Anyone believe in this? I refuse to. The entire situation changes when you are shown a free door.

Quote:Lemieux66http://en.m.wikipedia.org/wiki/Monty_Hall_problem

Anyone believe in this? I refuse to. The entire situation changes when you are shown a free door.

What, exactly, don't you believe?

In the show let's make a deal, the host would ALWAYS open a losing door that the player did not pick (there is always at least one such door available). Therefore it is always correct to switch.

If the host is opening an unpicked door at random then it makes no difference whether or not you switch. But Monty Hall was not picking at random -- he always picked a losing door.

Quote:Lemieux66http://en.m.wikipedia.org/wiki/Monty_Hall_problem

Anyone believe in this? I refuse to. The entire situation changes when you are shown a free door.

There are some good threads on this if you want to search the site. You should find them interesting. One of the threads was started by mkl54321 who was an excellent and very smart poster here.

Quote:

Originally Posted by Grumman

That is not inherently true. It could also mean the host is trying to trick you into moving away from the correct answer. Like I said, it's not about probabilities, it's about knowing how the host responds to information he can see that you can't, to work out what he knows.

Rules of the following game:

1) I think of a number between 1 and 100.

2) I ask you to guess a number between 1 and 100.

3) I tell you that I'm going to rule out 98 numbers, one of which is the number you chose. If you chose correctly in the first place, then I'll choose another number randomly not to rule out. If you chose incorrectly in the first place, then I won't rule out the correct number.

4) I give you the opportunity to switch to the other number, or stick with yours.

5) I reveal whether your final choice is the number I thought of in step 1.

Let's try it out; I'll fill in your guess more-or-less randomly in step 2.

1) I've got my number, but I'm not telling you what it is.

2) You guess that it's 47.

3) I tell you that it's not 1-46, 48-72, or 74-100. It's either 47 or 73.

4) You can now switch your guess to 73, or you can stick with 47.

5) If you stuck with 47, you'd be wrong. If you switched to 73, you'd be right.

Step 3 is important. I'm never going to rule out the correct number, any more than Monty is ever going to open the door with the prize and say, "Oops, too bad!" I'm ALWAYS going to rule out an incorrect choice. That move on my part is what skews the odds so heavily in favor of switching for you.

If you're still not convinced, play the game I described above with a friend. If you follow the rules correctly, if you switch every time, you'll win 99 times out of 100. If you stick with your original number every time, you'll win 1 time out of 100.

- That the Monty Hall problem exists in real live ? (well, at least some people have been contestants).

- That the Monty Hall problem is likely to produce some confusion ?

- That the Monty Hall is a solveable paradox ?

Quote:MangoJBelieve in what exactly ?

- That the Monty Hall problem exists in real live ? (well, at least some people have been contestants).

- That the Monty Hall problem is likely to produce some confusion ?

- That the Monty Hall is a solveable paradox ?

You pick one of three doors. One is shown not to be correct. So now one is correct and now one is not. If not shown a new door, obviously switch. If shown a new door, information now changes. So to switch seems strange to me.

Quote:Lemieux66You pick one of three doors. One is shown not to be correct. So now one is correct and now one is not. If not shown a new door, obviously switch. If shown a new door, information now changes. So to switch seems strange to me.

I'm not sure what you mean by "a new door".

There is a 1/3 probability that the right door is the one you picked, and a 2/3 probability that the right door is one of the other two.

At least one of the other two is guaranteed to be wrong, so the host (intentionally, every time) shows you a wrong door from the other two. Since he is always able to do this, you can't get any extra information about your door. It's STILL true that there is a 1/3 chance that your door is right, and a 2/3 chance that one of the other two doors is right. The difference is, now, if it's one of the other two, you know which of the two it is.

If you stay, you get the door that you picked (1/3 chance of being right). If you switch, you get BOTH of the other doors (2/3 chance of containing the right one). Obviously having two doors is better than having one.

Quote:AxiomOfChoiceSince he is always able to do this, you can't get any extra information about your door. It's STILL true that there is a 1/3 chance that your door is right

That is actually the most didactical solution I've read about Monty Hall problem. So if you don't gain information about your door, the probability of your door will still be 1/3. That is easy to understand. In consequence the probability of the other door left must be 2/3, so you better switch.

The beauty in this reasoning is the immediate value of information. If Monty would always open a random door (even one with a price), but it happens to be a goat the time you are in the show, then you would indeed gain information on your picked door (making it more likely to be the prize), eventually leading to 1/2 for each of the doors (and switching be indifferent).

