We've established the odds of successfully second-guessing are 66%. 2/3.
No. This is true in the standard MH problem, but you're assuming it's always the case in all variations when it's not. See my previous post for why it applies in the standard problem but not with ignorant Monty.
Johnny, Ignorant Monty only reveals the car if you didn't initially pick it. Therefore, although 2/3 of the time he does not reveal the car, of those 2/3 of the time you do not have a 1/3 chance of having picked the car initially. It is higher, and hence there is less benefit to switching.
You pick a goat initially 66% of the time, but Ignorant Monty causes you to instantly lose 50% of those times. Hence, you can only pick a goat and have a second choice 33% of the time.
You pick a car initially 33% of the time, and you always get a second choice.
In the cases where you get a second choice, you have an equal chance of having either a car or a goat, hence there is no benefit to switching.
With UNKNOWING monty, you win 66% of the cases where you are offered a choice.
No, you don't. Just list out the cases and see. There are six equally likely cases:
1) The car is behind door 1, and Monty reveals door 2
2) The car is behind door 1, and Monty reveals door 3
3) The car is behind door 2, and Monty reveals door 2
4) The car is behind door 2, and Monty reveals door 3
5) The car is behind door 3, and Monty reveals door 2
6) The car is behind door 3, and Monty reveals door 3
Of these, four cases offer you a choice (1, 2, 4 and 5) and two cases which offer you a choice let you win by switching. So if you always switch when offered a choice, you'll win two out of four, or 50% of the cases where you are offered a choice.
For anyone curious, since I'm the OP, this is the sentence that made the light switch flip on in my head: "It is more likely that you picked the wrong door first, so you should always switch doors."
Organichu on
0
Options
JohnnyCacheStarting DefensePlace at the tableRegistered Userregular
edited July 2008
There are three cases for decision A
You picked a car
you picked goat a
you picked goat b
Monty only offers information
Now you have nine trials where you may or may not get to make decision B
ASSUMING NO VARIENCE
you'll get three where monty reveals the car and ends the game
three where monty reveals goat a
three where monty reveals goat b
Now, in the latter six cases, should you switch or not? You know there was a 33% chance you picked a car, a 66% chance you didn't, and you've had one door eliminated. It's 2-1 for switching. 4 cases where you switch TOO the car, 2 cases where you switch FROM the car, 3 cases where monty wastes the car.
5 cases of loss, 4 cases of win.
As opposed to 3 cases of win, 6 cases of loss if you don't switch.
How does monty revealing the car 1/3 of the time change the odds that you chose wrong initially the other two times?
How can it?
What if you don't know what monty knows, only that this time, he didn't reveal the car? Should you switch then?
(Note: I realize that the answer to the ignorant monty version has been published at 50%, i'm being hardheaded on purpose, until I hear it in a way that makes me click)
You picked a car
you picked goat a
you picked goat b
Monty only offers information
Now you have nine trials where you may or may not get to make decision B
No, there are only six. You have 3 doors to choose from initially, and Ignorant Monty has the other 2 doors to choose from. 3 * 2 = 6.
ASSUMING NO VARIENCE
you'll get three where monty reveals the car and ends the game
three where monty reveals goat a
three where monty reveals goat b
No, you'll get two where Ignorant Monty reveals the car, 2 where he reveals Goat A, and 2 where he reveals Goat B.
Now, in the latter six cases, should you switch or not? You know there was a 33% chance you picked a car, a 66% chance you didn't, and you've had one door eliminated.
No, you do not know this. If you pick a goat initially, there's a 50% chance Ignorant Monty reveals the car, meaning that there is a 33% chance that you can pick a goat and make it to the second choice, just as there is a 33% chance that you can pick a car. You can't just throw out those possibilities.
How does monty revealing the car 1/3 of the time change the odds that you chose wrong initially the other two times?
How can it?
Imagine that 50% of the time you chose a goat initially, Monty said "That one doesn't count" and you lost. This would result in you having less chance of choosing a goat and getting to make the second choice. If there's less chance of having a goat, switching becomes less viable.
It's the exact same situation. Monty revealing the car 50% of the time you choose a goat does exactly the same thing as the above situation.
What if you don't know what monty knows, only that this time, he didn't reveal the car? Should you switch then?
You need to assume one way or the other if you want to follow a strategy.
Johnny, you simply aren't reading what we are saying. Or aren't comprehending it, I don't know.
The only times that Monty reveals the car and you lose without a second choice are when you picked a goat first. So half the time you pick the goat first you lose without a second choice, the other half you want to switch.
