability if there were two swaps?
probability proof-verification
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$begingroup$
can u simplify to give a final answer
$endgroup$
– K Split X
Jan 30 at 18:56
1
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Your denominators should be $26cdot21$, not $26cdot11$
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– Daniel Mathias
Jan 30 at 19:01
1
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Clearly the answer needs to be slightly less than $frac 12$
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– Ross Millikan
Jan 30 at 19:04
2
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Are you sure that's what "swap" means? I'd have thought that you extract one ball from each and put them back in the other bin (the difference being that your way you could swap the same ball twice).
$endgroup$
– lulu
Jan 30 at 19:06
1
$begingroup$
What do $10+1over22$ and $10over22$ represent in the final calculation? After the swap, there are $21$ balls in box 1, and there are three possible distributions of color: $10$ white and $11$ black (if the balls moved were the same color), $9$ white and $12$ black (if a white ball was moved from box 1 and a black ball was moved back), or $11$ white and $10$ black (if a black ball was moved from box 1 and a white ball was moved back). You need to find the probabilities ($p_1,p_2,p_3$) of my parenthetical events and find $p_1{10over21}+p_2{9over21}+p_3{11over21}$ (like jvdhooft did)
$endgroup$
– Steve Kass
Feb 1 at 3:40
|
show 2 more comments
$begingroup$
We have $2$ boxes, the first one contains $10$ white balls and $11$ black balls. The second box contain $12$ white balls and $13$ black balls.
We swap two balls between the boxes then we extract a ball from the first box. What is the probability that the ball is white?
A swap consists in taking a ball from the first box and put it in the second one, then taking a ball from the second box and put it in the first one.
$$P(text{white ball from box1})=frac{10}{21}$$
$$P(text{black ball from box1})=frac{11}{21}$$
$$P(text{white ball from box2})= frac{12+1}{26}frac{10}{21}+frac{12}{26}frac{11}{21}=frac{13cdot10+12cdot11}{26cdot 21}$$
$$P(text{black ball from box2})= frac{13}{26}frac{10}{21}+frac{13+1}{26}frac{11}{21}=frac{13cdot 10+14cdot 11}{26cdot 21}$$
$$P(text{white ball from box1 after swap})=frac{10+1}{22}cdotfrac{13cdot10+12cdot11}{26cdot 21}+frac{10}{22}cdotfrac{13cdot 10+14cdot 11}{26cdot 21}$$
$$=frac{2882}{12012}+frac{2840}{12012}=frac{5722}{12012}=0.4763$$
Have I done it correctly? I don't know how to verify myself with this kind of problems.
Also, is there a trick to find the probability if there were two swaps?
probability proof-verification
$endgroup$
We have $2$ boxes, the first one contains $10$ white balls and $11$ black balls. The second box contain $12$ white balls and $13$ black balls.
We swap two balls between the boxes then we extract a ball from the first box. What is the probability that the ball is white?
A swap consists in taking a ball from the first box and put it in the second one, then taking a ball from the second box and put it in the first one.
$$P(text{white ball from box1})=frac{10}{21}$$
$$P(text{black ball from box1})=frac{11}{21}$$
$$P(text{white ball from box2})= frac{12+1}{26}frac{10}{21}+frac{12}{26}frac{11}{21}=frac{13cdot10+12cdot11}{26cdot 21}$$
$$P(text{black ball from box2})= frac{13}{26}frac{10}{21}+frac{13+1}{26}frac{11}{21}=frac{13cdot 10+14cdot 11}{26cdot 21}$$
$$P(text{white ball from box1 after swap})=frac{10+1}{22}cdotfrac{13cdot10+12cdot11}{26cdot 21}+frac{10}{22}cdotfrac{13cdot 10+14cdot 11}{26cdot 21}$$
$$=frac{2882}{12012}+frac{2840}{12012}=frac{5722}{12012}=0.4763$$
Have I done it correctly? I don't know how to verify myself with this kind of problems.
Also, is there a trick to find the probability if there were two swaps?
probability proof-verification
probability proof-verification
edited Jan 30 at 19:37
asked Jan 30 at 18:54
user625055
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can u simplify to give a final answer
$endgroup$
– K Split X
Jan 30 at 18:56
1
$begingroup$
Your denominators should be $26cdot21$, not $26cdot11$
$endgroup$
– Daniel Mathias
Jan 30 at 19:01
1
$begingroup$
Clearly the answer needs to be slightly less than $frac 12$
$endgroup$
– Ross Millikan
Jan 30 at 19:04
2
$begingroup$
Are you sure that's what "swap" means? I'd have thought that you extract one ball from each and put them back in the other bin (the difference being that your way you could swap the same ball twice).
$endgroup$
– lulu
Jan 30 at 19:06
1
$begingroup$
What do $10+1over22$ and $10over22$ represent in the final calculation? After the swap, there are $21$ balls in box 1, and there are three possible distributions of color: $10$ white and $11$ black (if the balls moved were the same color), $9$ white and $12$ black (if a white ball was moved from box 1 and a black ball was moved back), or $11$ white and $10$ black (if a black ball was moved from box 1 and a white ball was moved back). You need to find the probabilities ($p_1,p_2,p_3$) of my parenthetical events and find $p_1{10over21}+p_2{9over21}+p_3{11over21}$ (like jvdhooft did)
$endgroup$
– Steve Kass
Feb 1 at 3:40
|
show 2 more comments
$begingroup$
can u simplify to give a final answer
$endgroup$
– K Split X
Jan 30 at 18:56
1
$begingroup$
Your denominators should be $26cdot21$, not $26cdot11$
$endgroup$
– Daniel Mathias
Jan 30 at 19:01
1
$begingroup$
Clearly the answer needs to be slightly less than $frac 12$
$endgroup$
– Ross Millikan
Jan 30 at 19:04
2
$begingroup$
Are you sure that's what "swap" means? I'd have thought that you extract one ball from each and put them back in the other bin (the difference being that your way you could swap the same ball twice).
$endgroup$
– lulu
Jan 30 at 19:06
1
$begingroup$
What do $10+1over22$ and $10over22$ represent in the final calculation? After the swap, there are $21$ balls in box 1, and there are three possible distributions of color: $10$ white and $11$ black (if the balls moved were the same color), $9$ white and $12$ black (if a white ball was moved from box 1 and a black ball was moved back), or $11$ white and $10$ black (if a black ball was moved from box 1 and a white ball was moved back). You need to find the probabilities ($p_1,p_2,p_3$) of my parenthetical events and find $p_1{10over21}+p_2{9over21}+p_3{11over21}$ (like jvdhooft did)
$endgroup$
– Steve Kass
Feb 1 at 3:40
$begingroup$
can u simplify to give a final answer
$endgroup$
– K Split X
Jan 30 at 18:56
$begingroup$
can u simplify to give a final answer
$endgroup$
– K Split X
Jan 30 at 18:56
1
1
$begingroup$
Your denominators should be $26cdot21$, not $26cdot11$
$endgroup$
– Daniel Mathias
Jan 30 at 19:01
$begingroup$
Your denominators should be $26cdot21$, not $26cdot11$
$endgroup$
– Daniel Mathias
Jan 30 at 19:01
1
1
$begingroup$
Clearly the answer needs to be slightly less than $frac 12$
$endgroup$
– Ross Millikan
Jan 30 at 19:04
$begingroup$
Clearly the answer needs to be slightly less than $frac 12$
$endgroup$
– Ross Millikan
Jan 30 at 19:04
2
2
$begingroup$
Are you sure that's what "swap" means? I'd have thought that you extract one ball from each and put them back in the other bin (the difference being that your way you could swap the same ball twice).