I really like this explanation.

Quote:AxiomOfChoiceI'm not sure what you mean by "a new door".

There is a 1/3 probability that the right door is the one you picked, and a 2/3 probability that the right door is one of the other two.

At least one of the other two is guaranteed to be wrong, so the host (intentionally, every time) shows you a wrong door from the other two. Since he is always able to do this, you can't get any extra information about your door. It's STILL true that there is a 1/3 chance that your door is right, and a 2/3 chance that one of the other two doors is right. The difference is, now, if it's one of the other two, you know which of the two it is.

If you stay, you get the door that you picked (1/3 chance of being right). If you switch, you get BOTH of the other doors (2/3 chance of containing the right one). Obviously having two doors is better than having one.

That's the best explanation I've seen so far, very easy to understand

Quote:Lemieux66http://en.m.wikipedia.org/wiki/Monty_Hall_problem

Anyone believe in this? I refuse to. The entire situation changes when you are shown a free door.

I just discovered the Monty Hall Problem yesterday. It is fascinating. I stumbled on why it works and wanted to share it in simple terms for those who might still be scratching their heads. Let's say we have three doors, and behind one is a car and behind the other two there are goats. Whatever door you choose another door with a goat behind it is revealed. If you choose a door and never switch you have a 1/3 chance of winning the car. Now, let's say you ALWAYS switch. Then there are only two possibilities. One: 1/3 of the time you choose the car and switch to a goat door. So 1/3 of the time you lose. Two: 2/3 of the time you choose a door where there is a goat. And every time you choose a goat you win the car because the other goat door is revealed leaving only the car door to switch to.

Quote:GreasyjohnI just discovered the Monty Hall Problem yesterday. It is fascinating. I stumbled on why it works and wanted to share it in simple terms for those who might still be scratching their heads. Let's say we have three doors, and behind one is a car and behind the other two there are goats. Whatever door you choose a door with a goat behind it is revealed. If you choose a door and never switch you have a 1/3 chance if winning the car. Now, let's say you ALWAYS switch. Then there are only two possibilities. One: 1/3 of the time you choose the car and switch to a goat door. So 1/3 of the time you lose. Two: 2/3 of the time you choose a door where there is a goat. And every time you choose a goat you win the car because the other goat door is revealed leaving only the car door to switch to.

Yes, but the math changes completely when you prefer to win the goat.

Quote:Lemieux66Yes, but the math changes completely when you prefer to win the goat.

I don't get what you mean.

Quote:GreasyjohnI don't get what you mean.

xkcd Monty Hall

Quote:GreasyjohnI don't get what you mean.

I want the goat. So I won't switch.

Quote:AxiomOfChoiceIf the host is opening an unpicked door at random then it makes no difference whether or not you switch. But Monty Hall was not picking at random -- he always picked a losing door.

He wouldn't always let you switch, though, and that also breaks the problem. In fact, I think you were never (not just in EV terms, but actually, p=0 never) better off switching.

Statistically, you will win about 2/3 of the time by switching, we've established that. However, that's given the benefit of repeated trials. If you're a contestant on "Let's Make a Deal", you only get to play the game once. Does this change your strategy at all?

I've had this discussion with people regarding "Deal or No Deal". Some suggest taking or rejecting a deal based on pot odds and expected value, even though there are no repeated trials. A contestant can only play the game once. In that situation, real life trumps the math.

Quote:DeucekiesI'll muddy the waters a little bit.

Statistically, you will win about 2/3 of the time by switching, we've established that. However, that's given the benefit of repeated trials. If you're a contestant on "Let's Make a Deal", you only get to play the game once. Does this change your strategy at all?

I've had this discussion with people regarding "Deal or No Deal". Some suggest taking or rejecting a deal based on pot odds and expected value, even though there are no repeated trials. A contestant can only play the game once. In that situation, real life trumps the math.

That makes some sense on Deal or No Deal, but no sense in the case of the Monty Hall problem, where there are only two possible outcomes. Why would you want to make it less likely that you get the better outcome, when that's all you're doing?

But also keep in mind that EV always adds, even in different games - so even if you'll only be playing a particular game once, it behooves you to play well in all the "games" of your life. (Although you may want to hedge in Deal or No Deal if the money gets higher than you're likely to see elsewhere.)

