Comments on: The Monty Hall problem

By: KK

KK — Fri, 30 Oct 2009 22:50:28 +0000

The original statement is that there are three unknown doors. When you “open” one door the original statement is false. There are two unknowns. In this case door a and c.
question should be
which statement is true
a=1
or
b=1
or
c=1
a+b+c=1
if
b = 0
then
a or b =1
Your chances are 50%

Therefore door and and c are equal.

With the common system if you choose a and they open door a and it is found to be correct you should still change because you odds are higher with two incorrect choices than the correct.
It goes back to the opening statement. If we have three unknowns. The moment you only have two unknowns this becomes false and you have an entirely new equation.

By: Philip Brocoum

Philip Brocoum — Wed, 07 Oct 2009 12:57:02 +0000

Here’s a good video explaining the problem in plain English:

http://www.youtube.com/watch?v=koPBkK_Ra-k

By: ob

ob — Tue, 21 Jul 2009 20:00:12 +0000

Heh, yeah. I read the previous one and couldn’t understand a word. This one makes sense.

By: Paul Tiemens

Paul Tiemens — Tue, 21 Jul 2009 19:06:06 +0000

Unfortunately the spacing in the last five lines above did not transmit as a table and looks like the ramblings of a mad scientist – how apt you might say?
There are three possible combinations for the three doors 1,2&3 ;- Lose,Lose,Win. Lose,Win,Lose, and Win,Lose,Lose.
What it is meant to show is that each door has the possibility of one Win and two Loses as for example your chosen door – say door 3. But two doors together as Monty has – say doors (1 & 2) have a total possibility of two wins and four loses. If as you say he always opens a losing door then he is left with two wins and only one lose between the two doors.

By: Paul Tiemens

Paul Tiemens — Tue, 21 Jul 2009 18:33:46 +0000

Sorry – Yes you are right. Monty is in fact swapping his two doors for your one door because in effect he is telling you which door of either of his two is a loser and is now giving you the option of opening the other one as well. So you do now have a 2:3 chance of winning.
If Monty has two doors (1 & 2) and you have door 3 ;-
With all doors closed. After opening one of Monty’s losing doors.
(1 & 2) 3 (1 & 2) 3
possible Win L L W – L W
or Lose L W L – W L
combinations W L L W – L

By: ob

ob — Mon, 20 Jul 2009 23:31:08 +0000

You would be correct if Monty was opening a door randomly. The difference occurs when he acts based on information that he possesses. If you don't believe it, run a little experiment and convince yourself. Or download this program and run the experiment.

By: Paul Tiemens

Paul Tiemens — Mon, 20 Jul 2009 23:17:28 +0000

When you first choose a door you have a 1:3 chance of winning.
Monty has a 2:3 chance of winning – he has 2 doors.
If you could swap with him your chances obviously are now 2:3.
When one of his doors is opened to reveal a goat his chances are now the same as yours 1:3.
it’s now 50/50 – swapping makes no difference. One door has gone and the 2:3 odds go with it.
Imagine if Monty had opened the door with the car behind it instead of one with a goat behind it, if it was your door you now have 100% chance of winning so long as you don’t swap.
If it’s one of his doors you have 100% chance of losing if you don’t swap.
Maths is simple and puzzles like this confuse people into thinking they don’t understand it so playing into the hands of those who try to use maths as a mystery language that can only be understood by them. Don’t be fooled.

By: Richard Hammerud

Richard Hammerud — Sun, 07 Jun 2009 18:05:57 +0000

The puzzle is tricky because it looks like there is a 50/50 chance that the money (or whatever prize is being offered) can be behind either door. And this is true. If you randomly choose a door you will win half the time. BUT if you do NOT randomly choose but always switch you will win two out of three times. So
ALWAYS SWITCH and you will win two out of three time. If you stick with your original choice you will win one out of three times.
Here are the options:
Always switch:; you will 2 out of 3 times.
Randomly choose: you win half the time.
Stick with original choice: you win 1 out of 3 times.

