Mixing
Expected number of times that two friends will appear in the same episode, given that the number of...
$begingroup$
Suppose that there are 4 episodes of a television series, and each episode requires 2 actors to appear in it. Suppose that there are 4 actors in total that will appear in the episodes: Amanda, Bert, Charlie, and Diana. Their contracts state that Amanda will appear in 3 episodes, Bert will appear in 2 episodes, Charlie will appear in 2 episodes, and Diana will appear in 1 episode. What is the expected value of episodes that Amanda and Bert will appear in together?
My Attempt:
There are 8 spots total that an actor could appear in over the course of the 4 episodes (4 episodes x 2 actors each = 8 spots). Since Amanda appears in 3 episodes, there are 5 spots remaining for Bert to appear in. Therefore, since Bert will appear in two episodes, the probability of Bert appearing in any one of the 5 remaining spots is (2/5). If you multiply that by 3 (the number of appearances Amanda makes), you get (6/5), which equals 1.2, so the expected number of episodes that Amanda and Bert should appear in together is 1.2.
Why this doesn't work:
You should get the same answer by calculating the probability of Amanda appearing in one of the remaining spots not occupied by Bert. If Bert appears in 2 episodes, there are 6 remaining spots for Amanda to appear in. Therefore, since Amanda will appear in 3 episodes, the probability of Amanda appearing in any one of the 6 remaining spots is (3/6). If you multiply that by 2 (the number of appearances Bert makes), you get (6/6), which equals 1, so the expected number of episodes that Amanda and Bert should appear in together is 1. However, this is not the expected value derived using the previous method.
What is the correct way of finding the expected value of episodes that Amanda and Bert will appear in together?
probability combinatorics
asked Jan 27 at 19:42
LukaszLukasz
32
$endgroup$
add a comment |
$begingroup$
Suppose that there are 4 episodes of a television series, and each episode requires 2 actors to appear in it. Suppose that there are 4 actors in total that will appear in the episodes: Amanda, Bert, Charlie, and Diana. Their contracts state that Amanda will appear in 3 episodes, Bert will appear in 2 episodes, Charlie will appear in 2 episodes, and Diana will appear in 1 episode. What is the expected value of episodes that Amanda and Bert will appear in together?
My Attempt:
There are 8 spots total that an actor could appear in over the course of the 4 episodes (4 episodes x 2 actors each = 8 spots). Since Amanda appears in 3 episodes, there are 5 spots remaining for Bert to appear in. Therefore, since Bert will appear in two episodes, the probability of Bert appearing in any one of the 5 remaining spots is (2/5). If you multiply that by 3 (the number of appearances Amanda makes), you get (6/5), which equals 1.2, so the expected number of episodes that Amanda and Bert should appear in together is 1.2.
Why this doesn't work:
You should get the same answer by calculating the probability of Amanda appearing in one of the remaining spots not occupied by Bert. If Bert appears in 2 episodes, there are 6 remaining spots for Amanda to appear in. Therefore, since Amanda will appear in 3 episodes, the probability of Amanda appearing in any one of the 6 remaining spots is (3/6). If you multiply that by 2 (the number of appearances Bert makes), you get (6/6), which equals 1, so the expected number of episodes that Amanda and Bert should appear in together is 1. However, this is not the expected value derived using the previous method.
What is the correct way of finding the expected value of episodes that Amanda and Bert will appear in together?
probability combinatorics
asked Jan 27 at 19:42
LukaszLukasz
32
$endgroup$
$begingroup$
I'm not sure the rules are clear. How are the actors assigned episodes? As your calculations show, two different ways of producing "random" assignments get to different probabilities. If you are saying "assignments are random in the sense that any two assignments consistent with the numbers are equally likely" then I think you might need to enumerate
$endgroup$
– lulu
Jan 27 at 20:32
add a comment |
$begingroup$
Suppose that there are 4 episodes of a television series, and each episode requires 2 actors to appear in it. Suppose that there are 4 actors in total that will appear in the episodes: Amanda, Bert, Charlie, and Diana. Their contracts state that Amanda will appear in 3 episodes, Bert will appear in 2 episodes, Charlie will appear in 2 episodes, and Diana will appear in 1 episode. What is the expected value of episodes that Amanda and Bert will appear in together?