Also, I don't think there are nine cases. There are only 2 cases for when you lose outright: when you pick goat A then he goes car, and when you pick goat B and then he goes car. 2 cases for when you want to switch: You pick goat A and he reveals goat B, you pick goat B and he reveals goat A. 2 cases for when you pick the car first: you pick the car he reveals goat A, you pick the car he reveals goat B.
Only 6 permutations.
Savant on
0
Options
JohnnyCacheStarting DefensePlace at the tableRegistered Userregular
edited July 2008
ah...I CHOSE nine trials, because nine divides by 3
You could do it with six trials or with 3 or with 33 I guess.
But how does the quality of what monty's revealing too you change the chances you chose right back when you were just picking 1 of 3 choices?
Your odds then were 33%
If he reveals a goat and KNOWS he's revealing a goat your odds of winning by switching are 66%
If he reveals a goat but wasn't sure what he was revealing your odds are now 50/50?
He gave you the same information. It should be the same decision
It shouldn't matter what you know about what he knows.
I mean, forget doing it 1000 times and counting the results: How can it be 50 percent likely one way, and 66% likely the other way, when the only difference is that monty knows it's a goat in one case, and in the other case it turns out to be a goat?
You made the same initial choices, you were presented with the same information, you're manipulating the same physical objects and face the same end game in both cases.
How can the odds be different?
Once again, I'm looking for an answer that makes some intuitive sense. I'm aware that it does come out 50/50, but it doesn't make intuitive sense to me.
You're completely ignoring the cases where you pick a goat and he reveals a car.
If you can only have a goat by picking a goat and then enduring a coinflip, how can you have a 66% chance of having a goat?
His Corkiness on
0
Options
Apothe0sisHave you ever questioned the nature of your reality?Registered Userregular
edited July 2008
So, I thought about it alllllll last night.
And I must apologise to ASimPerson, Smasher, et al.
I think I was making the same mistake as most make in the original problem. Changing conditions doesn't change the probability of the original choices that are made - so just like it can't reduce to 50/50 in the initial problem it can't rise in the case of Ignorant Monty.
As soon as I drew the probability trees for Monty and Ignorant Monty it was clear.
Apothe0sis on
0
Options
SmasherStarting to get dizzyRegistered Userregular
edited July 2008
Your chance of picking the car initially is 1/3 no matter what type of Monty we have. However, your chance of having picked the car given that Monty did not reveal the car does depend on which Monty we have.
That may seem counter-intuitive, but it's true. With normal Monty he will never reveal the car, so the fact that he doesn't reveal it doesn't change the odds that you picked it. With ignorant Monty on the other hand, all of the cases where he reveals the car are cases where you didn't pick the car. Therefore there are the same number of cases out of a given number where you pick the car and Ignorant Monty doesn't reveal it, but fewer cases where you don't pick the car and Ignorant Monty doesn't reveal it. That means your chances of having picked the car given that he revealed a goat are greater than those chances if he hasn't revealed anything yet (or than if you had normal Monty).
t apo: no worries. It's amazing how counter-intuitive such a seemingly simple problem can be; I was starting to doubt my logic for a while too.
I have here a standard 52 card deck of playing cards. I spread them out and ask you to take one. Got one? Okay, now I'm going to look at the deck of cards for a certain one... Ah, here it is! I toss the rest of the 50 cards.
Now, here's the deal. If you have the Ace of Spades, I'll give you $50 bucks. If I have the Ace of Spades, you give me $50 bucks. Want to play? :P
No? Why not! You're no fun.
Okay, fine, how about this game. We do the same thing, you pick a card randomly, and I'll randomly toss 50 cards and keep the last one. Same deal, if you have the Ace of Spades I give you $50, if I have it you give me $50. If neither of us have it, nothing happens.
(Depending on how loose you are with money, this isn't that bad of a game :P)
See the difference? In the first game it's practically a guarantee that I will have the AoS. But in the second game, we have the same chance. Why? Because in the first game I deliberately got rid of the losing cards, in the second game, by some miracle I might have gotten rid of the losing cards and left the AoS for myself, but the odds of that happening are the same as the odds of you picking the AoS from the start.
TechBoy on
0
Options
JohnnyCacheStarting DefensePlace at the tableRegistered Userregular
edited July 2008
Actually you know what? We chased over what's been fucking me up a couple posts ago and I just kept right on chugging: My fictive player, in my head, hasn't known if monty knows what he's revealing or not.
I've been approaching the problem the whole time from the angle of you, the player, are just presented with this choice, and it was infuriating me that the odds of the choice could vary. Knowing monty is basically cheating for you by actively killing a torpedo. Ignorant monty is spreading a "fair" game.