$endgroup$
– lulu
Jan 30 at 19:06
$begingroup$
Are you sure that's what "swap" means? I'd have thought that you extract one ball from each and put them back in the other bin (the difference being that your way you could swap the same ball twice).
$endgroup$
– lulu
Jan 30 at 19:06
1
1
$begingroup$
What do $10+1over22$ and $10over22$ represent in the final calculation? After the swap, there are $21$ balls in box 1, and there are three possible distributions of color: $10$ white and $11$ black (if the balls moved were the same color), $9$ white and $12$ black (if a white ball was moved from box 1 and a black ball was moved back), or $11$ white and $10$ black (if a black ball was moved from box 1 and a white ball was moved back). You need to find the probabilities ($p_1,p_2,p_3$) of my parenthetical events and find $p_1{10over21}+p_2{9over21}+p_3{11over21}$ (like jvdhooft did)
$endgroup$
– Steve Kass
Feb 1 at 3:40
$begingroup$
What do $10+1over22$ and $10over22$ represent in the final calculation? After the swap, there are $21$ balls in box 1, and there are three possible distributions of color: $10$ white and $11$ black (if the balls moved were the same color), $9$ white and $12$ black (if a white ball was moved from box 1 and a black ball was moved back), or $11$ white and $10$ black (if a black ball was moved from box 1 and a white ball was moved back). You need to find the probabilities ($p_1,p_2,p_3$) of my parenthetical events and find $p_1{10over21}+p_2{9over21}+p_3{11over21}$ (like jvdhooft did)
$endgroup$
– Steve Kass
Feb 1 at 3:40
|
show 2 more comments
4 Answers
4
active
oldest
votes
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Here’s another approach. [Added: Also, see more on this approach here]
Equivalently, we can do the following.
Reach into the first box and write “S” on one ball (the one to swap).
Now choose a ball from the first box. If it is not the “S” ball, that’s your ball. The probability of this happening is $20over21$, and the ball you choose will be white with probability $10over21$. If you do choose the “S” ball (you do this with probability $1over21$), discard it by throwing it into the second box and then choose a ball at random from the second box (now containing an extra ball), and that’s your ball. If you had to do this, the probability of a white result is ${12+{10over21}over26}$, because if we add a randomly-chosen ball from box 1 to box 2, the number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$ and the number of balls in box 2 increases to $26$. Therefore the total probability you want is
$$p = {20over21}cdot{10over21}+{1over21}cdot{12+{10over21}over26}={2731over5733}approx 0.4763649.$$
P.S. I don’t see an easy way to adapt this approach for two swaps.
answered Jan 30 at 19:40
Steve KassSteve Kass
11.4k11530
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I think this approach is awesome. I still have some little trouble understanding: The number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$.
$endgroup$
– user625055
Jan 30 at 20:09
1
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That’s the most hand-wavy part, but it can be justified. The expected value of the number of white balls in box 2 after a random one from box 1 is is added is $12+{10over21}$. (That’s not hard to show.) Does that help as a very quick attempt to make it seem like a valid approach?
$endgroup$
– Steve Kass
Jan 30 at 23:21
1
$begingroup$
That's a tough question. I just kind of thought it up and thought it was cool. (It was kind of new and cool to me, too.) Of course, I've solved a huge number of probability problems over time, and I know that one worthwhile thing to consider is whether the order in which things are explained to happen is important. Imagine drawing three balls without replacement from an urn and needing the probability that the third one is blue. The urn composition at the third draw depends on the first draws, so you can make it messy, or you can imagine that the third ball was drawn first. HTH
$endgroup$
– Steve Kass
Jan 31 at 21:28
1
$begingroup$
(I'm not sure I completely understand your previous comment with the "Is it okay" question, I'm afraid.) Maybe you can think of a way to ask a new question about this overall strategy? Not sure of the best way to use Stackexchange to continue. I'm definitely not a fan of moving extended discussions to chats or whatever is the accepted thing there.)
$endgroup$
– Steve Kass
Jan 31 at 21:30
1
$begingroup$
Your approach looks right for the way you are reading the problem. You have followed OP's definition of a swap at the end, which seems strange to me. I didn't go through the calculations carefully to find the discrepancy.
$endgroup$
– Ross Millikan
Jan 31 at 23:39
|
show 2 more comments
$begingroup$
The probability of ending up with an additional white ball equals:
$$frac{11}{21} frac{12}{26}$$
The probability of ending up with an additional black ball equals:
$$frac{10}{21} frac{13}{26}$$
We thus find an overall probability of:
$$frac{11}{21} frac{12}{26} frac{11}{21} + frac{10}{21} frac{13}{26} frac{9}{21} + left(1 - frac{11 cdot 12 + 10 cdot 13}{21 cdot 26}right) frac{10}{21} = frac{1452 + 1170 + 2840}{11466} = frac{5462}{11466} approx 0.4764$$
answered Jan 30 at 19:17
jvdhooftjvdhooft
5,65961641
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How did you found the probability of ending with a white ball?
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– user625055
Jan 30 at 19:31
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@Porogami In the first case, we end up with an additional ball, so the probability of getting a white ball equals $frac{11}{21}$. In the second case, we lose a ball, so the probability of getting a white ball equals $frac{9}{21}$. Finally, if we simply swap white for white or black for black, the probability of getting a white ball equals $frac{10}{21}$ as was initially the case.
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– jvdhooft
Jan 30 at 19:35
add a comment |
$begingroup$
Although this is already answered, I'll add my two cents. Perhaps someone will find it more intuitive.
Call probability of picking white ball from boxes $p_1$, and $p_2$. Since we're exchanging one ball, the probability will be between $p_1$ and $p_2$, closer to $p_1$, since we have $21$ balls in box 1.
$ p(W) = frac{20}{21}p_1 + frac1{21}p_2'$
notice we wrote $p_2'$. The dynamics of the problem will change $p_2$ mid process. But we can apply the same logic here, since 25 balls were in box 2 and now 1 more added.
$p_2' = frac{25}{26}p_2 + frac1{26}p_1 $
Combining the two equations will give
$ p(W) = frac{20}{21}p_1 + frac1{21}(frac{25}{26}p_2 + frac1{26}p_1)
$
Here again, the coefficients will tell the story. The final ball will come from box 1 with probability $frac{20}{21}+frac1{21}frac1{26}$, that is either directly or via box 2 detour. Or, with probability $frac1{21}frac{25}{26}$ from the second box. Therefore the final probability is the weighted average of the original probabilities with these weights (which themselves are probabilities).