My Attempt:
There are 8 spots total that an actor could appear in over the course of the 4 episodes (4 episodes x 2 actors each = 8 spots). Since Amanda appears in 3 episodes, there are 5 spots remaining for Bert to appear in. Therefore, since Bert will appear in two episodes, the probability of Bert appearing in any one of the 5 remaining spots is (2/5). If you multiply that by 3 (the number of appearances Amanda makes), you get (6/5), which equals 1.2, so the expected number of episodes that Amanda and Bert should appear in together is 1.2.
Why this doesn't work:
You should get the same answer by calculating the probability of Amanda appearing in one of the remaining spots not occupied by Bert. If Bert appears in 2 episodes, there are 6 remaining spots for Amanda to appear in. Therefore, since Amanda will appear in 3 episodes, the probability of Amanda appearing in any one of the 6 remaining spots is (3/6). If you multiply that by 2 (the number of appearances Bert makes), you get (6/6), which equals 1, so the expected number of episodes that Amanda and Bert should appear in together is 1. However, this is not the expected value derived using the previous method.
What is the correct way of finding the expected value of episodes that Amanda and Bert will appear in together?
probability combinatorics
asked Jan 27 at 19:42
LukaszLukasz
32
$endgroup$
Suppose that there are 4 episodes of a television series, and each episode requires 2 actors to appear in it. Suppose that there are 4 actors in total that will appear in the episodes: Amanda, Bert, Charlie, and Diana. Their contracts state that Amanda will appear in 3 episodes, Bert will appear in 2 episodes, Charlie will appear in 2 episodes, and Diana will appear in 1 episode. What is the expected value of episodes that Amanda and Bert will appear in together?
My Attempt:
There are 8 spots total that an actor could appear in over the course of the 4 episodes (4 episodes x 2 actors each = 8 spots). Since Amanda appears in 3 episodes, there are 5 spots remaining for Bert to appear in. Therefore, since Bert will appear in two episodes, the probability of Bert appearing in any one of the 5 remaining spots is (2/5). If you multiply that by 3 (the number of appearances Amanda makes), you get (6/5), which equals 1.2, so the expected number of episodes that Amanda and Bert should appear in together is 1.2.
Why this doesn't work:
You should get the same answer by calculating the probability of Amanda appearing in one of the remaining spots not occupied by Bert. If Bert appears in 2 episodes, there are 6 remaining spots for Amanda to appear in. Therefore, since Amanda will appear in 3 episodes, the probability of Amanda appearing in any one of the 6 remaining spots is (3/6). If you multiply that by 2 (the number of appearances Bert makes), you get (6/6), which equals 1, so the expected number of episodes that Amanda and Bert should appear in together is 1. However, this is not the expected value derived using the previous method.
What is the correct way of finding the expected value of episodes that Amanda and Bert will appear in together?
