Monty Hall

BuzzLY

There are a lot of very intelligent geeks on this board, so I'm hoping many of you are familiar with this problem -- seems it confounds even intelligent people. We'll see how you guys handle it.

This post has a two-fold purpose: One, it is intended to spark what I think will be an interesting conversation on math and logic. I believe even those not logically-minded will find this discussion interesting. Two, it may lead to our third coding contest. I haven't quite figured out how to apply it to a coding contest, so if any of you have any suggestions, please PM me (don't reply to this thread with contest suggestions, please).

Here it is:

You are a contestant on Let's Make a Deal. Monty Hall picks you for his next offer, and it's a good one: Behind one of three curtains is a brand new car! Unfortunately, the other two curtains lead to Hell -- no, just kidding, they each have a goat behind them.

Monty tells you to pick one. You do that. Then, before he reveals your curtain, he makes an offer. First, he opens one of the two remaining curtains, revealing a goat. He then offers to let you switch to the remaining curtain.

In considering this problem, you must know this: Monty Hall knows exactly which curtain contains the car, but he has no agenda. He does not care if you win or not. Your decision must be based solely on statistical probability, and not on any motivations or intuitions.

Herein lies your dilemma: What are the odds that the car is behind the curtain Monty is offering? Based on those odds, should you switch curtains, or stick with your original? Or does it not matter? And if you think you know the answer, what is your reasoning?

Bunkermaster

One question:
Whât ís thè cølør øf thè cûrtâín?

BuzzLY

Puce.

Merve

Let's say that you've got a fifty-fifty chance. You've got no way of knowing which one is going to make you a winner. Then you'd have to go with your gut instinct...right?

goldbug

How bout I just coldcock Monty with the box, and steal his wallet?

(He's always got that damn box as a third choice.)

GilesGuthrie

The real question here is this:

Does the information that an unchosen curtain reveals a goat change your perspective on the competition. It is one of statistics.

If you choose your curtain, from three possibilities, then you have a 33% chance of being right.

If you choose your curtain, from two possibilities, then you have a 50% chance of being right.

The philosophical question surrounds the timeline. If you make your choice before Monty reveals the goat (1 from 3, 33%), then Monty revealing the goat makes no real difference, because you have one hurdle still to cross.

However, if you withhold your choice until after Monty reveals the goat (or change choice subsequent to the revealing), then you have a 50% chance of winning.

My contention is that nothing has changed. The fact that Monty reveals a goat means that you effectively chose from two options anyway, because whichever you choose (prize, goat-a or goat-b), you will be shown a goat. Therefore, for the purposes of the quiz, goat-a = goat-b, and you are making a choice from two all along.

I think that the trick with these 'random' contests is to pick randomly, rather than developing an affinity for a choice. People faced with three objects in a line will tend to pick the same one each time, they will develop an affinity for one choice. By trying to apply this affinity to a set of data about which you have no information, you are opening yourself up to worries about being correct. You should pick assuming that you will not be correct (the pessimist's viewpoint), because you don't have any data that could contribute to a skilled choice.

So, if I choose a curtain, and am then shown that I have not picked a curtain that is definitely not correct, I would stick with my choice. It makes no difference, so you might as well not change.

frog

Who is Monty Hall?

stolzyboy

Originally posted by frog
Who is Monty Hall?

LOL, the host of "Let's Make A Deal"

http://www.potatodie.nl/Webimages/deal.gif

frog

Yeah, no way that hair is real right?

BuzzLY

Merve: It's not about knowing which is correct. It's about choosing the curtain that gives you the highest odds of being the car. You may still be wrong, but you want every advantage you can get, right? FWIW, your choice is NOT 50-50

GG: a very sound argument. In fact, the first half of your explanation, especially the part about it depending on whether your choice comes before Monty's reveal or after, is right on. However, it led you to the wrong conclusion. You say:

The fact that Monty reveals a goat means that you effectively chose from two options anyway, because whichever you choose (prize, goat-a or goat-b), you will be shown a goat. Therefore, for the purposes of the quiz, goat-a = goat-b, and you are making a choice from two all along.

That is not correct. The key here is that MONTY KNOWS WHERE THE CAR IS.

Look at it this way... Assume you chose curtain A. Instead of looking at it as a choice between curtain A, B, or C, let's look at it as A, not A, and not A. Your odds are A = 33%, not A = 66%. Agreed?

What if Monty told you, you can have either Curtain A, or I'll let you switch to both B and C (not A and not A). You would have 2 curtains, increasing your odds of winning the car to 66%. You know that at least one of those curtains contains a goat, right? Even if you won the car, the OTHER curtain contains the goat. It does not matter whether or not Monty SHOWED you that goat, you know it's there.

Therefore, if Monty offers you the choice to switch, you should defintely switch. You increase your odds to 66% of winning the car.

