What are the odds of sitting next to the same person on two flights?
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My wife left on a business trip this morning. 20 people from the same company caught two consecutive flights. Each person checked in independently, yet my wife ended up sitting next to the same colleague on both flights!
What are the odds?
Assume both aeroplanes had 150 seats, in 3+3 configuration, in 25 rows. It's not exactly right but will do for the purposes of the exercise.
Assume also that all other passengers checked in independently, so there are no couples choosing or being assigned seats next to each other, thus changing the odds. This isn't right either, but will also do for the purposes of the exercise.
My wife and colleague were sitting next to each other, not across the aisle from each other.
probability
asked Sep 3 at 8:27
Colin 't Hart
1234
add a comment |Â
up vote
4
down vote
favorite
My wife left on a business trip this morning. 20 people from the same company caught two consecutive flights. Each person checked in independently, yet my wife ended up sitting next to the same colleague on both flights!
What are the odds?
Assume both aeroplanes had 150 seats, in 3+3 configuration, in 25 rows. It's not exactly right but will do for the purposes of the exercise.
Assume also that all other passengers checked in independently, so there are no couples choosing or being assigned seats next to each other, thus changing the odds. This isn't right either, but will also do for the purposes of the exercise.
My wife and colleague were sitting next to each other, not across the aisle from each other.
probability
asked Sep 3 at 8:27
Colin 't Hart
1234
3
An important assumption is missing: That the seats are assigned uniformly randomly. (That may well not be the case, e.g. they may have ordered certain meals, or the colleague may have requested a window seat, etc.)
– joriki
Sep 3 at 9:14
Colin, you should familiarize yourself with the well-known "birthday problem", as one aspect of this type of issue.
– Fattie
Sep 3 at 11:52
1
@joriki: It's also conceivable that the seats were all booked at the same time, and as a result the airline assigned all the company's seats in the same part of the plane (since they shared a booking). As an extreme example, if the company's seats form a block in four rows of the plane, the odds go way up.
– Michael Seifert
Sep 3 at 14:31
1
@Fattie: I don't see the connection to the birthday problem.
– joriki
Sep 3 at 14:36
@MichaelSeifert Except in this case, we manually went in and chose my wife different seats. :-)
– Colin 't Hart
Sep 3 at 15:03
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
My wife left on a business trip this morning. 20 people from the same company caught two consecutive flights. Each person checked in independently, yet my wife ended up sitting next to the same colleague on both flights!
What are the odds?
Assume both aeroplanes had 150 seats, in 3+3 configuration, in 25 rows. It's not exactly right but will do for the purposes of the exercise.
Assume also that all other passengers checked in independently, so there are no couples choosing or being assigned seats next to each other, thus changing the odds. This isn't right either, but will also do for the purposes of the exercise.
My wife and colleague were sitting next to each other, not across the aisle from each other.
probability
asked Sep 3 at 8:27
Colin 't Hart
1234
My wife left on a business trip this morning. 20 people from the same company caught two consecutive flights. Each person checked in independently, yet my wife ended up sitting next to the same colleague on both flights!
What are the odds?
Assume both aeroplanes had 150 seats, in 3+3 configuration, in 25 rows. It's not exactly right but will do for the purposes of the exercise.
Assume also that all other passengers checked in independently, so there are no couples choosing or being assigned seats next to each other, thus changing the odds. This isn't right either, but will also do for the purposes of the exercise.
My wife and colleague were sitting next to each other, not across the aisle from each other.
probability
probability
asked Sep 3 at 8:27
Colin 't Hart
1234
asked Sep 3 at 8:27
Colin 't Hart
1234
asked Sep 3 at 8:27
Colin 't Hart
1234
asked Sep 3 at 8:27
Colin 't Hart
1234
asked Sep 3 at 8:27
Colin 't Hart
1234
1234
3
An important assumption is missing: That the seats are assigned uniformly randomly. (That may well not be the case, e.g. they may have ordered certain meals, or the colleague may have requested a window seat, etc.)
– joriki
Sep 3 at 9:14
Colin, you should familiarize yourself with the well-known "birthday problem", as one aspect of this type of issue.
– Fattie
Sep 3 at 11:52
1
@joriki: It's also conceivable that the seats were all booked at the same time, and as a result the airline assigned all the company's seats in the same part of the plane (since they shared a booking). As an extreme example, if the company's seats form a block in four rows of the plane, the odds go way up.
– Michael Seifert
Sep 3 at 14:31
1
@Fattie: I don't see the connection to the birthday problem.