probability combinatorics
probability combinatorics
asked Jan 27 at 19:42
LukaszLukasz
32
asked Jan 27 at 19:42
LukaszLukasz
32
asked Jan 27 at 19:42
LukaszLukasz
32
asked Jan 27 at 19:42
LukaszLukasz
32
asked Jan 27 at 19:42
LukaszLukasz
32
32
$begingroup$
I'm not sure the rules are clear. How are the actors assigned episodes? As your calculations show, two different ways of producing "random" assignments get to different probabilities. If you are saying "assignments are random in the sense that any two assignments consistent with the numbers are equally likely" then I think you might need to enumerate
$endgroup$
– lulu
Jan 27 at 20:32
add a comment |
$begingroup$
I'm not sure the rules are clear. How are the actors assigned episodes? As your calculations show, two different ways of producing "random" assignments get to different probabilities. If you are saying "assignments are random in the sense that any two assignments consistent with the numbers are equally likely" then I think you might need to enumerate
$endgroup$
– lulu
Jan 27 at 20:32
$begingroup$
I'm not sure the rules are clear. How are the actors assigned episodes? As your calculations show, two different ways of producing "random" assignments get to different probabilities. If you are saying "assignments are random in the sense that any two assignments consistent with the numbers are equally likely" then I think you might need to enumerate
$endgroup$
– lulu
Jan 27 at 20:32
$begingroup$
I'm not sure the rules are clear. How are the actors assigned episodes? As your calculations show, two different ways of producing "random" assignments get to different probabilities. If you are saying "assignments are random in the sense that any two assignments consistent with the numbers are equally likely" then I think you might need to enumerate
$endgroup$
– lulu
Jan 27 at 20:32
add a comment |
1 Answer
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$begingroup$
Since Bert appears in two episodes, he may share either one or two episodes with Amanda. (Since she appears in three, it is impossible that they share zero.)
The expected number of episodes that feature Amanda and Bert together is given by
$$ 1 cdot P(text{A and B share 1}) + 2 cdot P(text{ A and B share 2}).$$
To calculate these probabilities, we’ll consider ways to assign episodes to the actors. There are $binom{4}{3} = 4$ ways to choose which episodes Amanda appears in. (It turns out this isn’t very important though.) After assigning episodes to Amanda, we claim that there are $12$ ways of assigning episodes to the remaining actors. To see this, we take two cases.
Case 1: If Diana appears in the remaining episode, then there are $2$ ways to choose which of Bert or Charlie appears with her. The one who doesn't appear with Diana must appear with Amanda two times. There are $binom{3}{2} = 3$ ways to choose which of Amanda’s episodes he appears in. Multiplying, we see that there are $6$ assignments in this case.
Case 2: If Diana does not appear in the non-Amanda episode, then there are 3! ways to choose who acts in which of Amanda’s episodes, because the choice entails permuting Bert, Charlie, and Diana.
In all, after we have made a choice for which episodes Amanda acts in, there are $12$ ways to assign actors to the remaining episodes, so there are $48$ assignments of episodes possible overall.
Of these $48$, there are $12$ in which Bert and Amanda share two episodes - $3$ for each of the possible assignments of episodes to Amanda. This is similar to our work in Case 1, because Bert and Amanda share two episodes if and only if Charlie and Diana share the non-Amanda episode, in which case there are $binom{3}{2}$ ways to choose the episodes that Bert acts in.
Thus, we have
$$ P(text{ A and B share 2}) = frac{4(3)}{4(12)} = frac{1}{4}.$$
It follows that
$$P(text{ A and B share 1}) = 1 - frac{1}{4} = frac{3}{4}$$ and the expected number of episodes shared by Amanda and Bert is
$$ 1 left( frac{3}{4} right) + 2 left (frac{1}{4} right)= 1.25.$$
I’m adding an alternate solution that builds on (and corrects) intuition from the answer by Bram28, who wrote:
"Bert basically gets assigned $2$
out of $4$
episodes, so there are $binom{4}{2}=6$
possible assignments of $2$
episodes to Bert. Of those, $3$
will have Bert work Amanda both times, and $3$
will have Bert working with Amanda once.
So, the expected number is:
$$
frac{3}{6}⋅2+ frac{3}{6}⋅1=1.5."
$$
This is incorrect, because not all post-Amanda assignments of episodes to Bert are equally likely. Instead, some assignments to Bert create more options as to how we may assign the other actors.
If we assign episodes to Bert in a way that makes Amanda and Bert appear together twice, there is only one way to finish off the episode assignment - namely, having Charlie share Amanda’s remaining episode and having Charlie and Diana share the non-Amanda episode. However, if we assign episodes to Bert in a way that makes Amanda and Bert appear together once, then there are three ways to complete episode assignment: we could have Diana share the non-Amanda episode with Bert (1 way), or we could have Charlie share it, in which case there are two ways to decide which of Amanda’s episodes Charlie acts in (2 more ways).