With DonD it's not 0 or 1, but rather a value whose expected value can go up or down depending on your luck. If the value of the prize (or potential prizes) is insignificant to your total wealth then you would go for greatest expectancy (i.e. usually all the way). However if you're after a set amount, or something nice (like pay for a holiday) then the values aren't linear. For instance sums less than your target are probably linear, but values greater are not. Say you were after a £10k cruise you would take the offer even if the true value was greater - thus I guess it's better not to advertise your targets or expectations. If I recall, the first show I saw recorded had that and the elderly couple said it could buy them a holiday to Antarctica, so they took it.

Quote:DeucekiesStatistically, you will win about 2/3 of the time by switching, we've established that. However, that's given the benefit of repeated trials. If you're a contestant on "Let's Make a Deal", you only get to play the game once. Does this change your strategy at all?

Of course, in "let's make a deal" you can trade EV against variance. If the amounts are substantial - you would probably want to do that.

In the Monty Hall problem, whatever strategy you chose variance is always the same (as it is a 2-way outcome). The only thing changing with strategy is EV, and hence there is no reason to play anything else than for EV.

Quote:24BingoHe wouldn't always let you switch, though, and that also breaks the problem. In fact, I think you were never (not just in EV terms, but actually, p=0 never) better off switching.

I have no idea what happened in the actual show. However, the team "The Monty Hall Problem" is generally used to describe the math problem where the host always shows you a losing door and always offers you the chance to switch. The solution to that problem is to always switch.

Quote:AxiomOfChoiceI have no idea what happened in the actual show. However, the team "The Monty Hall Problem" is generally used to describe the math problem where the host always shows you a losing door and always offers you the chance to switch. The solution to that problem is to always switch.

Did the "Monty Hall Problem" (exactly as described) ever actually come up in Let's Make A Deal? Or was the term developed to frame the math question because people were familiar with Monty and the 3 door concept from LMAD?

I watched a lot of LMAD back in the day; even a few of the episodes they filmed in the Las Vegas Hilton ballroom. I can't ever recall a game that exactly reflected "The Monty Hall Problem."

There was the "Big Deal of the Day" where a contestant picked one of three doors, but they were never allowed to switch at any point. Plus the BDotD was usually 3 desirable prizes... maybe some were of less value than what the contestant traded away, but there were few, if any, "Zonks" in the BDotD, IIRC.

Quote:Lemieux66Yes, but the math changes completely when you prefer to win the goat.

As an aside, I always wondered if the contestants ever got to keep the Zonk prizes or were offered cash equivalents. I mean, livestock has value... giant rocking chairs have value... even plush shag carpeted junkyard appliances have value.

Quote:JoemanDid the "Monty Hall Problem" (exactly as described) ever actually come up in Let's Make A Deal? Or was the term developed to frame the math question because people were familiar with Monty and the 3 door concept from LMAD?

No. Not exactly anyway.

"Doors" were only used for the Big Deal, and there were never any options once you chose a door.

The reveal with option to switch was often used with the three curtains. However, the curtains never had more than one zonk. Often it was with three prizes of varying value.

Also note that Monty had discretion on his side. He would often do the reveal and option only if you picked the zonk/low value prize, which further increased the EV of the switch.

At tribal counsel, Deshawn had to play a game where he would either me immediately eliminated or granted immunity at Tribal Counsel. The game turned out to be the Monty Hall problem. Imagine my reaction when after Jeff revealed a "goat" and Deshawn was given the option to switch, he said "no." I was beside myself, screaming at TV. I prayed the gods of math would punish him by getting the other "goat" and being eliminated. Alas, it was not to be. As they say, sometimes it is better to be lucky than right.

Here is a link to the episode. The Monty Hall portion starts at the 37:00 point. Be warned, you'll have to suffer through some ads to get to it.

Quote:WizardI was screaming at the television last night, watching the latest episode of Survivor.

At tribal counsel, Deshawn had to play a game where he would either me immediately eliminated or granted immunity at Tribal Counsel. The game turned out to be the Monty Hall problem. Imagine my reaction when after Jeff revealed a "goat" and Deshawn was given the option to switch, he said "no." I was beside myself, screaming at TV. I prayed the gods of math would punish him by getting the other "goat" and being eliminated. Alas, it was not to be. As they say, sometimes it is better to be lucky than right.

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Yes, and one of the contestants (I believe it was Xander) called out "its the Monty Hall problem" before Deshawn made his decision. So, this was not lost on the at least one of the contestants.

