Monty Hall Problem

Anything you want to talk about.

Is it to your advantage to switch your choice?

Your odds of winning are increased by switching.
38
76%
Your odds of winning are increased by keeping your original choice.
3
6%
Your odds of winning are the same (50/50) whether you switch or not.
9
18%
 
Total votes : 50

Re: Monty Hall Problem

Postby Tarsier » 17 Nov 2017, 05:57

@NoPun:

Why?

I'm not proposing this as a new one to discuss, but thought it was worth posting what is considered the hardest logic problem ever:

Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly and False always speaks falsely. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which. Whether Random says ja or da should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he says ja; if tails, he says da.*

*Note that this last sentence is a modification from the original publication of the problem, forcing a more complex answer.


This one was well over my head. I remember working through some published answers to it and having to create charts to follow the logic.
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Re: Monty Hall Problem

Postby NoPunIn10Did » 17 Nov 2017, 06:29

Whoops! Got the airplane one backward, I think. This seems like it's a misdirection problem. The last "person" to board is not one of the passengers, but rather one of the crew, who have designated seats. So the chance is 100%
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Re: Monty Hall Problem

Postby Jegpeg » 17 Nov 2017, 09:50

NoPunIn10Did wrote:
Whoops! Got the airplane one backward, I think. This seems like it's a misdirection problem. The last "person" to board is not one of the passengers, but rather one of the crew, who have designated seats. So the chance is 100%


Sorry just to clarify the last "person" is a passenger (whenever I have travelled by plane the crew got on board before the passengers but I should have been more specific with the question). Neither answer given so far is correct.
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Re: Monty Hall Problem

Postby Tarsier » 17 Nov 2017, 15:32

The airplane one doesn't require fancy probability mathematics, but pretend it did. For what event(s) would you attempt to calculate probability?

The odds of X (and Y?) happening when the second passenger boards are _______________?
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Re: Monty Hall Problem

Postby sevenbrokenbricks » 17 Nov 2017, 21:11

Re: the airplane problem:
50% chance the final passenger sits in their own seat.

The first passenger preemptively bounces someone out of their seat. That person, in choosing randomly among unfilled seats does one of the following:
    sits in the first passenger's seat (and nobody else gets bounced, hence the final passenger sits in their own seat),
    sits in the final passenger's seat (and the final passenger gets bounced),
    or sits in someone else's seat entirely (in which case they're bouncing someone else out of their seat, and the decision gets passed off to them).
No matter how many times the decision gets passed off, one seat is a winner, and one seat is a loser. The others just ask the question again with fewer seats.

And so help me if you tell me that it's 100% because the final passenger is always sitting in his own seat because it's his seat by definition because he's the one sitting in it, I will strangle someone :twisted:
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Re: Monty Hall Problem

Postby NoPunIn10Did » 17 Nov 2017, 22:41

sevenbrokenbricks wrote:Re: the airplane problem:
50% chance the final passenger sits in their own seat.

The first passenger preemptively bounces someone out of their seat. That person, in choosing randomly among unfilled seats does one of the following:
    sits in the first passenger's seat (and nobody else gets bounced, hence the final passenger sits in their own seat),
    sits in the final passenger's seat (and the final passenger gets bounced),
    or sits in someone else's seat entirely (in which case they're bouncing someone else out of their seat, and the decision gets passed off to them).
No matter how many times the decision gets passed off, one seat is a winner, and one seat is a loser. The others just ask the question again with fewer seats.

And so help me if you tell me that it's 100% because the final passenger is always sitting in his own seat because it's his seat by definition because he's the one sitting in it, I will strangle someone :twisted:


Naw, I tried the problem with 3 people, and the answer for 3 was 1/4.

Person A is supposed to sit in seat a, B in b, C in c.

Person A boards and takes seat b or c (50% chance for either). If A took seat b, then there's a 50% chance that B will take seat a, leaving seat c available for C. This is the only scenario that gives C seat c.