This shows that it is three times as likely for Amanda and Bert to share one episode as it is for them to share two episodes. Thus,
$ P(text{ A and B share 2}) = frac{4(3)}{4(12)} = frac{1}{4}$ and
$P(text{ A and B share 1}) = 1 - frac{1}{4} = frac{3}{4}$.
answered Jan 27 at 20:52
Jordan GreenJordan Green
1,146410
$endgroup$
add a comment |
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$begingroup$
Since Bert appears in two episodes, he may share either one or two episodes with Amanda. (Since she appears in three, it is impossible that they share zero.)
The expected number of episodes that feature Amanda and Bert together is given by
$$ 1 cdot P(text{A and B share 1}) + 2 cdot P(text{ A and B share 2}).$$
To calculate these probabilities, we’ll consider ways to assign episodes to the actors. There are $binom{4}{3} = 4$ ways to choose which episodes Amanda appears in. (It turns out this isn’t very important though.) After assigning episodes to Amanda, we claim that there are $12$ ways of assigning episodes to the remaining actors. To see this, we take two cases.
Case 1: If Diana appears in the remaining episode, then there are $2$ ways to choose which of Bert or Charlie appears with her. The one who doesn't appear with Diana must appear with Amanda two times. There are $binom{3}{2} = 3$ ways to choose which of Amanda’s episodes he appears in. Multiplying, we see that there are $6$ assignments in this case.
Case 2: If Diana does not appear in the non-Amanda episode, then there are 3! ways to choose who acts in which of Amanda’s episodes, because the choice entails permuting Bert, Charlie, and Diana.
In all, after we have made a choice for which episodes Amanda acts in, there are $12$ ways to assign actors to the remaining episodes, so there are $48$ assignments of episodes possible overall.
Of these $48$, there are $12$ in which Bert and Amanda share two episodes - $3$ for each of the possible assignments of episodes to Amanda. This is similar to our work in Case 1, because Bert and Amanda share two episodes if and only if Charlie and Diana share the non-Amanda episode, in which case there are $binom{3}{2}$ ways to choose the episodes that Bert acts in.
Thus, we have
$$ P(text{ A and B share 2}) = frac{4(3)}{4(12)} = frac{1}{4}.$$
It follows that
$$P(text{ A and B share 1}) = 1 - frac{1}{4} = frac{3}{4}$$ and the expected number of episodes shared by Amanda and Bert is
$$ 1 left( frac{3}{4} right) + 2 left (frac{1}{4} right)= 1.25.$$
I’m adding an alternate solution that builds on (and corrects) intuition from the answer by Bram28, who wrote:
"Bert basically gets assigned $2$
out of $4$
episodes, so there are $binom{4}{2}=6$
possible assignments of $2$
episodes to Bert. Of those, $3$
will have Bert work Amanda both times, and $3$
will have Bert working with Amanda once.
So, the expected number is:
$$
frac{3}{6}⋅2+ frac{3}{6}⋅1=1.5."
$$
This is incorrect, because not all post-Amanda assignments of episodes to Bert are equally likely. Instead, some assignments to Bert create more options as to how we may assign the other actors.
If we assign episodes to Bert in a way that makes Amanda and Bert appear together twice, there is only one way to finish off the episode assignment - namely, having Charlie share Amanda’s remaining episode and having Charlie and Diana share the non-Amanda episode. However, if we assign episodes to Bert in a way that makes Amanda and Bert appear together once, then there are three ways to complete episode assignment: we could have Diana share the non-Amanda episode with Bert (1 way), or we could have Charlie share it, in which case there are two ways to decide which of Amanda’s episodes Charlie acts in (2 more ways).