– joriki
Sep 3 at 14:36
@MichaelSeifert Except in this case, we manually went in and chose my wife different seats. :-)
– Colin 't Hart
Sep 3 at 15:03
add a comment |Â
3
An important assumption is missing: That the seats are assigned uniformly randomly. (That may well not be the case, e.g. they may have ordered certain meals, or the colleague may have requested a window seat, etc.)
– joriki
Sep 3 at 9:14
Colin, you should familiarize yourself with the well-known "birthday problem", as one aspect of this type of issue.
– Fattie
Sep 3 at 11:52
1
@joriki: It's also conceivable that the seats were all booked at the same time, and as a result the airline assigned all the company's seats in the same part of the plane (since they shared a booking). As an extreme example, if the company's seats form a block in four rows of the plane, the odds go way up.
– Michael Seifert
Sep 3 at 14:31
1
@Fattie: I don't see the connection to the birthday problem.
– joriki
Sep 3 at 14:36
@MichaelSeifert Except in this case, we manually went in and chose my wife different seats. :-)
– Colin 't Hart
Sep 3 at 15:03
3
3
An important assumption is missing: That the seats are assigned uniformly randomly. (That may well not be the case, e.g. they may have ordered certain meals, or the colleague may have requested a window seat, etc.)
– joriki
Sep 3 at 9:14
An important assumption is missing: That the seats are assigned uniformly randomly. (That may well not be the case, e.g. they may have ordered certain meals, or the colleague may have requested a window seat, etc.)
– joriki
Sep 3 at 9:14
Colin, you should familiarize yourself with the well-known "birthday problem", as one aspect of this type of issue.
– Fattie
Sep 3 at 11:52
Colin, you should familiarize yourself with the well-known "birthday problem", as one aspect of this type of issue.
– Fattie
Sep 3 at 11:52
1
1
@joriki: It's also conceivable that the seats were all booked at the same time, and as a result the airline assigned all the company's seats in the same part of the plane (since they shared a booking). As an extreme example, if the company's seats form a block in four rows of the plane, the odds go way up.
– Michael Seifert
Sep 3 at 14:31
@joriki: It's also conceivable that the seats were all booked at the same time, and as a result the airline assigned all the company's seats in the same part of the plane (since they shared a booking). As an extreme example, if the company's seats form a block in four rows of the plane, the odds go way up.
– Michael Seifert
Sep 3 at 14:31
1
1
@Fattie: I don't see the connection to the birthday problem.
– joriki
Sep 3 at 14:36
@Fattie: I don't see the connection to the birthday problem.
– joriki
Sep 3 at 14:36
@MichaelSeifert Except in this case, we manually went in and chose my wife different seats. :-)
– Colin 't Hart
Sep 3 at 15:03
@MichaelSeifert Except in this case, we manually went in and chose my wife different seats. :-)
– Colin 't Hart
Sep 3 at 15:03
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
8
down vote
accepted
Your wife has probability $frac23$ to sit in a chair that has only $1$ chair next to it, and has probability $frac13$ to sit in a chair that has $2$ other chairs next to it.
The probability that at the first flight your wife had no colleagues sitting next to her is:$$p_0:=frac23fracbinom14819binom10binom14919+frac13fracbinom14719binom20binom14919=frac23frac130149+frac13frac130cdot129149cdot148=frac55250
66156approx0.835147228
$$
The probability that at the first flight your wife had $2$ colleagues sitting next to her is: $$p_2:=frac13fracbinom14717binom22binom14919=frac13frac19cdot18149cdot148=frac34266156approx0.005169599
$$
The probability that at the first flight your wife had exactly $1$ colleague sitting next to her is:$$p_1:=1-p_0-p_2=frac1056466156approx0.159683173
$$
The probability that at the first flight your wife had a colleague sitting next to her, and that this same person was sitting next to her the second flight is:$$p_1timesleft(frac23frac1149+frac13frac2149right)+p_2timesleft(frac23frac2149+frac13left(1-frac147149frac146148right)right)=$$$$frac1056466156frac4447+frac34266156frac118266156=frac6658132
4376616336
approx0.001521$$
answered Sep 3 at 9:05
drhab
89k541122
But the people need some decimals.
– Pedro Tamaroff♦
Sep 3 at 9:07
1
You're neglecting the small probability that she had two colleagues sitting next to her on the first flight.
– joriki
Sep 3 at 9:12
@joriki thank you for attending me. I will try to fix that.
– drhab
Sep 3 at 9:15
@joriki Could you give it a second look please Hawkeye?