(50%)*(50%) = 1/4.

It then gets more complex with 4 people, and it yields 1/3 as the answer. At 5 people I was getting a result of 35/96, though that may be a math error.
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Re: Monty Hall Problem

Postby Jegpeg » 17 Nov 2017, 23:38

NoPunIn10Did wrote:
sevenbrokenbricks wrote:Re: the airplane problem:
50% chance the final passenger sits in their own seat.

The first passenger preemptively bounces someone out of their seat. That person, in choosing randomly among unfilled seats does one of the following:
    sits in the first passenger's seat (and nobody else gets bounced, hence the final passenger sits in their own seat),
    sits in the final passenger's seat (and the final passenger gets bounced),
    or sits in someone else's seat entirely (in which case they're bouncing someone else out of their seat, and the decision gets passed off to them).
No matter how many times the decision gets passed off, one seat is a winner, and one seat is a loser. The others just ask the question again with fewer seats.

And so help me if you tell me that it's 100% because the final passenger is always sitting in his own seat because it's his seat by definition because he's the one sitting in it, I will strangle someone :twisted:


Naw, I tried the problem with 3 people, and the answer for 3 was 1/4.

Person A is supposed to sit in seat a, B in b, C in c.

Person A boards and takes seat b or c (50% chance for either). If A took seat b, then there's a 50% chance that B will take seat a, leaving seat c available for C. This is the only scenario that gives C seat c.

(50%)*(50%) = 1/4.

It then gets more complex with 4 people, and it yields 1/3 as the answer. At 5 people I was getting a result of 35/96, though that may be a math error.


I'll be honest I mis-remembered the problem and the in the original format the first passenger pick a seat at random (not a different seat) in that case the solution is 50%.

As a posted however the solution is only one step more complicated to solve for a general number of people. 1/3 for 4 is correct, 35/96 is not correct for 5 people.
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Re: Monty Hall Problem

Postby Tarsier » 18 Nov 2017, 00:53

50 percent is correct.
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Re: Monty Hall Problem

Postby Jegpeg » 18 Nov 2017, 02:06

Tarsier wrote:50 percent is correct.


50% is correct if the 1st person selects any seat at random (including their own) as Nopunin showed if the first person only chooses from the seats that are not his own the odds are not 50%
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Re: Monty Hall Problem

Postby Tarsier » 18 Nov 2017, 03:47

Jegpeg wrote:
Tarsier wrote:50 percent is correct.


50% is correct if the 1st person selects any seat at random (including their own) as Nopunin showed if the first person only chooses from the seats that are not his own the odds are not 50%


Yes they are. Seven nailed it.

When someone boards the plane, he will either find his seat occupied or not.

If not, he'll sit in his seat. That has no bearing on the last passenger, as it doesn't take his seat.

If his seat is occupied, he'll take another random seat.

None of the seats on the plane are consequential except for two. If anyone sits in the original passenger's seat, the cycle is closed. Every following passenger will have his own seat available, including the last passenger. OR if anyone sits in the seat assigned to the last passenger, the conclusion is also resolved.

The odds that a passenger will choose the first passenger's vacant seat are equal to the odds that he will choose the last passenger's vacant seat. Since each has equal odds, and those are the only two seats that will affect the outcome of the puzzle, the probability is 50/50.

Let's try it on a smaller scale, with four passengers (with seat a being assigned to the first passenger and seat d assigned to the last):

Passenger A sits in seat b.

If Passenger B sits in seat c, then we simply move on to the next passenger in line. Passenger C has a 50/50 choice of the remaining two seats, a or d.

If Passenger B doesn't sit in seat c, then he has a 50/50 shot of either a or d.

You could do this with five passengers, (a and e being the first and last seat). Seats c and d simply pass the buck to the next passenger. However, the first person who does not select one of the non-consequential seats has a choice of seat a or e.
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