In general let number of balls in box 1 and 2 be equal to $n$ and $m$ respectively. Define $alpha = frac{m}{n(m+1)}$. Also, let $p_i(s)$ shows the probability of drawing white in box $i$ after $s$ swaps. Where $p_i=p_i(0)$. Then, we can write
$$ p_1(s+1) = (1-alpha) p_1(s) + alpha p_2(s) $$
and
$$ p_2(s+1) = p_2(s) + frac{n}{m}(p_1(s) - p_1(s+1)) $$
second equation is due to conservation of total white balls. Now you can compute probabilities after any number of swaps from this iterative process.
answered Feb 1 at 20:44
karakfakarakfa
2,025811
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Thank you alot! Does it matter so much if the balls from the first urn are fewer than in the second one? // What if we had the same number of balls in both urns?
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– user625055
Feb 1 at 21:42
1
$begingroup$
Well, same number of odds in both boxes will give you the same odds since you're taking the weighted average of the same number. Doesn't matter how many in each box. Perhaps I misunderstood the question, here the first box already has fewer balls.
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– karakfa
Feb 1 at 22:02
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Yes, of course, I just asked about another case. Thanks, I have checked and it matches with Steve answer.
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– user625055
Feb 1 at 22:18
add a comment |
$begingroup$
So... for this, I'll need a lot of names for events that sound similar, so bear with me.
Since you have defined a "swap" to be where a ball is first moved from the first box into the second box, and then a ball is chosen from the second box which now includes the ball just moved and this is moved back to the first box.
Let $b_1$ be the event that a black ball was moved from the first into the second. Let $w_1$ be the event a white ball was moved from the first into the second.
Let $b_2$ be the event that a black ball was moved from the second into the first. Let $w_2$ be the event that a white ball was moved from the second into the first.
Let $W$ be the event that a white ball was selected from the first box after the swap occurred.
We are interested in finding $Pr(W)$
By total probability we have:
$$Pr(W)=Pr(Wmid b_1b_2)Pr(b_2mid b_1)Pr(b_1)+Pr(Wmid b_1w_2)Pr(w_2mid b_1)Pr(b_1)\
+Pr(Wmid w_1b_2)Pr(b_2mid w_1)Pr(w_1)+Pr(Wmid w_1w_2) Pr(w_2mid w_1)Pr(w_1)$$
Each of these terms can be found with relatively little effort, though it is tedious to find so many. I'll do one pair of them for now and leave the rest for you.
$Pr(Wmid b_1b_2) = frac{10}{21}$ and $Pr(b_2mid b_1)=frac{14}{26}$ and $Pr(b_1)=frac{11}{21}$
answered Jan 30 at 19:06
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here’s another approach. [Added: Also, see more on this approach here]
Equivalently, we can do the following.
Reach into the first box and write “S” on one ball (the one to swap).
Now choose a ball from the first box. If it is not the “S” ball, that’s your ball. The probability of this happening is $20over21$, and the ball you choose will be white with probability $10over21$. If you do choose the “S” ball (you do this with probability $1over21$), discard it by throwing it into the second box and then choose a ball at random from the second box (now containing an extra ball), and that’s your ball. If you had to do this, the probability of a white result is ${12+{10over21}over26}$, because if we add a randomly-chosen ball from box 1 to box 2, the number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$ and the number of balls in box 2 increases to $26$. Therefore the total probability you want is
$$p = {20over21}cdot{10over21}+{1over21}cdot{12+{10over21}over26}={2731over5733}approx 0.4763649.$$
P.S. I don’t see an easy way to adapt this approach for two swaps.
answered Jan 30 at 19:40
Steve KassSteve Kass
11.4k11530
$endgroup$
$begingroup$
I think this approach is awesome. I still have some little trouble understanding: The number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$.
$endgroup$
– user625055
Jan 30 at 20:09
1
$begingroup$
That’s the most hand-wavy part, but it can be justified. The expected value of the number of white balls in box 2 after a random one from box 1 is is added is $12+{10over21}$. (That’s not hard to show.) Does that help as a very quick attempt to make it seem like a valid approach?
$endgroup$
– Steve Kass
Jan 30 at 23:21
1
$begingroup$
That's a tough question. I just kind of thought it up and thought it was cool. (It was kind of new and cool to me, too.) Of course, I've solved a huge number of probability problems over time, and I know that one worthwhile thing to consider is whether the order in which things are explained to happen is important. Imagine drawing three balls without replacement from an urn and needing the probability that the third one is blue. The urn composition at the third draw depends on the first draws, so you can make it messy, or you can imagine that the third ball was drawn first. HTH
$endgroup$
– Steve Kass
Jan 31 at 21:28
1
$begingroup$
(I'm not sure I completely understand your previous comment with the "Is it okay" question, I'm afraid.) Maybe you can think of a way to ask a new question about this overall strategy? Not sure of the best way to use Stackexchange to continue. I'm definitely not a fan of moving extended discussions to chats or whatever is the accepted thing there.)
$endgroup$
– Steve Kass
Jan 31 at 21:30
1
$begingroup$
Your approach looks right for the way you are reading the problem. You have followed OP's definition of a swap at the end, which seems strange to me. I didn't go through the calculations carefully to find the discrepancy.
$endgroup$
– Ross Millikan
Jan 31 at 23:39
|
show 2 more comments
$begingroup$
Here’s another approach. [Added: Also, see more on this approach here]
Equivalently, we can do the following.
Reach into the first box and write “S” on one ball (the one to swap).
Now choose a ball from the first box. If it is not the “S” ball, that’s your ball. The probability of this happening is $20over21$, and the ball you choose will be white with probability $10over21$. If you do choose the “S” ball (you do this with probability $1over21$), discard it by throwing it into the second box and then choose a ball at random from the second box (now containing an extra ball), and that’s your ball. If you had to do this, the probability of a white result is ${12+{10over21}over26}$, because if we add a randomly-chosen ball from box 1 to box 2, the number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$ and the number of balls in box 2 increases to $26$. Therefore the total probability you want is
$$p = {20over21}cdot{10over21}+{1over21}cdot{12+{10over21}over26}={2731over5733}approx 0.4763649.$$
P.S. I don’t see an easy way to adapt this approach for two swaps.
answered Jan 30 at 19:40
Steve KassSteve Kass
11.4k11530
$endgroup$
$begingroup$
I think this approach is awesome. I still have some little trouble understanding: The number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$.
$endgroup$
– user625055
Jan 30 at 20:09
1
$begingroup$
That’s the most hand-wavy part, but it can be justified. The expected value of the number of white balls in box 2 after a random one from box 1 is is added is $12+{10over21}$. (That’s not hard to show.) Does that help as a very quick attempt to make it seem like a valid approach?