This shows that it is three times as likely for Amanda and Bert to share one episode as it is for them to share two episodes. Thus,
$ P(text{ A and B share 2}) = frac{4(3)}{4(12)} = frac{1}{4}$ and
$P(text{ A and B share 1}) = 1 - frac{1}{4} = frac{3}{4}$.
answered Jan 27 at 20:52
Jordan GreenJordan Green
1,146410
$endgroup$
add a comment |
$begingroup$
Since Bert appears in two episodes, he may share either one or two episodes with Amanda. (Since she appears in three, it is impossible that they share zero.)
The expected number of episodes that feature Amanda and Bert together is given by
$$ 1 cdot P(text{A and B share 1}) + 2 cdot P(text{ A and B share 2}).$$
To calculate these probabilities, we’ll consider ways to assign episodes to the actors. There are $binom{4}{3} = 4$ ways to choose which episodes Amanda appears in. (It turns out this isn’t very important though.) After assigning episodes to Amanda, we claim that there are $12$ ways of assigning episodes to the remaining actors. To see this, we take two cases.
Case 1: If Diana appears in the remaining episode, then there are $2$ ways to choose which of Bert or Charlie appears with her. The one who doesn't appear with Diana must appear with Amanda two times. There are $binom{3}{2} = 3$ ways to choose which of Amanda’s episodes he appears in. Multiplying, we see that there are $6$ assignments in this case.
Case 2: If Diana does not appear in the non-Amanda episode, then there are 3! ways to choose who acts in which of Amanda’s episodes, because the choice entails permuting Bert, Charlie, and Diana.
In all, after we have made a choice for which episodes Amanda acts in, there are $12$ ways to assign actors to the remaining episodes, so there are $48$ assignments of episodes possible overall.
Of these $48$, there are $12$ in which Bert and Amanda share two episodes - $3$ for each of the possible assignments of episodes to Amanda. This is similar to our work in Case 1, because Bert and Amanda share two episodes if and only if Charlie and Diana share the non-Amanda episode, in which case there are $binom{3}{2}$ ways to choose the episodes that Bert acts in.
Thus, we have
$$ P(text{ A and B share 2}) = frac{4(3)}{4(12)} = frac{1}{4}.$$
It follows that
$$P(text{ A and B share 1}) = 1 - frac{1}{4} = frac{3}{4}$$ and the expected number of episodes shared by Amanda and Bert is
$$ 1 left( frac{3}{4} right) + 2 left (frac{1}{4} right)= 1.25.$$
I’m adding an alternate solution that builds on (and corrects) intuition from the answer by Bram28, who wrote:
"Bert basically gets assigned $2$
out of $4$
episodes, so there are $binom{4}{2}=6$
possible assignments of $2$
episodes to Bert. Of those, $3$
will have Bert work Amanda both times, and $3$
will have Bert working with Amanda once.
So, the expected number is:
$$
frac{3}{6}⋅2+ frac{3}{6}⋅1=1.5."
$$
This is incorrect, because not all post-Amanda assignments of episodes to Bert are equally likely. Instead, some assignments to Bert create more options as to how we may assign the other actors.
If we assign episodes to Bert in a way that makes Amanda and Bert appear together twice, there is only one way to finish off the episode assignment - namely, having Charlie share Amanda’s remaining episode and having Charlie and Diana share the non-Amanda episode. However, if we assign episodes to Bert in a way that makes Amanda and Bert appear together once, then there are three ways to complete episode assignment: we could have Diana share the non-Amanda episode with Bert (1 way), or we could have Charlie share it, in which case there are two ways to decide which of Amanda’s episodes Charlie acts in (2 more ways).
This shows that it is three times as likely for Amanda and Bert to share one episode as it is for them to share two episodes. Thus,
$ P(text{ A and B share 2}) = frac{4(3)}{4(12)} = frac{1}{4}$ and
$P(text{ A and B share 1}) = 1 - frac{1}{4} = frac{3}{4}$.
answered Jan 27 at 20:52
Jordan GreenJordan Green
1,146410
$endgroup$
add a comment |
$begingroup$
Since Bert appears in two episodes, he may share either one or two episodes with Amanda. (Since she appears in three, it is impossible that they share zero.)