Quote:gordonm888Yes, and one of the contestants (I believe it was Xander) called out "its the Monty Hall problem" before Deshawn made his decision. So, this was not lost on the at least one of the contestants.

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Yes, I caught that too. Earlier in the season they had a Prisoner's Dilemma situation and somebody made a similar comment. They must have a logician on staff this season.

Who are you rooting for? Who do you think currently has the edge?

I am pulling for Xander. I like him as a player a lot. I feel like Ricard is right now the strongest player in the game and most likely to win at this point.

To get back on topic, Deshawn did not switch and was right! I'm super confused on how switching is better than staying. You had a 1/3 shot when you first chose and still a 1/3 shot after the first incorrect was shown. What makes one box better than the other?

You have a 1/3 chance of being right. So, you have a 2/3 chance of being wrong.Quote:mwalz9… I'm super confused on how switching is better than staying. You had a 1/3 shot when you first chose and still a 1/3 shot after the first incorrect was shown. What makes one box better than the other?

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That doesn’t change once you’re shown that one of the 2/3 choices was also wrong.

What DOES change is, once shown a bad choice, you’re still 2/3 wrong, so the other choice becomes 2/3 right.

But all that is strictly math. As pointed out, Monty had options of his own. Maybe Jeff had options as well. Monty didn’t always offer the switch.

Would Jeff have made the offer if a wrong choice was made?

Quote:DJTeddyBearYou have a 1/3 chance of being right. So, you have a 2/3 chance of being wrong.Quote:mwalz9… I'm super confused on how switching is better than staying. You had a 1/3 shot when you first chose and still a 1/3 shot after the first incorrect was shown. What makes one box better than the other?

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That doesn’t change once you’re shown that one of the 2/3 choices was also wrong.

What DOES change is, once shown a bad choice, you’re still 2/3 wrong, so the other choice becomes 2/3 right.

But all that is strictly math. As pointed out, Monty had options of his own. Maybe Jeff had options as well. Monty didn’t always offer the switch.

Would Jeff have made the offer if a wrong choice was made?

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I disagree. Looking at the other choice, as a single choice, it also is only 1/3 right or 2/3 wrong.

Quote:mwalz9Quote:DJTeddyBearYou have a 1/3 chance of being right. So, you have a 2/3 chance of being wrong.Quote:mwalz9… I'm super confused on how switching is better than staying. You had a 1/3 shot when you first chose and still a 1/3 shot after the first incorrect was shown. What makes one box better than the other?

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That doesn’t change once you’re shown that one of the 2/3 choices was also wrong.

What DOES change is, once shown a bad choice, you’re still 2/3 wrong, so the other choice becomes 2/3 right.

But all that is strictly math. As pointed out, Monty had options of his own. Maybe Jeff had options as well. Monty didn’t always offer the switch.

Would Jeff have made the offer if a wrong choice was made?

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I disagree. Looking at the other choice, as a single choice, it also is only 1/3 right or 2/3 wrong.

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That’s not correct under the assumption that the host will only open a door with a goat, and never the winning door.

Imagine there are 100 doors. You pick one. Then the host opens 98 doors with a goat and asks if you want to switch.

If the host opens all but 2, my odds were still the same if I had selected the one I selected or if I had selected the one he is trying to make me switch to!

Quote:mwalz9I don't care if there are 3 doors, 100 doors or 1,000,000 doors.

If the host opens all but 2, my odds were still the same if I had selected the one I selected or if I had selected the one he is trying to make me switch to!

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No. You are assuming the host is randomly opening doors. He’s not. He knows what door has the prize and is intentionally opening all the other doors.

You pick door 1. The host asks if you want to keep door one and switch to both doors 2 and 3. Would you rather have just one door or two?

That’s the Month Hall problem. The fact that the host opens 1 or 98 or 999,998 doors that he already knows we’re empty doesn’t change the offer.

Switching let’s you win EVERY time unless you picked the right door originally. (Reread the last sentence. It’s the key. Switching makes you win every single time, unless you got lucky and picked the winner originally.)

Since with three doors you pick right 1/3 of the time, switching let’s you win 1 - 1/3 = 2/3.

Quote:mwalz9If the host opens all but 2, my odds were still the same if I had selected the one I selected or if I had selected the one he is trying to make me switch to!

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You are wrong. As to why, I have an entire article on the Monty Hall Problem.