$endgroup$
– Steve Kass
Jan 30 at 23:21
1
$begingroup$
That's a tough question. I just kind of thought it up and thought it was cool. (It was kind of new and cool to me, too.) Of course, I've solved a huge number of probability problems over time, and I know that one worthwhile thing to consider is whether the order in which things are explained to happen is important. Imagine drawing three balls without replacement from an urn and needing the probability that the third one is blue. The urn composition at the third draw depends on the first draws, so you can make it messy, or you can imagine that the third ball was drawn first. HTH
$endgroup$
– Steve Kass
Jan 31 at 21:28
1
$begingroup$
(I'm not sure I completely understand your previous comment with the "Is it okay" question, I'm afraid.) Maybe you can think of a way to ask a new question about this overall strategy? Not sure of the best way to use Stackexchange to continue. I'm definitely not a fan of moving extended discussions to chats or whatever is the accepted thing there.)
$endgroup$
– Steve Kass
Jan 31 at 21:30
1
$begingroup$
Your approach looks right for the way you are reading the problem. You have followed OP's definition of a swap at the end, which seems strange to me. I didn't go through the calculations carefully to find the discrepancy.
$endgroup$
– Ross Millikan
Jan 31 at 23:39
|
show 2 more comments
$begingroup$
Here’s another approach. [Added: Also, see more on this approach here]
Equivalently, we can do the following.
Reach into the first box and write “S” on one ball (the one to swap).
Now choose a ball from the first box. If it is not the “S” ball, that’s your ball. The probability of this happening is $20over21$, and the ball you choose will be white with probability $10over21$. If you do choose the “S” ball (you do this with probability $1over21$), discard it by throwing it into the second box and then choose a ball at random from the second box (now containing an extra ball), and that’s your ball. If you had to do this, the probability of a white result is ${12+{10over21}over26}$, because if we add a randomly-chosen ball from box 1 to box 2, the number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$ and the number of balls in box 2 increases to $26$. Therefore the total probability you want is
$$p = {20over21}cdot{10over21}+{1over21}cdot{12+{10over21}over26}={2731over5733}approx 0.4763649.$$
P.S. I don’t see an easy way to adapt this approach for two swaps.
answered Jan 30 at 19:40
Steve KassSteve Kass
11.4k11530
$endgroup$
Here’s another approach. [Added: Also, see more on this approach here]
Equivalently, we can do the following.
Reach into the first box and write “S” on one ball (the one to swap).
Now choose a ball from the first box. If it is not the “S” ball, that’s your ball. The probability of this happening is $20over21$, and the ball you choose will be white with probability $10over21$. If you do choose the “S” ball (you do this with probability $1over21$), discard it by throwing it into the second box and then choose a ball at random from the second box (now containing an extra ball), and that’s your ball. If you had to do this, the probability of a white result is ${12+{10over21}over26}$, because if we add a randomly-chosen ball from box 1 to box 2, the number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$ and the number of balls in box 2 increases to $26$. Therefore the total probability you want is
$$p = {20over21}cdot{10over21}+{1over21}cdot{12+{10over21}over26}={2731over5733}approx 0.4763649.$$
P.S. I don’t see an easy way to adapt this approach for two swaps.
answered Jan 30 at 19:40
Steve KassSteve Kass
11.4k11530
edited Feb 2 at 1:38
answered Jan 30 at 19:40
Steve KassSteve Kass
11.4k11530
answered Jan 30 at 19:40
Steve KassSteve Kass
11.4k11530
answered Jan 30 at 19:40
Steve KassSteve Kass
11.4k11530
11.4k11530
$begingroup$
I think this approach is awesome. I still have some little trouble understanding: The number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$.
$endgroup$
– user625055
Jan 30 at 20:09
1
$begingroup$
That’s the most hand-wavy part, but it can be justified. The expected value of the number of white balls in box 2 after a random one from box 1 is is added is $12+{10over21}$. (That’s not hard to show.) Does that help as a very quick attempt to make it seem like a valid approach?
$endgroup$
– Steve Kass
Jan 30 at 23:21
1
$begingroup$
That's a tough question. I just kind of thought it up and thought it was cool. (It was kind of new and cool to me, too.) Of course, I've solved a huge number of probability problems over time, and I know that one worthwhile thing to consider is whether the order in which things are explained to happen is important. Imagine drawing three balls without replacement from an urn and needing the probability that the third one is blue. The urn composition at the third draw depends on the first draws, so you can make it messy, or you can imagine that the third ball was drawn first. HTH
$endgroup$
– Steve Kass
Jan 31 at 21:28
1
$begingroup$
(I'm not sure I completely understand your previous comment with the "Is it okay" question, I'm afraid.) Maybe you can think of a way to ask a new question about this overall strategy? Not sure of the best way to use Stackexchange to continue. I'm definitely not a fan of moving extended discussions to chats or whatever is the accepted thing there.)
$endgroup$
– Steve Kass
Jan 31 at 21:30
1
$begingroup$
Your approach looks right for the way you are reading the problem. You have followed OP's definition of a swap at the end, which seems strange to me. I didn't go through the calculations carefully to find the discrepancy.
$endgroup$
– Ross Millikan
Jan 31 at 23:39
|
show 2 more comments
$begingroup$
I think this approach is awesome. I still have some little trouble understanding: The number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$.
$endgroup$
– user625055
Jan 30 at 20:09
1
$begingroup$
That’s the most hand-wavy part, but it can be justified. The expected value of the number of white balls in box 2 after a random one from box 1 is is added is $12+{10over21}$. (That’s not hard to show.) Does that help as a very quick attempt to make it seem like a valid approach?
$endgroup$
– Steve Kass
Jan 30 at 23:21
1
$begingroup$
That's a tough question. I just kind of thought it up and thought it was cool. (It was kind of new and cool to me, too.) Of course, I've solved a huge number of probability problems over time, and I know that one worthwhile thing to consider is whether the order in which things are explained to happen is important. Imagine drawing three balls without replacement from an urn and needing the probability that the third one is blue. The urn composition at the third draw depends on the first draws, so you can make it messy, or you can imagine that the third ball was drawn first. HTH
$endgroup$
– Steve Kass
Jan 31 at 21:28
1
$begingroup$
(I'm not sure I completely understand your previous comment with the "Is it okay" question, I'm afraid.) Maybe you can think of a way to ask a new question about this overall strategy? Not sure of the best way to use Stackexchange to continue. I'm definitely not a fan of moving extended discussions to chats or whatever is the accepted thing there.)
$endgroup$
– Steve Kass
Jan 31 at 21:30
1
$begingroup$
Your approach looks right for the way you are reading the problem. You have followed OP's definition of a swap at the end, which seems strange to me. I didn't go through the calculations carefully to find the discrepancy.
$endgroup$
– Ross Millikan
Jan 31 at 23:39
$begingroup$
I think this approach is awesome. I still have some little trouble understanding: The number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$.
