Crazy guy on the airplane
The answer is exactly 50%.
It is clear enough to realize that when the crazy person comes on he has an even chance of choosing seat 1, which ensures that passenger 100 will get their seat, or seat 100 which will ensure that passenger 100 will never get it.
What requires the jump is to realize that if the crazy passenger choses another seat he is simply passing the original choice further down the line. At some stage before passenger 100 boards the plane either seat 1 or seat 100 will have to be taken.
Thursday, October 24, 2002
The chances are 50%.
Let us denote passengers as using Pi and seats using Si, where i = 1..100. Crazy guy is P1.
case 1) P1 takes S1. P100 gets S100 with Prob = 1/100
case 2) P1 takes Si (for any i between 1 and 99).
Prob = 1/100 * 1/2.
consider P1 taking S99. S1 and S100 are left after P2 through P98 board the plane.
Prob(P100 gets S100) = Prob(P99 takes S1) = 1/2 = result1 (say)
consider P1 taking S98. S1, S99 and S100 are left for P98.
Prob(P100 gets S100) = Prob(P98 takes S1) + Prob(P98 takes S99)*Prob(S100 gets P100 given that P99 finds S99 occupied) = 1/3 + 1/3*result1 = 1/2 = result2 (say)
Consider P1 taking S97. S1, S98, S99 and S100 are left for P97.
Prob(P100 gets S100) = Prob(P97 takes S1) +
Prob(P97 takes S98)*result2 + Prob(P97 takes S99)*result1 = 1/4 + 1/4*1/2 + 1/4*1/2 = 1/2.
If P1 takes P96 we can show that the probability is
1/5 + 1/5*1/2 + 1/5*1/2 + 1/5*1/2 = 1/2
In general this probability can be written as
1/n + 1/n*1/2 *(n-2) = 1/2.
The overall probability for this case is then
1/100*1/2*98 = 49/100
case 3) P1 takes S100. In this case Prob(P100 gets S100) is zero.
So the total probability is
1/100 + 49/100 = 0.5
Wednesday, December 4, 2002
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