An interview question about probability. You enter a stadium with 1000 seats. you are told that under one of thechairs is a prize. You choose a seat randomly.q1, what's the probability your seat has a prizeq2, now 990 seats are removed, not including your seat or the one with theprize under it. There are 10 seats left. What's the probability that yourseat contains the prize now?My answer to q1: 1/1000to q2: 1/10. right ? Thanks a lot!

Since the seat you are sitting in will never be removed, the probability does not change. The answer is 1/1000.

Look up the "Monty Hall" problem - same thing, different numbers.

q1 . .. . 1 /1000 q2 . . . . . 0the way the question is written the initial guess would be 1/1000 however it says neither your chair OR the one with the prize will be removed. . .therefore i would argue that the question is implying your chair does not have the prize , hence 0

q1 = 1/1000q2 = 99/1000

I am sorry. q2=1/10 of course.

- prebenbang
**Posts:**1**Joined:**

What bwarren said.q1: 1/1000q2: 1/1000The experiment is: pick a seat, watch as 990 seats are removed, and then look under your chair.This is no different than the experiment: pick a seat, then look under your chair.You will find a prize no more often in the first one than in the second.

Last edited by prebenbang on March 12th, 2012, 11:00 pm, edited 1 time in total.

- katastrofa
**Posts:**8944**Joined:****Location:**Alpha Centauri

The probability doesn't change. The Monty Hall problem is about switching your choice when the probability space shrinks after removing some losing events, and you get a better chance to find the award elsewhere. Here you cannot switch, so the chance of finding the prize under all the remaining seats grows, but your chance remains the same.a

Last edited by katastrofa on May 9th, 2012, 10:00 pm, edited 1 time in total.

- gauravquant
**Posts:**38**Joined:**

QuoteOriginally posted by: goekhanI am sorry. q2=1/10 of course.You could have edited your earlier reply. Just a suggestion.

I think both are 1/1000. Similar to the "Monty Hall" problem, the probability of getting the prize does not change after the host deliberately removed one loser, which is 1/3. However, in "Monty Hall", you can choose to switch; hence, if you choose a loser at the beginning, you will win the prize after the switch. The prob of choosing a loser is 2/3, which does not change either. As a result, in "Monty Hall", the probability of winning with switch is 2/3.Both this problem and "Monty Hall" do not change the probability after the remove.

Yes, no doubt. The probability does not change at all. Both are 1/1000.

The correct answer depends on the process used to remove the 990 chairs. There are two reasonable possibilities:1. The chairs were removed by someone who knows the location of the prize, and they are deliberately not removing the prize chair.2. The chairs were removed at random (or by someone who doesn't know the location of the prize), and you observed that there was no prize in any of the removed chairs. In this case, the fact that the prize chair has not been removed is a lucky accident.In case 1, p=1/1000 you have the prize. In case 2, p=1/10 you have the prize.

Last edited by yegulalp on July 10th, 2012, 10:00 pm, edited 1 time in total.

With slight re-wording this would be a generalization of the Monte Hall

problem. E.g. There are N chairs with 1 prize randomly placed beneath.

You pick a chair at random. K chairs are removed--including neither your

choice nor the winning chair.

Q1: What is the probability your chair is the winner after removing K chairs?

Q2: What is the probability a given chair other than yours is the winner?

Q3: Do you switch?

A1: Everyone agrees that the probability your chair wins is still 1/N.

A2: Hence, the probability a chair other than yours wins is (N-1)/N

This probability is distributed over (N-K-1) chairs, so the probability

any one of them wins is [(N-1)/(N)] * [1/(N-K-1)]

A3: Yes iff A2 > A1

Ex: (Monte Hall)

N=3, K=1

A1 = 1/3

A2 = 2/3

A2 > A1 so switch

If you phrase Q2 more indirectly, the problem seems harder.

problem. E.g. There are N chairs with 1 prize randomly placed beneath.

You pick a chair at random. K chairs are removed--including neither your

choice nor the winning chair.

Q1: What is the probability your chair is the winner after removing K chairs?

Q2: What is the probability a given chair other than yours is the winner?

Q3: Do you switch?

A1: Everyone agrees that the probability your chair wins is still 1/N.

A2: Hence, the probability a chair other than yours wins is (N-1)/N

This probability is distributed over (N-K-1) chairs, so the probability

any one of them wins is [(N-1)/(N)] * [1/(N-K-1)]

A3: Yes iff A2 > A1

Ex: (Monte Hall)

N=3, K=1

A1 = 1/3

A2 = 2/3

A2 > A1 so switch

If you phrase Q2 more indirectly, the problem seems harder.

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