$endgroup$
– user625055
Jan 30 at 20:09
$begingroup$
I think this approach is awesome. I still have some little trouble understanding: The number of white balls in box 2 effectively increases from $12$ to $12+{10over21}$.
$endgroup$
– user625055
Jan 30 at 20:09
1
1
$begingroup$
That’s the most hand-wavy part, but it can be justified. The expected value of the number of white balls in box 2 after a random one from box 1 is is added is $12+{10over21}$. (That’s not hard to show.) Does that help as a very quick attempt to make it seem like a valid approach?
$endgroup$
– Steve Kass
Jan 30 at 23:21
$begingroup$
That’s the most hand-wavy part, but it can be justified. The expected value of the number of white balls in box 2 after a random one from box 1 is is added is $12+{10over21}$. (That’s not hard to show.) Does that help as a very quick attempt to make it seem like a valid approach?
$endgroup$
– Steve Kass
Jan 30 at 23:21
1
1
$begingroup$
That's a tough question. I just kind of thought it up and thought it was cool. (It was kind of new and cool to me, too.) Of course, I've solved a huge number of probability problems over time, and I know that one worthwhile thing to consider is whether the order in which things are explained to happen is important. Imagine drawing three balls without replacement from an urn and needing the probability that the third one is blue. The urn composition at the third draw depends on the first draws, so you can make it messy, or you can imagine that the third ball was drawn first. HTH
$endgroup$
– Steve Kass
Jan 31 at 21:28
$begingroup$
That's a tough question. I just kind of thought it up and thought it was cool. (It was kind of new and cool to me, too.) Of course, I've solved a huge number of probability problems over time, and I know that one worthwhile thing to consider is whether the order in which things are explained to happen is important. Imagine drawing three balls without replacement from an urn and needing the probability that the third one is blue. The urn composition at the third draw depends on the first draws, so you can make it messy, or you can imagine that the third ball was drawn first. HTH
$endgroup$
– Steve Kass
Jan 31 at 21:28
1
1
$begingroup$
(I'm not sure I completely understand your previous comment with the "Is it okay" question, I'm afraid.) Maybe you can think of a way to ask a new question about this overall strategy? Not sure of the best way to use Stackexchange to continue. I'm definitely not a fan of moving extended discussions to chats or whatever is the accepted thing there.)
$endgroup$
– Steve Kass
Jan 31 at 21:30
$begingroup$
(I'm not sure I completely understand your previous comment with the "Is it okay" question, I'm afraid.) Maybe you can think of a way to ask a new question about this overall strategy? Not sure of the best way to use Stackexchange to continue. I'm definitely not a fan of moving extended discussions to chats or whatever is the accepted thing there.)
$endgroup$
– Steve Kass
Jan 31 at 21:30
1
1
$begingroup$
Your approach looks right for the way you are reading the problem. You have followed OP's definition of a swap at the end, which seems strange to me. I didn't go through the calculations carefully to find the discrepancy.
$endgroup$
– Ross Millikan
Jan 31 at 23:39
$begingroup$
Your approach looks right for the way you are reading the problem. You have followed OP's definition of a swap at the end, which seems strange to me. I didn't go through the calculations carefully to find the discrepancy.
$endgroup$
– Ross Millikan
Jan 31 at 23:39
|
show 2 more comments
$begingroup$
The probability of ending up with an additional white ball equals:
$$frac{11}{21} frac{12}{26}$$
The probability of ending up with an additional black ball equals:
$$frac{10}{21} frac{13}{26}$$
We thus find an overall probability of:
$$frac{11}{21} frac{12}{26} frac{11}{21} + frac{10}{21} frac{13}{26} frac{9}{21} + left(1 - frac{11 cdot 12 + 10 cdot 13}{21 cdot 26}right) frac{10}{21} = frac{1452 + 1170 + 2840}{11466} = frac{5462}{11466} approx 0.4764$$
answered Jan 30 at 19:17
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
How did you found the probability of ending with a white ball?
$endgroup$
– user625055
Jan 30 at 19:31
$begingroup$
@Porogami In the first case, we end up with an additional ball, so the probability of getting a white ball equals $frac{11}{21}$. In the second case, we lose a ball, so the probability of getting a white ball equals $frac{9}{21}$. Finally, if we simply swap white for white or black for black, the probability of getting a white ball equals $frac{10}{21}$ as was initially the case.
$endgroup$
– jvdhooft
Jan 30 at 19:35
add a comment |
$begingroup$
The probability of ending up with an additional white ball equals:
$$frac{11}{21} frac{12}{26}$$
The probability of ending up with an additional black ball equals:
$$frac{10}{21} frac{13}{26}$$
We thus find an overall probability of:
$$frac{11}{21} frac{12}{26} frac{11}{21} + frac{10}{21} frac{13}{26} frac{9}{21} + left(1 - frac{11 cdot 12 + 10 cdot 13}{21 cdot 26}right) frac{10}{21} = frac{1452 + 1170 + 2840}{11466} = frac{5462}{11466} approx 0.4764$$
answered Jan 30 at 19:17
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
How did you found the probability of ending with a white ball?
$endgroup$
– user625055
Jan 30 at 19:31
$begingroup$
@Porogami In the first case, we end up with an additional ball, so the probability of getting a white ball equals $frac{11}{21}$. In the second case, we lose a ball, so the probability of getting a white ball equals $frac{9}{21}$. Finally, if we simply swap white for white or black for black, the probability of getting a white ball equals $frac{10}{21}$ as was initially the case.
$endgroup$
– jvdhooft
Jan 30 at 19:35
add a comment |
$begingroup$
The probability of ending up with an additional white ball equals:
$$frac{11}{21} frac{12}{26}$$
The probability of ending up with an additional black ball equals:
$$frac{10}{21} frac{13}{26}$$
We thus find an overall probability of:
$$frac{11}{21} frac{12}{26} frac{11}{21} + frac{10}{21} frac{13}{26} frac{9}{21} + left(1 - frac{11 cdot 12 + 10 cdot 13}{21 cdot 26}right) frac{10}{21} = frac{1452 + 1170 + 2840}{11466} = frac{5462}{11466} approx 0.4764$$
answered Jan 30 at 19:17
jvdhooftjvdhooft
5,65961641
$endgroup$
The probability of ending up with an additional white ball equals:
$$frac{11}{21} frac{12}{26}$$
The probability of ending up with an additional black ball equals:
$$frac{10}{21} frac{13}{26}$$
We thus find an overall probability of:
$$frac{11}{21} frac{12}{26} frac{11}{21} + frac{10}{21} frac{13}{26} frac{9}{21} + left(1 - frac{11 cdot 12 + 10 cdot 13}{21 cdot 26}right) frac{10}{21} = frac{1452 + 1170 + 2840}{11466} = frac{5462}{11466} approx 0.4764$$
answered Jan 30 at 19:17
jvdhooftjvdhooft
5,65961641
answered Jan 30 at 19:17
jvdhooftjvdhooft
5,65961641
answered Jan 30 at 19:17
jvdhooftjvdhooft
5,65961641
answered Jan 30 at 19:17
jvdhooftjvdhooft
5,65961641
5,65961641
$begingroup$
How did you found the probability of ending with a white ball?