By: Dr. Decay

Dr. Decay — Mon, 11 May 2009 11:15:26 +0000

On the pedantic footnote 1: its REALLY important. If your were playing the game and Monty really was trying to screw you. He would only open another door if you had chosen the one with the car. Otherwise, he would just stop the game. And the smarter you are about probability theory, the more easily he will screw you.

Of course Monty wasn’t trying to screw the contestants. He wanted them to succeed so that he could launch into a paean about that great new Pontiac. That may amount to screwing the viewers, but we were rapt.

By: Bill Aylward

Bill Aylward — Thu, 07 May 2009 16:37:16 +0000

A simpler way to look at this is to consider the result of two strategies, sticking with your original choice, and switching, for each of the three possibilities for what is behind the originally chosen door;

The following table shows what you will get for each strategy;

Car Goat Goat

Stick Car Goat Goat
Switch Goat Car Car

Since each column is equally likely (your first choice being truly random), the ‘Stick’ strategy has a 1/3 chance of winning the care, and the ‘Switch’ strategy 2/3rds

By: Jens

Jens — Wed, 06 May 2009 17:52:42 +0000

Thanks. I actually forgot to paste the link to github. Here it is: http://github.com/Jens-n/Monty-Hall-Problem

By: ob

ob — Wed, 06 May 2009 14:42:49 +0000

> Consider the other states of Monty. He can want you to win, or not care if you win. The
> odds change.

Actually, the odds don’t change. Assume Monty cannot open the door with the car and win himself, and let’s consider just the case where you picked door A (Sa) and Monty is deciding whether to open door B (B_m), since door C is equivalent.

If Monty wants you to win, he chooses P(B_m|Sa) = 1/2, P(B_m|Sb) = 0, and P(B_m|Sc) = 1. Exactly as if he wants you to lose! What if he doesn’t care? Again, the only time he has a choice about which door to open is when you chose the winning door (in other words, Sa).

As for giving you a chance to switch, we’re assuming that is part of the problem statement. They do it to build up some suspense ;)

By: Steve

Steve — Wed, 06 May 2009 13:20:31 +0000

From the problem statement:

…and the host, who knows what’s behind the doors…

By: Layperson

Layperson — Wed, 06 May 2009 12:18:38 +0000

Monty removes one of the other two doors from consideration, and in effect we have to re-allocate the probability we had formally accorded to that door before Monty revealed it was not the correct one.

Your solution appears to always reallocate 100% of its probability to the other as-yet-unselected door. That does not seem correct to my layman’s appreciation of math.

Look at it from Monty’s point of view. When we arrive at his decision of which door to reveal, there is a 1/3 probability that our door choice makes his reveal meaningless and effectively random (because we chose the door with the car). There’s a 2/3 chance that our door choice forces his hand, and he really has no choice in what to do (because one of the reamining doors holds the car, and he will never choose that door).

Do we not reallocate the probabilities based on that? So, 1/3 * 1/3 is reallocated to the door I chose, and 2/3 * 1/3 is reallocated to the as-yet-untouched door.

Final probabilties:

My original door: 1/3 + (1/3 * 1/3) = approx. 44%
The other door: 1/3 + (2/3 * 1/3) = approx 56%
Monty’s chosen door: 0% (revealed as a goat)

Can someone help me understand if/why my solution is not correct?

By: Simonsays

Simonsays — Wed, 06 May 2009 09:31:16 +0000

If this assumption:
“Monty is, of course, the host of the game show. And he is trying to screw you. He doesn’t want you to get the car. ”
is correct, then Monty would only ever give you a chance to switch in the case you chose correctly the first time. If you didn’t choose correctly, he’d just say nope, wrong, get off my stage. So, you would never switch.

Consider the other states of Monty. He can want you to win, or not care if you win. The odds change. What are the odds that Monty is in any particular state? Cannot be calculated. Thus the actual answer to this problem is “not enough information to determine”