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Exitonly
10-15-2005, 05:25 AM
Anyone feel like explaining this to me? i found a website that tried, but it did a sucky job, because i still can't decide if it's 50% of 66%, despite knowing the answer.

pzhon
10-15-2005, 05:31 AM

Exitonly
10-15-2005, 05:36 AM
my bad, forgot that the search actually does work sometimes.

10-15-2005, 06:18 PM
One way I've seen explained that makes LOTS of sense is this:

Pretend there are 100 curtains, one of which has "the prize" behind it(instead of the normal 3 curtains)

You pick one, and Monty takes away 98 of the unpicked ones, showing no prize in them.

Now, are you seriously considering keeping your original curtain?

Exitonly
10-15-2005, 09:40 PM
sorry, that doesnt make any more sense than the orginal 3 curtain thing to me.

i can rationalize both answers to myself, and can't decide which i want to believe.

UATrewqaz
10-15-2005, 11:18 PM
My statistics professor explained this to me perfectly.

3 doors (1 car, 2 goats).

Thus you have a 1 in 3 chance (33%) of picking the right door.

A door is then revealed to you showing you 1 of the goats. So there are two doors remaining.

YOU SHOULD ALWAYS SWITCH (assuming they will always show you a goat, because they always can).

If you do not switch you have your 33% chance of picking correctly.

If you switch:
33.3% of the time you have the car and switch to the goat.
66.6% of the time you have a goat and are switching to the car.

Thus you have a 66% chance of winning the car if you switch.

Some people had a hard time understanding this so he had a great example.

He pulled out a deck of cards and said, you have to select the ace of spades, and a student selected 1 card (face down). He then took the remaining deck and began showing him cards, saying "Ok 4 of clubs" discarded it, etc etc etc

Until there was 1 only card left in his "deck"

So now there are two cards, the one the student picked and the one he has left (he has shown him 50 cards from the 51 remaining that aer NOT the ace of spades).

He then asked the student, "Do you wanna switch"
and of course the studen twas like "hell ya"

The odds of him having the ace of spades were nto 50% even though only 2 cards remained.

KJL
10-15-2005, 11:34 PM
The way that I think it is best explained is by looking at all the possible outcomes.

Assume the prize is behind door 3
If you never switch the outcomes are:
You pick door 1, the host shows you door 2, you stay with door 1. Outcome: lose
You pick door 2, the host shows you door 1, you stay with door 2. Outcome: lose
You pick door 3, the host shows you door 1 or 2, you stay with door 3. Outcome: win

If you always switch the outcomes are:
You pick door 1, the host shows you door 2, you switch to door 3. Outcome: win
You pick door 2, the host shows you door 1, you switch to door 3. Outcome: win
You pick door 3, the host shows you door 1 or 2, you switch to door 1 or 2. Outcome: lose.

Therefore by not switching you win 1/3 times. By switiching you win 2/3.

pzhon
10-16-2005, 12:02 AM
[ QUOTE ]
My statistics professor explained this to me perfectly.

[/ QUOTE ]
That may be, but you left out crucial assumptions about the problem. See the writeup in the archives of rec.puzzles, or past discussions in the archives of this forum.

UATrewqaz
10-16-2005, 12:34 AM
Yes I left them out because they would confuse the OP I assumed.

Read King Yao's book "Weighing the Odds in Hold'Em Poker" for a detailed explination of the assumptions.

Basically, we are assuming that Monty Hall is not trying to trick us and that he was going to show us a goat no matter what we choose originally.

If Monty Hall WANTS us to lose he may only show us a goat if we originally picked the car, in a tricky attempt to get us to switch.

This scenario is more appllicable to an opponent in poker, but as it applies to the Monty Hall it is safe to assume that Monty Hall always shows a goat no matter what we picked, it makes for better tv drama, a better show, etc.

KenProspero
10-16-2005, 12:44 AM
The solution only works (from a pure math point of view) if Monte always shows you a non-car door and is always offers you the chance to switch. These are usually implicitely assumed in the problem. But ....

Some years ago, Monte himself commented and said that there were a lot of factors that went into whether he offered the switch, so in the real world, the Monte Hall problem never met the requisite conditions.

If the conditions above aren't met, it becomes an interesting game theory problem, with certain unanswerable questions, such as what are Monte's motivations and how many cars does he want to give away in a week.