$endgroup$
– user625055
Jan 30 at 19:31
$begingroup$
@Porogami In the first case, we end up with an additional ball, so the probability of getting a white ball equals $frac{11}{21}$. In the second case, we lose a ball, so the probability of getting a white ball equals $frac{9}{21}$. Finally, if we simply swap white for white or black for black, the probability of getting a white ball equals $frac{10}{21}$ as was initially the case.
$endgroup$
– jvdhooft
Jan 30 at 19:35
add a comment |
$begingroup$
How did you found the probability of ending with a white ball?
$endgroup$
– user625055
Jan 30 at 19:31
$begingroup$
@Porogami In the first case, we end up with an additional ball, so the probability of getting a white ball equals $frac{11}{21}$. In the second case, we lose a ball, so the probability of getting a white ball equals $frac{9}{21}$. Finally, if we simply swap white for white or black for black, the probability of getting a white ball equals $frac{10}{21}$ as was initially the case.
$endgroup$
– jvdhooft
Jan 30 at 19:35
$begingroup$
How did you found the probability of ending with a white ball?
$endgroup$
– user625055
Jan 30 at 19:31
$begingroup$
How did you found the probability of ending with a white ball?
$endgroup$
– user625055
Jan 30 at 19:31
$begingroup$
@Porogami In the first case, we end up with an additional ball, so the probability of getting a white ball equals $frac{11}{21}$. In the second case, we lose a ball, so the probability of getting a white ball equals $frac{9}{21}$. Finally, if we simply swap white for white or black for black, the probability of getting a white ball equals $frac{10}{21}$ as was initially the case.
$endgroup$
– jvdhooft
Jan 30 at 19:35
$begingroup$
@Porogami In the first case, we end up with an additional ball, so the probability of getting a white ball equals $frac{11}{21}$. In the second case, we lose a ball, so the probability of getting a white ball equals $frac{9}{21}$. Finally, if we simply swap white for white or black for black, the probability of getting a white ball equals $frac{10}{21}$ as was initially the case.
$endgroup$
– jvdhooft
Jan 30 at 19:35
add a comment |
$begingroup$
Although this is already answered, I'll add my two cents. Perhaps someone will find it more intuitive.
Call probability of picking white ball from boxes $p_1$, and $p_2$. Since we're exchanging one ball, the probability will be between $p_1$ and $p_2$, closer to $p_1$, since we have $21$ balls in box 1.
$ p(W) = frac{20}{21}p_1 + frac1{21}p_2'$
notice we wrote $p_2'$. The dynamics of the problem will change $p_2$ mid process. But we can apply the same logic here, since 25 balls were in box 2 and now 1 more added.
$p_2' = frac{25}{26}p_2 + frac1{26}p_1 $
Combining the two equations will give
$ p(W) = frac{20}{21}p_1 + frac1{21}(frac{25}{26}p_2 + frac1{26}p_1)
$
Here again, the coefficients will tell the story. The final ball will come from box 1 with probability $frac{20}{21}+frac1{21}frac1{26}$, that is either directly or via box 2 detour. Or, with probability $frac1{21}frac{25}{26}$ from the second box. Therefore the final probability is the weighted average of the original probabilities with these weights (which themselves are probabilities).
In general let number of balls in box 1 and 2 be equal to $n$ and $m$ respectively. Define $alpha = frac{m}{n(m+1)}$. Also, let $p_i(s)$ shows the probability of drawing white in box $i$ after $s$ swaps. Where $p_i=p_i(0)$. Then, we can write
$$ p_1(s+1) = (1-alpha) p_1(s) + alpha p_2(s) $$
and
$$ p_2(s+1) = p_2(s) + frac{n}{m}(p_1(s) - p_1(s+1)) $$
second equation is due to conservation of total white balls. Now you can compute probabilities after any number of swaps from this iterative process.
answered Feb 1 at 20:44
karakfakarakfa
2,025811
$endgroup$
$begingroup$
Thank you alot! Does it matter so much if the balls from the first urn are fewer than in the second one? // What if we had the same number of balls in both urns?
$endgroup$
– user625055
Feb 1 at 21:42
1
$begingroup$
Well, same number of odds in both boxes will give you the same odds since you're taking the weighted average of the same number. Doesn't matter how many in each box. Perhaps I misunderstood the question, here the first box already has fewer balls.
$endgroup$
– karakfa
Feb 1 at 22:02
$begingroup$
Yes, of course, I just asked about another case. Thanks, I have checked and it matches with Steve answer.
$endgroup$
– user625055
Feb 1 at 22:18
add a comment |
$begingroup$
Although this is already answered, I'll add my two cents. Perhaps someone will find it more intuitive.
Call probability of picking white ball from boxes $p_1$, and $p_2$. Since we're exchanging one ball, the probability will be between $p_1$ and $p_2$, closer to $p_1$, since we have $21$ balls in box 1.
$ p(W) = frac{20}{21}p_1 + frac1{21}p_2'$
notice we wrote $p_2'$. The dynamics of the problem will change $p_2$ mid process. But we can apply the same logic here, since 25 balls were in box 2 and now 1 more added.
$p_2' = frac{25}{26}p_2 + frac1{26}p_1 $
Combining the two equations will give
$ p(W) = frac{20}{21}p_1 + frac1{21}(frac{25}{26}p_2 + frac1{26}p_1)
$
Here again, the coefficients will tell the story. The final ball will come from box 1 with probability $frac{20}{21}+frac1{21}frac1{26}$, that is either directly or via box 2 detour. Or, with probability $frac1{21}frac{25}{26}$ from the second box. Therefore the final probability is the weighted average of the original probabilities with these weights (which themselves are probabilities).
In general let number of balls in box 1 and 2 be equal to $n$ and $m$ respectively. Define $alpha = frac{m}{n(m+1)}$. Also, let $p_i(s)$ shows the probability of drawing white in box $i$ after $s$ swaps. Where $p_i=p_i(0)$. Then, we can write
$$ p_1(s+1) = (1-alpha) p_1(s) + alpha p_2(s) $$
and
$$ p_2(s+1) = p_2(s) + frac{n}{m}(p_1(s) - p_1(s+1)) $$
second equation is due to conservation of total white balls. Now you can compute probabilities after any number of swaps from this iterative process.
answered Feb 1 at 20:44
karakfakarakfa
2,025811
$endgroup$
$begingroup$
Thank you alot! Does it matter so much if the balls from the first urn are fewer than in the second one? // What if we had the same number of balls in both urns?
$endgroup$
– user625055
Feb 1 at 21:42
1
$begingroup$
Well, same number of odds in both boxes will give you the same odds since you're taking the weighted average of the same number. Doesn't matter how many in each box. Perhaps I misunderstood the question, here the first box already has fewer balls.