– drhab
Sep 3 at 10:05
Yes, looks great :-)
– joriki
Sep 3 at 10:12
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Your wife has probability $frac23$ to sit in a chair that has only $1$ chair next to it, and has probability $frac13$ to sit in a chair that has $2$ other chairs next to it.
The probability that at the first flight your wife had no colleagues sitting next to her is:$$p_0:=frac23fracbinom14819binom10binom14919+frac13fracbinom14719binom20binom14919=frac23frac130149+frac13frac130cdot129149cdot148=frac55250
66156approx0.835147228
$$
The probability that at the first flight your wife had $2$ colleagues sitting next to her is: $$p_2:=frac13fracbinom14717binom22binom14919=frac13frac19cdot18149cdot148=frac34266156approx0.005169599
$$
The probability that at the first flight your wife had exactly $1$ colleague sitting next to her is:$$p_1:=1-p_0-p_2=frac1056466156approx0.159683173
$$
The probability that at the first flight your wife had a colleague sitting next to her, and that this same person was sitting next to her the second flight is:$$p_1timesleft(frac23frac1149+frac13frac2149right)+p_2timesleft(frac23frac2149+frac13left(1-frac147149frac146148right)right)=$$$$frac1056466156frac4447+frac34266156frac118266156=frac6658132
4376616336
approx0.001521$$
answered Sep 3 at 9:05
drhab
89k541122
But the people need some decimals.
– Pedro Tamaroff♦
Sep 3 at 9:07
1
You're neglecting the small probability that she had two colleagues sitting next to her on the first flight.
– joriki
Sep 3 at 9:12
@joriki thank you for attending me. I will try to fix that.
– drhab
Sep 3 at 9:15
@joriki Could you give it a second look please Hawkeye?
– drhab
Sep 3 at 10:05
Yes, looks great :-)
– joriki
Sep 3 at 10:12
 |Â
show 1 more comment
up vote
8
down vote
accepted
Your wife has probability $frac23$ to sit in a chair that has only $1$ chair next to it, and has probability $frac13$ to sit in a chair that has $2$ other chairs next to it.
The probability that at the first flight your wife had no colleagues sitting next to her is:$$p_0:=frac23fracbinom14819binom10binom14919+frac13fracbinom14719binom20binom14919=frac23frac130149+frac13frac130cdot129149cdot148=frac55250
66156approx0.835147228
$$
The probability that at the first flight your wife had $2$ colleagues sitting next to her is: $$p_2:=frac13fracbinom14717binom22binom14919=frac13frac19cdot18149cdot148=frac34266156approx0.005169599
$$
The probability that at the first flight your wife had exactly $1$ colleague sitting next to her is:$$p_1:=1-p_0-p_2=frac1056466156approx0.159683173
$$
The probability that at the first flight your wife had a colleague sitting next to her, and that this same person was sitting next to her the second flight is:$$p_1timesleft(frac23frac1149+frac13frac2149right)+p_2timesleft(frac23frac2149+frac13left(1-frac147149frac146148right)right)=$$$$frac1056466156frac4447+frac34266156frac118266156=frac6658132
4376616336
approx0.001521$$
answered Sep 3 at 9:05
drhab
89k541122
But the people need some decimals.
– Pedro Tamaroff♦
Sep 3 at 9:07
1
You're neglecting the small probability that she had two colleagues sitting next to her on the first flight.
– joriki
Sep 3 at 9:12
@joriki thank you for attending me. I will try to fix that.
– drhab
Sep 3 at 9:15
@joriki Could you give it a second look please Hawkeye?
– drhab
Sep 3 at 10:05
Yes, looks great :-)
– joriki
Sep 3 at 10:12
 |Â
show 1 more comment
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Your wife has probability $frac23$ to sit in a chair that has only $1$ chair next to it, and has probability $frac13$ to sit in a chair that has $2$ other chairs next to it.
The probability that at the first flight your wife had no colleagues sitting next to her is:$$p_0:=frac23fracbinom14819binom10binom14919+frac13fracbinom14719binom20binom14919=frac23frac130149+frac13frac130cdot129149cdot148=frac55250
66156approx0.835147228
$$
The probability that at the first flight your wife had $2$ colleagues sitting next to her is: $$p_2:=frac13fracbinom14717binom22binom14919=frac13frac19cdot18149cdot148=frac34266156approx0.005169599
$$
The probability that at the first flight your wife had exactly $1$ colleague sitting next to her is:$$p_1:=1-p_0-p_2=frac1056466156approx0.159683173
$$
The probability that at the first flight your wife had a colleague sitting next to her, and that this same person was sitting next to her the second flight is:$$p_1timesleft(frac23frac1149+frac13frac2149right)+p_2timesleft(frac23frac2149+frac13left(1-frac147149frac146148right)right)=$$$$frac1056466156frac4447+frac34266156frac118266156=frac6658132
4376616336
approx0.001521$$
answered Sep 3 at 9:05
drhab
89k541122
Your wife has probability $frac23$ to sit in a chair that has only $1$ chair next to it, and has probability $frac13$ to sit in a chair that has $2$ other chairs next to it.
