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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.

Colin 't Hart
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    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 03 '18 at 09:14
  • Colin, you should familiarize yourself with the well-known "birthday problem", as one aspect of this type of issue. – Fattie Sep 03 '18 at 11:52
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    @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 03 '18 at 14:31
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    @Fattie: I don't see the connection to the birthday problem. – joriki Sep 03 '18 at 14:36
  • @MichaelSeifert Except in this case, we manually went in and chose my wife different seats. :-) – Colin 't Hart Sep 03 '18 at 15:03

1 Answers1

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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:$$p_0:=\frac23\frac{\binom{148}{19}\binom10}{\binom{149}{19}}+\frac13\frac{\binom{147}{19}\binom{2}{0}}{\binom{149}{19}}=\frac23\frac{130}{149}+\frac13\frac{130\cdot129}{149\cdot148}=\frac{55250 }{66156}\approx0.835147228 $$

The probability that at the first flight your wife had $2$ colleagues sitting next to her is: $$p_2:=\frac13\frac{\binom{147}{17}\binom22}{\binom{149}{19}}=\frac13\frac{19\cdot18}{149\cdot148}=\frac{342}{66156}\approx0.005169599 $$

The probability that at the first flight your wife had exactly $1$ colleague sitting next to her is:$$p_1:=1-p_0-p_2=\frac{10564}{66156}\approx0.159683173 $$

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:$$p_1\times\left(\frac23\frac1{149}+\frac13\frac2{149}\right)+p_2\times\left(\frac23\frac2{149}+\frac13\left(1-\frac{147}{149}\frac{146}{148}\right)\right)=$$$$\frac{10564}{66156}\frac{4}{447}+\frac{342}{66156}\frac{1182}{66156}=\frac{6658132 }{4376616336 }\approx0.001521$$

drhab
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