$endgroup$
– karakfa
Feb 1 at 22:02
$begingroup$
Yes, of course, I just asked about another case. Thanks, I have checked and it matches with Steve answer.
$endgroup$
– user625055
Feb 1 at 22:18
add a comment |
$begingroup$
Although this is already answered, I'll add my two cents. Perhaps someone will find it more intuitive.
Call probability of picking white ball from boxes $p_1$, and $p_2$. Since we're exchanging one ball, the probability will be between $p_1$ and $p_2$, closer to $p_1$, since we have $21$ balls in box 1.
$ p(W) = frac{20}{21}p_1 + frac1{21}p_2'$
notice we wrote $p_2'$. The dynamics of the problem will change $p_2$ mid process. But we can apply the same logic here, since 25 balls were in box 2 and now 1 more added.
$p_2' = frac{25}{26}p_2 + frac1{26}p_1 $
Combining the two equations will give
$ p(W) = frac{20}{21}p_1 + frac1{21}(frac{25}{26}p_2 + frac1{26}p_1)
$
Here again, the coefficients will tell the story. The final ball will come from box 1 with probability $frac{20}{21}+frac1{21}frac1{26}$, that is either directly or via box 2 detour. Or, with probability $frac1{21}frac{25}{26}$ from the second box. Therefore the final probability is the weighted average of the original probabilities with these weights (which themselves are probabilities).
In general let number of balls in box 1 and 2 be equal to $n$ and $m$ respectively. Define $alpha = frac{m}{n(m+1)}$. Also, let $p_i(s)$ shows the probability of drawing white in box $i$ after $s$ swaps. Where $p_i=p_i(0)$. Then, we can write
$$ p_1(s+1) = (1-alpha) p_1(s) + alpha p_2(s) $$
and
$$ p_2(s+1) = p_2(s) + frac{n}{m}(p_1(s) - p_1(s+1)) $$
second equation is due to conservation of total white balls. Now you can compute probabilities after any number of swaps from this iterative process.
answered Feb 1 at 20:44
karakfakarakfa
2,025811
$endgroup$
Although this is already answered, I'll add my two cents. Perhaps someone will find it more intuitive.
Call probability of picking white ball from boxes $p_1$, and $p_2$. Since we're exchanging one ball, the probability will be between $p_1$ and $p_2$, closer to $p_1$, since we have $21$ balls in box 1.
$ p(W) = frac{20}{21}p_1 + frac1{21}p_2'$
notice we wrote $p_2'$. The dynamics of the problem will change $p_2$ mid process. But we can apply the same logic here, since 25 balls were in box 2 and now 1 more added.
$p_2' = frac{25}{26}p_2 + frac1{26}p_1 $
Combining the two equations will give
$ p(W) = frac{20}{21}p_1 + frac1{21}(frac{25}{26}p_2 + frac1{26}p_1)
$
Here again, the coefficients will tell the story. The final ball will come from box 1 with probability $frac{20}{21}+frac1{21}frac1{26}$, that is either directly or via box 2 detour. Or, with probability $frac1{21}frac{25}{26}$ from the second box. Therefore the final probability is the weighted average of the original probabilities with these weights (which themselves are probabilities).
In general let number of balls in box 1 and 2 be equal to $n$ and $m$ respectively. Define $alpha = frac{m}{n(m+1)}$. Also, let $p_i(s)$ shows the probability of drawing white in box $i$ after $s$ swaps. Where $p_i=p_i(0)$. Then, we can write
$$ p_1(s+1) = (1-alpha) p_1(s) + alpha p_2(s) $$
and
$$ p_2(s+1) = p_2(s) + frac{n}{m}(p_1(s) - p_1(s+1)) $$
second equation is due to conservation of total white balls. Now you can compute probabilities after any number of swaps from this iterative process.
answered Feb 1 at 20:44
karakfakarakfa
2,025811
edited Feb 2 at 1:57
answered Feb 1 at 20:44
karakfakarakfa
2,025811
answered Feb 1 at 20:44
karakfakarakfa
2,025811
answered Feb 1 at 20:44
karakfakarakfa
2,025811
2,025811
$begingroup$
Thank you alot! Does it matter so much if the balls from the first urn are fewer than in the second one? // What if we had the same number of balls in both urns?
$endgroup$
– user625055
Feb 1 at 21:42
1
$begingroup$
Well, same number of odds in both boxes will give you the same odds since you're taking the weighted average of the same number. Doesn't matter how many in each box. Perhaps I misunderstood the question, here the first box already has fewer balls.
$endgroup$
– karakfa
Feb 1 at 22:02
$begingroup$
Yes, of course, I just asked about another case. Thanks, I have checked and it matches with Steve answer.
$endgroup$
– user625055
Feb 1 at 22:18
add a comment |
$begingroup$
Thank you alot! Does it matter so much if the balls from the first urn are fewer than in the second one? // What if we had the same number of balls in both urns?
$endgroup$
– user625055
Feb 1 at 21:42
1
$begingroup$
Well, same number of odds in both boxes will give you the same odds since you're taking the weighted average of the same number. Doesn't matter how many in each box. Perhaps I misunderstood the question, here the first box already has fewer balls.
$endgroup$
– karakfa
Feb 1 at 22:02
$begingroup$
Yes, of course, I just asked about another case. Thanks, I have checked and it matches with Steve answer.
$endgroup$
– user625055
Feb 1 at 22:18
$begingroup$
Thank you alot! Does it matter so much if the balls from the first urn are fewer than in the second one? // What if we had the same number of balls in both urns?
$endgroup$
– user625055
Feb 1 at 21:42
$begingroup$
Thank you alot! Does it matter so much if the balls from the first urn are fewer than in the second one? // What if we had the same number of balls in both urns?
$endgroup$
– user625055
Feb 1 at 21:42
1
1
$begingroup$
Well, same number of odds in both boxes will give you the same odds since you're taking the weighted average of the same number. Doesn't matter how many in each box. Perhaps I misunderstood the question, here the first box already has fewer balls.
$endgroup$
– karakfa
Feb 1 at 22:02
$begingroup$
Well, same number of odds in both boxes will give you the same odds since you're taking the weighted average of the same number. Doesn't matter how many in each box. Perhaps I misunderstood the question, here the first box already has fewer balls.
$endgroup$
– karakfa
Feb 1 at 22:02
$begingroup$
Yes, of course, I just asked about another case. Thanks, I have checked and it matches with Steve answer.
$endgroup$
– user625055
Feb 1 at 22:18
$begingroup$
Yes, of course, I just asked about another case. Thanks, I have checked and it matches with Steve answer.
$endgroup$
– user625055
Feb 1 at 22:18
add a comment |
$begingroup$
So... for this, I'll need a lot of names for events that sound similar, so bear with me.
Since you have defined a "swap" to be where a ball is first moved from the first box into the second box, and then a ball is chosen from the second box which now includes the ball just moved and this is moved back to the first box.