The probability that at the first flight your wife had no colleagues sitting next to her is:$$p_0:=frac23fracbinom14819binom10binom14919+frac13fracbinom14719binom20binom14919=frac23frac130149+frac13frac130cdot129149cdot148=frac55250
66156approx0.835147228
$$
The probability that at the first flight your wife had $2$ colleagues sitting next to her is: $$p_2:=frac13fracbinom14717binom22binom14919=frac13frac19cdot18149cdot148=frac34266156approx0.005169599
$$
The probability that at the first flight your wife had exactly $1$ colleague sitting next to her is:$$p_1:=1-p_0-p_2=frac1056466156approx0.159683173
$$
The probability that at the first flight your wife had a colleague sitting next to her, and that this same person was sitting next to her the second flight is:$$p_1timesleft(frac23frac1149+frac13frac2149right)+p_2timesleft(frac23frac2149+frac13left(1-frac147149frac146148right)right)=$$$$frac1056466156frac4447+frac34266156frac118266156=frac6658132
4376616336
approx0.001521$$
answered Sep 3 at 9:05
drhab
89k541122
edited Sep 3 at 11:15
answered Sep 3 at 9:05
drhab
89k541122
answered Sep 3 at 9:05
drhab
89k541122
answered Sep 3 at 9:05
drhab
89k541122
89k541122
But the people need some decimals.
– Pedro Tamaroff♦
Sep 3 at 9:07
1
You're neglecting the small probability that she had two colleagues sitting next to her on the first flight.
– joriki
Sep 3 at 9:12
@joriki thank you for attending me. I will try to fix that.
– drhab
Sep 3 at 9:15
@joriki Could you give it a second look please Hawkeye?
– drhab
Sep 3 at 10:05
Yes, looks great :-)
– joriki
Sep 3 at 10:12
 |Â
show 1 more comment
But the people need some decimals.
– Pedro Tamaroff♦
Sep 3 at 9:07
1
You're neglecting the small probability that she had two colleagues sitting next to her on the first flight.
– joriki
Sep 3 at 9:12
@joriki thank you for attending me. I will try to fix that.
– drhab
Sep 3 at 9:15
@joriki Could you give it a second look please Hawkeye?
– drhab
Sep 3 at 10:05
Yes, looks great :-)
– joriki
Sep 3 at 10:12
But the people need some decimals.
– Pedro Tamaroff♦
Sep 3 at 9:07
But the people need some decimals.
– Pedro Tamaroff♦
Sep 3 at 9:07
1
1
You're neglecting the small probability that she had two colleagues sitting next to her on the first flight.
– joriki
Sep 3 at 9:12
You're neglecting the small probability that she had two colleagues sitting next to her on the first flight.
– joriki
Sep 3 at 9:12
@joriki thank you for attending me. I will try to fix that.
– drhab
Sep 3 at 9:15
@joriki thank you for attending me. I will try to fix that.
– drhab
Sep 3 at 9:15
@joriki Could you give it a second look please Hawkeye?
– drhab
Sep 3 at 10:05
@joriki Could you give it a second look please Hawkeye?
– drhab
Sep 3 at 10:05
Yes, looks great :-)
– joriki
Sep 3 at 10:12
Yes, looks great :-)
– joriki
Sep 3 at 10:12
 |Â
show 1 more comment
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3
An important assumption is missing: That the seats are assigned uniformly randomly. (That may well not be the case, e.g. they may have ordered certain meals, or the colleague may have requested a window seat, etc.)
– joriki
Sep 3 at 9:14
Colin, you should familiarize yourself with the well-known "birthday problem", as one aspect of this type of issue.
– Fattie
Sep 3 at 11:52
1
@joriki: It's also conceivable that the seats were all booked at the same time, and as a result the airline assigned all the company's seats in the same part of the plane (since they shared a booking). As an extreme example, if the company's seats form a block in four rows of the plane, the odds go way up.
– Michael Seifert
Sep 3 at 14:31
1
@Fattie: I don't see the connection to the birthday problem.
– joriki
Sep 3 at 14:36
@MichaelSeifert Except in this case, we manually went in and chose my wife different seats. :-)
– Colin 't Hart
Sep 3 at 15:03