Anyway, for anyone who doubts the solution and can't follow the math, try this.

You be the contestant and get a friend to be Monte.

Have your friend take two black cards and one red card. Mix them up (so you can't see) and determine where the red card is. Your friend will offer you the choice of cards,. You 'win' if you get the red card.

Before you look, your friend turns over one black card and offers you the chance to switch to the other card that you didn't choose.

Now -- for this experiment, either always accept the choice or always decline it.

If you run the experiment enough times, you'll find one of the following:

1. If you always accept you win 2/3 of the time or

2. If you always decline you win 1/3 of the time.

Exitonly
10-16-2005, 02:14 AM
[ QUOTE ]
The way that I think it is best explained is by looking at all the possible outcomes.

Assume the prize is behind door 3
If you never switch the outcomes are:
You pick door 1, the host shows you door 2, you stay with door 1. Outcome: lose
You pick door 2, the host shows you door 1, you stay with door 2. Outcome: lose
You pick door 3, the host shows you door 1 or 2, you stay with door 3. Outcome: win

If you always switch the outcomes are:
You pick door 1, the host shows you door 2, you switch to door 3. Outcome: win
You pick door 2, the host shows you door 1, you switch to door 3. Outcome: win
You pick door 3, the host shows you door 1 or 2, you switch to door 1 or 2. Outcome: lose.

Therefore by not switching you win 1/3 times. By switiching you win 2/3.

[/ QUOTE ]

This did it for me, i understood all the other answers, even with that deck or cards and the A of spades i felt like there was evidencee for staying or switching..

but all the outcomes written that way.

Top notch, really glad i posted this.

Thanks to those that responded.

10-18-2005, 07:07 AM
This monte think is starting to get to me /images/graemlins/smile.gif

[ QUOTE ]
The way that I think it is best explained is by looking at all the possible outcomes.

Assume the prize is behind door 3
If you never switch the outcomes are:
You pick door 1, the host shows you door 2, you stay with door 1. Outcome: lose
You pick door 2, the host shows you door 1, you stay with door 2. Outcome: lose
You pick door 3, the host shows you door 1 or 2, you stay with door 3. Outcome: win &amp; W

If you always switch the outcomes are:
You pick door 1, the host shows you door 2, you switch to door 3. Outcome: win
You pick door 2, the host shows you door 1, you switch to door 3. Outcome: win &amp; win (!)
You pick door 3, the host shows you door 1 or 2, you switch to door 1 or 2. Outcome: lose and lose (!).

Therefore by not switching you win 2/4 times. By switiching you win 2/4.

[/ QUOTE ]

FYP

SonofJen
10-18-2005, 04:44 PM
I think this a pretty solid explanation. The only thing I would add to those still struggling to see where the 33.3% and 66.7% figures come from is to remember that those are the odds that you'll pick either goat or car (respectively) at the beggining of the experiment when you have three unknown doors to choose from.

KJL
10-18-2005, 08:35 PM
No. The host will pick one or the other. It makes no difference which one. You can't count them as seperate events, because you are not making the descion. The host could always choose to only show you door 1.

10-19-2005, 07:19 PM
[ QUOTE ]
This monte think is starting to get to me /images/graemlins/smile.gif

[ QUOTE ]
The way that I think it is best explained is by looking at all the possible outcomes.

Assume the prize is behind door 3
If you never switch the outcomes are:
You pick door 1, the host shows you door 2, you stay with door 1. Outcome: lose
You pick door 2, the host shows you door 1, you stay with door 2. Outcome: lose
You pick door 3, the host shows you another door, you stay with door 3. Outcome: win

If you always switch the outcomes are:
You pick door 1, the host shows you door 2, you switch to door 3. Outcome: win
You pick door 2, the host shows you door 1, you switch to door 3. Outcome: win
You pick door 3, the host shows you another door, you switch to the other door. Outcome: lose

Therefore by not switching you win 1/3 times. By switiching you win 2/3.

[/ QUOTE ]

FYP

[/ QUOTE ]

FYP.

You pick door 3 one time out of three. Half of those times, you are shown door one, half you are shown door 2. This means that when you always stay, you win [(0/3)+(0/3)+(1/6)+(1/6)]=1/3 of the time. When you always switch, you win [(1/3)+(1/3)+(0/6)+(0/6)]=2/3 of the time. Therefore, you should always switch.