Not convinced? What if Monty had 100 curtains, and told you to choose one. you have a 1% chance of getting it right. Keep in mind that Monty knows where the car is, so no matter what curtain you choose (99 of them contain goats, remember), he can show you 98 curtains that have goats behind them. Every time. Are you going to tell me that after he shows you the 98 goats, your odds of winning jump from 1% to 50%? No way. You chose BEFORE he revealed those goats. Therefore, your 1% odds don't change. The car is 99% sure of being behind that final curtain.

ahundiak

http://www.phpbuilder.com/board/showthread.php?s=&threadid=10223223&highlight=goat

BuzzLY

doh! LOL... thanks, ahundiak. I guess you know the answer to this one then, don't you?

Elizabeth

Thanks ahundiak, I was thinking I was in some sort of parallel "dejavu" universe. Glad you cleared that up.

Buzz- enough with the peroxide on the ol' hair, k buddy? 😃

-Elizabeth

Sxooter

Sorry buzzly, but I'll have to disagree here.

After Monty has opened the curtain to show the goat, there are only two possibilities, you either have the curtain with a goat, or with a car.

Chance has no memory. The fact that until a few seconds ago you had a 33% chance of being right has no effect on the chance that you now have a goat or a car.

There are two curtains. One has a goat, one has a car. 50/50 chance.

You state:

That is not correct. The key here is that MONTY KNOWS WHERE THE CAR IS.

but you never tell us why it's actually important. Again, if Monty doesn't care if you win or lose, then he will ALWAYS show you the curtain with the goat in it.

Now, if he did care, and his choice was predicated on whether you'd chosen the goat or the car to start, then hell yes, you'd pay attention when he offered to let you change curtains.

But chance has no memory, and Monty doesn't care, therefore, 50/50

jebster

I think the real question is, would you rather have the car then the goat...

ahundiak

Sxooter,
It's easy to test with a deck of cards and a partner. Instead of a car the prize is the Ace of Hearts. Shuffle the deck and slide out one card. Have the partner turn over all but one of the remaining cards being careful not to reveal the Ace of Hearts.

Do you really think that the removed card has a 50/50 chance of being the Ace of Hearts?

bad76

Originally posted by ahundiak
Do you really think that the removed card has a 50/50 chance of being the Ace of Hearts?

Sure !
The base of stat is "The probability has not memory":
each extraction don't depend from previous.

BuzzLY

Ah... but therein lies the answer. Choice DOES have a memory -- Monty Hall.

Let's look at the options for just one iteration of this.

Assume the car is behind curtain A. Also assume that Monty will always know where the car is, and will always show you a goat.

#1: You pick A. Monty can show you either B or C in this case, and if you switch, you lose.

#2. You pick B. Monty shows you curtain C (A has the car, so he can't show you that one). If you switch to A, you win.

#3. You pick C. Monty shows you curtain B (A has the car, so he can't show you that one). If you switch to A, you win.

You can clearly see from this that if you ran this as a program 1000 times, switching curtains make you a winner 2 out of 3 times.

BuzzLY

Originally posted by Elizabeth
Thanks ahundiak, I was thinking I was in some sort of parallel "dejavu" universe. Glad you cleared that up.

Buzz- enough with the peroxide on the ol' hair, k buddy? 😃

-Elizabeth

Just trying to emulate my favorite LARTer😃

bad76

Originally posted by BuzzLY
...
You can clearly see from this that if you ran this as a program 1000 times, switching curtains make you a winner 2 out of 3 times.

This is another problem. And this has 2/3 prob. It's the initial problem with a condition, to always switch. This say what is better to do.

But "probability" of a problem is not "probability under condition. Mathematically saying, the prob of problem, for me, as above mentioned, is 50/50

Because Monty say "Ok this curtain is not right!", and look that ALWAYS can say this => prob=1 => not prob=0, you have to choice beetween only two. And prob are 50/50

Try to look the attach. Is an winzozz executable that make this.
At bottom there is three radiobutton. The first is what i say, second is what you say, third show that also switching randomly give your problem 50/50 solution.

The code:

1) Play random:

bool function calc1()
{   

    $wcell=random(3);
    $scell=random(3);
    $ocell=0; $rcell=0;

while(($ocell==$wcell)||($ocell==$scell)) $ocell++;
$rcell=random(3); while($rcell==$ocell) $rcell=random(3);

return ($wcell==$rcell);
}

2) always switch

bool function calc2()
{   $wcell=random(3);
    $scell=random(3);
    $ocell=0; $rcell=0;

while(($ocell==$wcell)||($ocell==$scell)) $ocell++;
while(($rcell==$ocell)||($rcell==$scell)) rcell++;

return ($wcell==$rcell);
}

3) random switch

bool function calc3()
{   

    $wcell=random(3);
    $scell=random(3);
    $ocell=0; $rcell=0;

while(($ocell==$wcell)||($ocell==$scell)) $ocell++;

$s=random(2); 
if($s==1) while(($rcell==$ocell)||($rcell==$scell)) $rcell++;
else $rcell=$scell;

return ($wcell==$rcell);
}