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MidwestAP
MidwestAP
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March 19th, 2018 at 10:00:19 AM permalink
Quote: MidwestAP

The probability that the seat assigned to the last person (Passenger 100) is available when he/she enters the plane (using Wiz's scenario) is 1 minus the sum of the probabilities that Passenger 100's seat is selected on any previous round (a round being a new passenger entering the plane):

Passenger 1 --> 1/100 = .001
Passenger 2 --> (1- prob of Passenger 1 already sat down in Passenger 100 seat) x (1/99) = .001
Passenger 3 -- > [1-(prob of Passenger 1 + prob Passenger 2) already sat down in Passenger 100 seat] x (1/98) = .001
Passenger 4 --> [1-(prob of Passenger 1 + prob Passenger 2+prob of Passenger 3) already sat down in Passenger 100 seat] x (1/97) = .001
.
.
.
Passenger 99 --> 1-(prob of Passenger 1 +... +prob of Passenger 98) already sat down in Passenger 100 seat] x (1/2) = .001

The sum of the probabilities that the seat is previously selected = 0.99

Therefore the chance that it is available = 1 - 0.99 = 1%

But this can't be correct, as there has to be a greater than 1% chance since if the first person entering the plane sits down in their assigned seat, then by rule, the last person will as well.

Therefore, I think we need to add in the 1% chance that Passenger 1 sits down in the correct seat, therefore the answer to the question is 2%.

But, I may be way off here.

Doc
Doc
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March 19th, 2018 at 10:12:20 AM permalink
Comments on MidwestAP's guess:

Quote: MidwestAP

...Passenger 2 --> (1- prob of Passenger 1 already sat down in Passenger 100 seat) x (1/99) = .001

I think you are treating every passenger as if they are making a random selection. They only do that if their own seat is already occupied.
MidwestAP
MidwestAP
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March 19th, 2018 at 10:23:43 AM permalink
Quote: Doc

Comments on MidwestAP's guess:

I think you are treating every passenger as if they are making a random selection. They only do that if their own seat is already occupied.



Yes, you are correct, let me rethink this.
Venthus
Venthus
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March 19th, 2018 at 11:05:39 AM permalink
There's a limit to how long ago we can edit a post now? Anyhow--

Now that it's morning, I'm pretty sure that 1% is a trap answer, and the key component to that is that people favor their own seats. Now I'm leaning towards 50% based on the possible combinations when person 1 is assigned s98 or s99...
Wizard
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Wizard
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March 19th, 2018 at 11:17:26 AM permalink
Quote: Romes

I'm gonna go with 1% off the top of my head, without putting too much thought in to it.



It must be more than that. There is a 1% chance passenger 1 picks the correct seat and then everybody else will also get their correct seat. However, he could also pick the wrong seat but nobody ever picks seat 100 so passenger 100 gets his seat that way. For example, if passenger 1 picks seat 2, and passenger 2 picks seat 1, then the other 98 passengers will get the correct seat.
It's not whether you win or lose; it's whether or not you had a good bet.
gamerfreak
gamerfreak
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March 19th, 2018 at 1:40:22 PM permalink
I know some people prefer answers via proofs rather than brute force simulation, but I couldn't help myself.

50% - 10 million simulations of the scenario came out to a probability of 0.5000151 which agrees with Doc’s hypothesis


I didn't spend a lot of time stepping through everything to make sure it's correct, so anyone is welcome to take a look at the code --
https://pastebin.com/uLUqdzHM
Last edited by: gamerfreak on Mar 19, 2018
SOOPOO
SOOPOO
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March 19th, 2018 at 2:55:07 PM permalink
Quote: Doc

Here are my thoughts, minus a calculated final answer:

For clarification number the passengers 1-100 based on the order they enter the plane. Number the seats 1-100 based on the passenger that each was originally assigned to.

There is a 1% chance that passenger 1 randomly picks seat 1 that he was assigned. Everything works fine, and passenger 100 gets his assigned seat.

There is a 1% chance that passenger 1 randomly picks seat 100, and passenger 100 will NOT get his assigned seat.

If passenger 1 picks seat 99 (1% chance), then all goes well until passenger 99 must randomly pick between the available seats 1 and 100, which means a 50% chance that passenger 100 gets his own seat.

It progressively gets more complicated. If passenger 1 picks seat 98 (again 1% chance), all goes well until passenger 98 makes a random selection between seats 1, 99, and 100. If he chooses 1, all is fine; if he chooses 100, there is failure; if he chooses 99, then passenger 99 makes a random selection between the remaining two seats. I believe this gives (1-(1/3)*(1+0+.5))=0.5 probability that neither of them sits in seat 100.

I think this leads to a summation of terms, each including a 1% chance of passenger 1 choosing seat number n, combined with a product of probabilities that passengers n through 99 sequentially do not randomly choose seat 100 if it is still available when they board.

I further think (1) that I don't know how to draw that summation equation on this forum, and (2) I am too lazy to calculate the total.

I'll take a wild guess that 98 of the 100 terms wind up with a (.01*0.5) probability with the other two being (.01*1) and (.01*0) for a grand total of 50% chance that passenger 100 gets his own seat.



I agree with Doc.

This happens on Southwest all the time.
gordonm888
gordonm888
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March 19th, 2018 at 3:17:38 PM permalink
A tricky question until you realize something:



For the parade of persons who randomly select a seat (including the first person), the question is: Will some person choose to sit in the first person's assigned seat before they choose to sit in the last person's assigned seat? Once a person does randomly select the assigned seat of the first person, then all the remaining open seats will be filled by the people who were assigned to them. This is because the act of randomly picking the assigned seat of the first person ends the need for any subsequent person to randomly select a seat.

On the other hand, if a person who is randomly selecting a seat picks the last person's assigned seat then that resolves the immediate question -the last person will not get their seat.

SO, the chance that the last person will sit in his own seat must be 0.5 -the probability that people picking at random will pick the first person's assigned seat before they pick the last person's assigned seat.
Doc
Doc
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March 19th, 2018 at 4:02:43 PM permalink
Quote: SOOPOO

I agree with Doc.

Although I didn't present the equation, calculate all of its terms, nor compute a final answer, I did make my "wild guess" as to the values of the terms and their sum. My guess is extremely close to gamefreak's gamerfreak's simulation result, so my confidence is growing. Too bad that I am too lazy to complete the intermediate steps.
Last edited by: Doc on Mar 19, 2018
RS
RS
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March 19th, 2018 at 4:12:48 PM permalink
Quote: gordonm888

A tricky question until you realize something:



For the parade of persons who randomly select a seat (including the first person), the question is: Will some person choose to sit in the first person's assigned seat before they choose to sit in the last person's assigned seat? Once a person does randomly select the assigned seat of the first person, then all the remaining open seats will be filled by the people who were assigned to them. This is because the act of randomly picking the assigned seat of the first person ends the need for any subsequent person to randomly select a seat.

On the other hand, if a person who is randomly selecting a seat picks the last person's assigned seat then that resolves the immediate question -the last person will not get their seat.

SO, the chance that the last person will sit in his own seat must be 0.5 -the probability that people picking at random will pick the first person's assigned seat before they pick the last person's assigned seat.


I think this is the best description. The fact a seat is eliminated every turn doesn’t change the maffs.

One seat is “good”, one is “bad”, and every other seat acts like a “push”. Like if you have a die, if you keep re-rolling it, you have a 50% chance of rolling a 1 before a 6 (and rolling 2-5 results in a push for a re-roll). Same thing happens here.
нет сговор. нет непроходимость. полный освобождение от ответственности.

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