I’m taking a second crack at FiveThirtyEight’s Riddler problem. I thought I solved the problem two weeks ago, but my answer wasn’t quite right.

Here is this week’s problem:

There’s an airplane with 100 seats, and there are 100 ticketed passengers each with an assigned seat. They line up to board in some random order. However, the first person to board is the worst person alive, and just sits in a random seat, without even looking at his boarding pass. Each subsequent passenger sits in his or her own assigned seat if it’s empty, but sits in a random open seat if the assigned seat is occupied. What is the probability that you, the hundredth passenger to board, finds your seat unoccupied

I’m not even going to try and figure out the math for this puzzle. My statistics skills aren’t nearly strong enough. But my coding skills are. This is an easy problem to solve with a Monte Carlo simulation.

Here’s my python solution.

And the output charted.

The last passenger has a 50% chance of getting their assigned seat.

The number of passengers who get their assigned seats is distributed around 95, with a 1% chance of everyone getting their assigned seat (in the event the first passenger randomly chooses his/her own seat).

Interestingly, it doesn’t matter how many seats are on the plane. If there are 10 or 200 seats, the last passenger still has a 50% chance of getting their assigned seat. The second last passenger has 67% chance, and the 3rd last passenger has a 75% chance.

It’s easy to derive the equation for the Xth last passenger from this, even though I can’t figure out the math to prove it. The Xth last passenger has a X/X+1 chance of getting their assigned seat. So the 5th last passenger, has a 5/6 = 83.3% chance of getting their seat.

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Well okay, spoken like a true math nerd. Me I am just happy to find out there are 100 seats and 100 passengers.

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