The expected number of episodes that feature Amanda and Bert together is given by
$$ 1 cdot P(text{A and B share 1}) + 2 cdot P(text{ A and B share 2}).$$
To calculate these probabilities, we’ll consider ways to assign episodes to the actors. There are $binom{4}{3} = 4$ ways to choose which episodes Amanda appears in. (It turns out this isn’t very important though.) After assigning episodes to Amanda, we claim that there are $12$ ways of assigning episodes to the remaining actors. To see this, we take two cases.
Case 1: If Diana appears in the remaining episode, then there are $2$ ways to choose which of Bert or Charlie appears with her. The one who doesn't appear with Diana must appear with Amanda two times. There are $binom{3}{2} = 3$ ways to choose which of Amanda’s episodes he appears in. Multiplying, we see that there are $6$ assignments in this case.
Case 2: If Diana does not appear in the non-Amanda episode, then there are 3! ways to choose who acts in which of Amanda’s episodes, because the choice entails permuting Bert, Charlie, and Diana.
In all, after we have made a choice for which episodes Amanda acts in, there are $12$ ways to assign actors to the remaining episodes, so there are $48$ assignments of episodes possible overall.
Of these $48$, there are $12$ in which Bert and Amanda share two episodes - $3$ for each of the possible assignments of episodes to Amanda. This is similar to our work in Case 1, because Bert and Amanda share two episodes if and only if Charlie and Diana share the non-Amanda episode, in which case there are $binom{3}{2}$ ways to choose the episodes that Bert acts in.
Thus, we have
$$ P(text{ A and B share 2}) = frac{4(3)}{4(12)} = frac{1}{4}.$$
It follows that
$$P(text{ A and B share 1}) = 1 - frac{1}{4} = frac{3}{4}$$ and the expected number of episodes shared by Amanda and Bert is
$$ 1 left( frac{3}{4} right) + 2 left (frac{1}{4} right)= 1.25.$$
I’m adding an alternate solution that builds on (and corrects) intuition from the answer by Bram28, who wrote:
"Bert basically gets assigned $2$
out of $4$
episodes, so there are $binom{4}{2}=6$
possible assignments of $2$
episodes to Bert. Of those, $3$
will have Bert work Amanda both times, and $3$
will have Bert working with Amanda once.
So, the expected number is:
$$
frac{3}{6}⋅2+ frac{3}{6}⋅1=1.5."
$$
This is incorrect, because not all post-Amanda assignments of episodes to Bert are equally likely. Instead, some assignments to Bert create more options as to how we may assign the other actors.
If we assign episodes to Bert in a way that makes Amanda and Bert appear together twice, there is only one way to finish off the episode assignment - namely, having Charlie share Amanda’s remaining episode and having Charlie and Diana share the non-Amanda episode. However, if we assign episodes to Bert in a way that makes Amanda and Bert appear together once, then there are three ways to complete episode assignment: we could have Diana share the non-Amanda episode with Bert (1 way), or we could have Charlie share it, in which case there are two ways to decide which of Amanda’s episodes Charlie acts in (2 more ways).
This shows that it is three times as likely for Amanda and Bert to share one episode as it is for them to share two episodes. Thus,
$ P(text{ A and B share 2}) = frac{4(3)}{4(12)} = frac{1}{4}$ and
$P(text{ A and B share 1}) = 1 - frac{1}{4} = frac{3}{4}$.
answered Jan 27 at 20:52
Jordan GreenJordan Green
1,146410
$endgroup$
Since Bert appears in two episodes, he may share either one or two episodes with Amanda. (Since she appears in three, it is impossible that they share zero.)