Let $b_1$ be the event that a black ball was moved from the first into the second. Let $w_1$ be the event a white ball was moved from the first into the second.
Let $b_2$ be the event that a black ball was moved from the second into the first. Let $w_2$ be the event that a white ball was moved from the second into the first.
Let $W$ be the event that a white ball was selected from the first box after the swap occurred.
We are interested in finding $Pr(W)$
By total probability we have:
$$Pr(W)=Pr(Wmid b_1b_2)Pr(b_2mid b_1)Pr(b_1)+Pr(Wmid b_1w_2)Pr(w_2mid b_1)Pr(b_1)\
+Pr(Wmid w_1b_2)Pr(b_2mid w_1)Pr(w_1)+Pr(Wmid w_1w_2) Pr(w_2mid w_1)Pr(w_1)$$
Each of these terms can be found with relatively little effort, though it is tedious to find so many. I'll do one pair of them for now and leave the rest for you.
$Pr(Wmid b_1b_2) = frac{10}{21}$ and $Pr(b_2mid b_1)=frac{14}{26}$ and $Pr(b_1)=frac{11}{21}$
answered Jan 30 at 19:06
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
So... for this, I'll need a lot of names for events that sound similar, so bear with me.
Since you have defined a "swap" to be where a ball is first moved from the first box into the second box, and then a ball is chosen from the second box which now includes the ball just moved and this is moved back to the first box.
Let $b_1$ be the event that a black ball was moved from the first into the second. Let $w_1$ be the event a white ball was moved from the first into the second.
Let $b_2$ be the event that a black ball was moved from the second into the first. Let $w_2$ be the event that a white ball was moved from the second into the first.
Let $W$ be the event that a white ball was selected from the first box after the swap occurred.
We are interested in finding $Pr(W)$
By total probability we have:
$$Pr(W)=Pr(Wmid b_1b_2)Pr(b_2mid b_1)Pr(b_1)+Pr(Wmid b_1w_2)Pr(w_2mid b_1)Pr(b_1)\
+Pr(Wmid w_1b_2)Pr(b_2mid w_1)Pr(w_1)+Pr(Wmid w_1w_2) Pr(w_2mid w_1)Pr(w_1)$$
Each of these terms can be found with relatively little effort, though it is tedious to find so many. I'll do one pair of them for now and leave the rest for you.
$Pr(Wmid b_1b_2) = frac{10}{21}$ and $Pr(b_2mid b_1)=frac{14}{26}$ and $Pr(b_1)=frac{11}{21}$
answered Jan 30 at 19:06
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
So... for this, I'll need a lot of names for events that sound similar, so bear with me.
Since you have defined a "swap" to be where a ball is first moved from the first box into the second box, and then a ball is chosen from the second box which now includes the ball just moved and this is moved back to the first box.
Let $b_1$ be the event that a black ball was moved from the first into the second. Let $w_1$ be the event a white ball was moved from the first into the second.
Let $b_2$ be the event that a black ball was moved from the second into the first. Let $w_2$ be the event that a white ball was moved from the second into the first.
Let $W$ be the event that a white ball was selected from the first box after the swap occurred.
We are interested in finding $Pr(W)$
By total probability we have:
$$Pr(W)=Pr(Wmid b_1b_2)Pr(b_2mid b_1)Pr(b_1)+Pr(Wmid b_1w_2)Pr(w_2mid b_1)Pr(b_1)\
+Pr(Wmid w_1b_2)Pr(b_2mid w_1)Pr(w_1)+Pr(Wmid w_1w_2) Pr(w_2mid w_1)Pr(w_1)$$
Each of these terms can be found with relatively little effort, though it is tedious to find so many. I'll do one pair of them for now and leave the rest for you.
$Pr(Wmid b_1b_2) = frac{10}{21}$ and $Pr(b_2mid b_1)=frac{14}{26}$ and $Pr(b_1)=frac{11}{21}$
answered Jan 30 at 19:06
JMoravitzJMoravitz
48.7k43988
$endgroup$
So... for this, I'll need a lot of names for events that sound similar, so bear with me.
Since you have defined a "swap" to be where a ball is first moved from the first box into the second box, and then a ball is chosen from the second box which now includes the ball just moved and this is moved back to the first box.
Let $b_1$ be the event that a black ball was moved from the first into the second. Let $w_1$ be the event a white ball was moved from the first into the second.
Let $b_2$ be the event that a black ball was moved from the second into the first. Let $w_2$ be the event that a white ball was moved from the second into the first.
Let $W$ be the event that a white ball was selected from the first box after the swap occurred.
We are interested in finding $Pr(W)$
By total probability we have:
$$Pr(W)=Pr(Wmid b_1b_2)Pr(b_2mid b_1)Pr(b_1)+Pr(Wmid b_1w_2)Pr(w_2mid b_1)Pr(b_1)\
+Pr(Wmid w_1b_2)Pr(b_2mid w_1)Pr(w_1)+Pr(Wmid w_1w_2) Pr(w_2mid w_1)Pr(w_1)$$
Each of these terms can be found with relatively little effort, though it is tedious to find so many. I'll do one pair of them for now and leave the rest for you.
$Pr(Wmid b_1b_2) = frac{10}{21}$ and $Pr(b_2mid b_1)=frac{14}{26}$ and $Pr(b_1)=frac{11}{21}$
answered Jan 30 at 19:06
JMoravitzJMoravitz
48.7k43988
edited Jan 30 at 20:18
user625055
answered Jan 30 at 19:06
JMoravitzJMoravitz
48.7k43988
answered Jan 30 at 19:06
JMoravitzJMoravitz
48.7k43988
answered Jan 30 at 19:06
JMoravitzJMoravitz
48.7k43988
48.7k43988
add a comment |
add a comment |
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$begingroup$
can u simplify to give a final answer
$endgroup$
– K Split X
Jan 30 at 18:56
1
$begingroup$
Your denominators should be $26cdot21$, not $26cdot11$
$endgroup$
– Daniel Mathias
Jan 30 at 19:01
1
$begingroup$
Clearly the answer needs to be slightly less than $frac 12$
$endgroup$
– Ross Millikan
Jan 30 at 19:04
2
$begingroup$
Are you sure that's what "swap" means? I'd have thought that you extract one ball from each and put them back in the other bin (the difference being that your way you could swap the same ball twice).
$endgroup$
– lulu
Jan 30 at 19:06
1
$begingroup$
What do $10+1over22$ and $10over22$ represent in the final calculation? After the swap, there are $21$ balls in box 1, and there are three possible distributions of color: $10$ white and $11$ black (if the balls moved were the same color), $9$ white and $12$ black (if a white ball was moved from box 1 and a black ball was moved back), or $11$ white and $10$ black (if a black ball was moved from box 1 and a white ball was moved back). You need to find the probabilities ($p_1,p_2,p_3$) of my parenthetical events and find $p_1{10over21}+p_2{9over21}+p_3{11over21}$ (like jvdhooft did)
$endgroup$
– Steve Kass
Feb 1 at 3:40