I pick Door #1. Monty opens Door #2 and it is a goat. I can switch doors if I like. I introduce a new twist. I will flip a coin. There are two doors so each is a 50-50 prop.

heads means keep my door, tails means switch. The coin flip is 50-50. So if I let the coin decide, it is 50-50, but if switch on my own, it's 66%?

Quote:billryanI will accept I am wrong, but would love to have it explained.

I pick Door #1. Monty opens Door #2 and it is a goat. I can switch doors if I like. I introduce a new twist. I will flip a coin. There are two doors so each is a 50-50 prop.

heads means keep my door, tails means switch. The coin flip is 50-50. So if I let the coin decide, it is 50-50, but if switch on my own, it's 66%?

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Yes. Look at it this way: suppose you are playing a game where somebody rolls a 6-sided die, keeps the number hidden, and you have to decide whether or not the number is at least 3. If you toss a coin, you have a 50% chance of being right; if you choose "yes, the number is at least 3," you have a 2/3 chance of being right.

Going back to the original problem:

There is a 1/3 chance you chose the winning door; Monty will always open a door with a goat behind it.

There is a 2/3 chance you chose a door with a goat; Monty will always open the other door with a goat behind it.

In every case, Monty opens a door and shows you a goat.

If you switch, then, if you originally chose the winning door, you now have a goat, and if you originally chose a goat, you now have a winning door. The probability of originally choosing a door with a goat was 2/3, so that is the probability of you winning if you always switch.

Assume the player picks his door randomly and if the player picks the car, the host will pick another door to open randomly.

Then, the player picks door B and the host opens door C. What is the probability the car is behind door A?

Pr(Car in A given Player picks B and host reveals goat in C) =

Pr(Car in A and Player picks B and host reveals goat in C) / Pr(Player picks B and host reveals goat in C) =

Pr(Player picks B and Car in A and host reveals goat in C) / [Pr(Player picks B and car in A and host reveals goat in C) + Pr(Player picks B and car in B and host reveals goat in C)]=

(1/3)*(1/3) / [(1/3)*(1/3) + (1/3)*(1/3)*(1/2)] =

(1/3) / [(1/3) + (1/3)*(1/2)] =

(1/3) / (3/6) =

(2/6) / (3/6) =

2/3

In, fact let's do it again but without the part about the player randomizing his choice.

Pr(Car in A given host reveals goat in C) =

Pr(Car in A and host reveals goat in C) / Pr(Host reveals goat in C) =

Pr(Car in A and host reveals goat in C) / [Pr(car in A and host reveals goat in C) + Pr(car in B and host reveals goat in C)]=

(1/3) / [(1/3) + (1/3)*(1/2)] =

(1/3) / (3/6) =

(2/6) / (3/6) =

2/3

Before you pick a door you are told this: "After you pick a door, Monty Hall will knowingly choose one of the other two doors and reveal that the prize was not behind that door. So go ahead and choose a door."

When you go ahead and choose a door, will your odds be 1 in 3 of picking the correct door, or will they be 1 in 2?

I hope you said "1 in 3." Now when Monty Hall does exactly what you were told he would do, the odds that you originally picked the correct door are still 1 in 3. Which means that the odds of the only remaining door being the correct door must be whatever is left, namely 2 in 3.

Quote:gordonm888Let's try this as an explanation:

Before you pick a door you are told this: "After you pick a door, Monty Hall will knowingly choose one of the other two doors and reveal that the prize was not behind that door. So go ahead and choose a door."

When you go ahead and choose a door, will your odds be 1 in 3 of picking the correct door, or will they be 1 in 2?

I hope you said "1 in 3." Now when Monty Hall does exactly what you were told he would do, the odds that you originally picked the correct door are still 1 in 3. Which means that the odds of the only remaining door being the correct door must be whatever is left, namely 2 in 3.

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I like that.

I think it helps to label the goats A and B. As mentioned above, all you know for sure is that after you select a door Monty will open a different door and show you a goat. There are three scenarios...

1) You pick goat A and get shown goat B. You switch and win the car.

2) You pick goat B and get shown goat A. You switch and win the car.

3) You pick the car and get shown either goat. You switch and lose the car.

So 2 of the 3 switch scenarios win. But what if you don't switch?

1) You pick goat A and get shown goat B. You stay and lose the car.

2) You pick goat B and get shown goat A. You stay and lose the car.

3) You pick the car and get shown either goat. You stay and win the car.

So 2 of the 3 stay scenarios lose.