The expected number of episodes that feature Amanda and Bert together is given by
$$ 1 cdot P(text{A and B share 1}) + 2 cdot P(text{ A and B share 2}).$$
To calculate these probabilities, we’ll consider ways to assign episodes to the actors. There are $binom{4}{3} = 4$ ways to choose which episodes Amanda appears in. (It turns out this isn’t very important though.) After assigning episodes to Amanda, we claim that there are $12$ ways of assigning episodes to the remaining actors. To see this, we take two cases.
Case 1: If Diana appears in the remaining episode, then there are $2$ ways to choose which of Bert or Charlie appears with her. The one who doesn't appear with Diana must appear with Amanda two times. There are $binom{3}{2} = 3$ ways to choose which of Amanda’s episodes he appears in. Multiplying, we see that there are $6$ assignments in this case.
Case 2: If Diana does not appear in the non-Amanda episode, then there are 3! ways to choose who acts in which of Amanda’s episodes, because the choice entails permuting Bert, Charlie, and Diana.
In all, after we have made a choice for which episodes Amanda acts in, there are $12$ ways to assign actors to the remaining episodes, so there are $48$ assignments of episodes possible overall.
Of these $48$, there are $12$ in which Bert and Amanda share two episodes - $3$ for each of the possible assignments of episodes to Amanda. This is similar to our work in Case 1, because Bert and Amanda share two episodes if and only if Charlie and Diana share the non-Amanda episode, in which case there are $binom{3}{2}$ ways to choose the episodes that Bert acts in.
Thus, we have
$$ P(text{ A and B share 2}) = frac{4(3)}{4(12)} = frac{1}{4}.$$
It follows that
$$P(text{ A and B share 1}) = 1 - frac{1}{4} = frac{3}{4}$$ and the expected number of episodes shared by Amanda and Bert is
$$ 1 left( frac{3}{4} right) + 2 left (frac{1}{4} right)= 1.25.$$
I’m adding an alternate solution that builds on (and corrects) intuition from the answer by Bram28, who wrote:
"Bert basically gets assigned $2$
out of $4$
episodes, so there are $binom{4}{2}=6$
possible assignments of $2$
episodes to Bert. Of those, $3$
will have Bert work Amanda both times, and $3$
will have Bert working with Amanda once.
So, the expected number is:
$$
frac{3}{6}⋅2+ frac{3}{6}⋅1=1.5."
$$
This is incorrect, because not all post-Amanda assignments of episodes to Bert are equally likely. Instead, some assignments to Bert create more options as to how we may assign the other actors.
If we assign episodes to Bert in a way that makes Amanda and Bert appear together twice, there is only one way to finish off the episode assignment - namely, having Charlie share Amanda’s remaining episode and having Charlie and Diana share the non-Amanda episode. However, if we assign episodes to Bert in a way that makes Amanda and Bert appear together once, then there are three ways to complete episode assignment: we could have Diana share the non-Amanda episode with Bert (1 way), or we could have Charlie share it, in which case there are two ways to decide which of Amanda’s episodes Charlie acts in (2 more ways).
This shows that it is three times as likely for Amanda and Bert to share one episode as it is for them to share two episodes. Thus,
$ P(text{ A and B share 2}) = frac{4(3)}{4(12)} = frac{1}{4}$ and
$P(text{ A and B share 1}) = 1 - frac{1}{4} = frac{3}{4}$.
answered Jan 27 at 20:52
Jordan GreenJordan Green
1,146410
edited Jan 27 at 22:15
answered Jan 27 at 20:52
Jordan GreenJordan Green
1,146410
answered Jan 27 at 20:52
Jordan GreenJordan Green
1,146410
answered Jan 27 at 20:52
Jordan GreenJordan Green
1,146410
1,146410
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$begingroup$
I'm not sure the rules are clear. How are the actors assigned episodes? As your calculations show, two different ways of producing "random" assignments get to different probabilities. If you are saying "assignments are random in the sense that any two assignments consistent with the numbers are equally likely" then I think you might need to enumerate
$endgroup$
– lulu
Jan 27 at 20:32