Interesting math puzzle.

Miscellaneous and Personal Stuff I Must Share
21

Whenever the crazy passenger sits down, he’s going to cause one of three outcomes.

  1. He randomly sits in his own seat. Your seat will be available when you board.

  2. He randomly sits in your seat. This has no effect on the other passengers. But it leaves his seat empty. So you will sit in his seat when you board.

  3. He randomly sits in another passenger’s seat. When that passenger boards, they will find their seat occupied and will seek out another seat. Effectively, they will now become another crazy passenger. If they sit in the original crazy passenger’s seat, it breaks the chain and your seat will be available. If they sit in your seat, it will leave the original crazy passenger’s seat available for you. And if they sit in another passenger’s seat, they will just pass the craziness on to that passenger.

So the only way the chain of seat changes gets broken is when somebody sits in your seat or the original crazy passenger’s seat. Which means you’re going to end up sitting in either your seat or the original crazy passenger’s seat, with it being a 50/50 chance which one it will be.

22

My simulation says 50-50

I think Little Nemo is correct.

The first 2 outcomes are equally likely, and the 3rd outcome passes on the crazy and leaves the situation unresolved.

23

You have to confiders the possibility that if there is a crazy person on the manifest, there is also a sane one, or even that, by default, the rest are all sane. A sane person would just ask Mr. C what his assigned seat number is, and go and sit there. Nothing else is disrupted. There is 1/100 chance that that will be you, the only way you would not get your assigned seat.

24

I did by drawing out branching possibility trees and summing the chances of each final outcome, and found it comes out to 50/50 each time.

If there are n assigned seats when Random Dude walks in, the chance you get your seat becomes:

1/n + (1/n(n-1) + 1/n(n-2) + … 1/2n) + (1/n(n-1)(n-2) + 1/n(n-1)(n-3) + … 1/2n(n-1))…and so on down to 1/n!

All these fractions can be written in terms of n! and they will sum to give 1/2.
So no matter when Random Dude comes in, your chance of getting your seat is one half.

25

But what if United drags out somebody (not crazy person) after he sits down?

26

No Car Talk fans here? This was the puzzler from, wow, Oct. 4, 2004! Time flies! And it doesn’t even have a boarding pass!
http://www.cartalk.com/content/one-seat-left-it-yours

27

That’s a different problem. In that one, it’s always the first person who may end up in the wrong seat. In the problem I posted, the crazy person (or ticket dropper) could be anywhere in line from first place to ninety ninth. It may work out the same mathematically, but it’s two different problems.

28

That was the problem referred to here:

29

Isn’t that the same problem then, just the latter is a general case?

30

The randomness part is what’s throwing me. Instead of the problem as given, think of it like this:

You have 100 marbles: 1 blue, 1 red and 98 yellow. You have a box with 100 slots: 1 blue, 1 red and 98 yellow. When the marbles are dropped into the box the blue one can land in any slot, yellow must land on yellow unless blue has previously been dropped AND the blue spot is open or contains a yellow marble, in which case they can land anywhere, and red must go last and land on whatever is left.

Now, drop the blue marble first. There is a 1 in 100 chance it lands in the blue (correct) slot, 1 in 100 that it lands in the red slot and 98 in 100 that it lands in any yellow spot.
*Scenario 1 and 2 resolve to 50/50 - in one you are assured of red being open at the end, in the other there is no chance.
*In the other 98 scenarios blue lands in a random yellow spot. Now drop the 98 yellow marbles one at a time to randomly fill the remaining 99 slots (97 yellow, 1 blue and 1 red). The odds of red being the only slot open after all yellow marbles are dropped is 1 in 98. Given this, the overall odds can’t be 50%

If the blue marble is dropped as the 99th marble then the odds of red being open are 50%, since by the rules all 98 yellows will be appropriately filled and only blue and red remain. Otherwise the odds change with the drop order. Or what am I missing?

31

This isn’t what the problem says, though. The 98 yellow marbles aren’t interchangeable. A yellow marble only falls randomly if its own spot is already filled.

32

Ah, that’s what I missed!. The 50/50 answer makes more sense now.

33

I agree with the 50% answer. Here’s how I did it:

Step 1
Three scenarios
A: Crazy sits in Crazy’s Seat (1% Chance) and if this happens I WIN (because everyone sits in their correct seat)
B: Crazy sits in MY Seat (1% chance) and if this happens I LOSE (obvious reason)
C: Crazy sits in “OTHER” Seat (Let’s call the other seat “Jack’s Seat”: (98% chance, and we still don’t know if I win)

Step 2
I: Jack sits in Crazy’s Seat (50% chance x 98% from above = 49% of all outcomes)–I WIN
II: Jack sits in MY Seat (50% chance x 98% from above = 49% of all outcomes)–I LOSE

Add up the “I WIN” scenarios = 50%