Techboy gets a cookie for using the right metaphor for me.
I was treating the problem like a hand of poker - if you don't know if a card is live or dead, you assume its live, in an effort to find the best way to play the hand in lieu of information. But Knowing Monty is going to take some of my opponent's outs out of the deck for me. Ignorant monty is just going to deal. The problem was I was assuming neither monty would tell you his dealing strategy.
I just illustrated the reason poker players on TV have different odds in their minds at any given time then the actual odds in the little box on the television...
How can they be different from the odds when you're presented with identical outcomes and information?
The information isn't identical. In the case of Ignorant Monty, you know that if he didn't reveal the car, it's less likely that you chose a goat to begin with. He never reveals the car if you chose the car, and he reveals the car 50% of the time that you chose a goat. Hence if he doesn't reveal the car, your chance of having chosen a goat go down.
It may help if you change the numbers, ignoring the practicalities. If Monty reveals the car 99% of the time you choose a goat, and in a particular case he does not reveal the car, would you bet that you had a goat or the car?
His Corkiness on
0
Options
JohnnyCacheStarting DefensePlace at the tableRegistered Userregular
edited July 2008
The information only becomes non-identical when your player knows if monty is ignorant or not. That was my disconnect. I get it now.
So now I repeat my followup question:
If you don't know which monty you're dealing with, and he shows you a duck, what do you do?
The information only becomes non-identical when your player knows if monty is ignorant or not. That was my disconnect. I get it now.
So now I repeat my followup question:
If you don't know which monty you're dealing with, and he shows you a duck, what do you do?
Switch. The only time switching is actively harmful is when you have a malevolent Monty who only offers you the choice to switch when you picked the prize, but that possibility is balanced by the benevolent Monty who only offers the switch when you didn't.
If it's just between normal and ignorant Monty, switching won't hurt and it might help.
A friend pointed out to me recently that this can be used to great effect on Who Wants To Be A Millionaire - if you are completely unaware of the answer, mentally choose one, then use your 50/50 lifeline, and your choice is still available, then switching will yield a 75% chance that you will get it right.
I just had a thought about this, and I don't think this offers any advantage.
You choose an incorrect answer at the start (3/4 chance), use your 50:50, and there is a 1/3 chance that the answer you chose is not eliminated. You then switch to the correct answer. 3/4 * 1/3 = 1/4.
If your answer is eliminated, you pick one of the two remaining choices at random: 3/4 * 2/3 * 1/2 = 1/4. There's no other way to win, so the strategy offers a 1/4 + 1/4 = 50% chance.
Using your 50:50 and then picking one at random gives you, obviously, a 50% chance.
This situation is fundamentally different from the Monty Hall problem in that it's quite possible for your original choice to be eliminated by the 50:50. Because of this, your first choice really has no impact at all.
Edit: This doesn't take into account an asshole host who asks for your best guess, then ensures that guess is not eliminated, whether it is correct or not.
Posts
No. This is true in the standard MH problem, but you're assuming it's always the case in all variations when it's not. See my previous post for why it applies in the standard problem but not with ignorant Monty.
You pick a goat initially 66% of the time, but Ignorant Monty causes you to instantly lose 50% of those times. Hence, you can only pick a goat and have a second choice 33% of the time.
You pick a car initially 33% of the time, and you always get a second choice.
In the cases where you get a second choice, you have an equal chance of having either a car or a goat, hence there is no benefit to switching.
No, you don't. Just list out the cases and see. There are six equally likely cases:
1) The car is behind door 1, and Monty reveals door 2
2) The car is behind door 1, and Monty reveals door 3
3) The car is behind door 2, and Monty reveals door 2
4) The car is behind door 2, and Monty reveals door 3
5) The car is behind door 3, and Monty reveals door 2
6) The car is behind door 3, and Monty reveals door 3
Of these, four cases offer you a choice (1, 2, 4 and 5) and two cases which offer you a choice let you win by switching. So if you always switch when offered a choice, you'll win two out of four, or 50% of the cases where you are offered a choice.
You picked a car
you picked goat a
you picked goat b
Monty only offers information
Now you have nine trials where you may or may not get to make decision B
ASSUMING NO VARIENCE
you'll get three where monty reveals the car and ends the game
three where monty reveals goat a
three where monty reveals goat b
Now, in the latter six cases, should you switch or not? You know there was a 33% chance you picked a car, a 66% chance you didn't, and you've had one door eliminated. It's 2-1 for switching. 4 cases where you switch TOO the car, 2 cases where you switch FROM the car, 3 cases where monty wastes the car.
5 cases of loss, 4 cases of win.
As opposed to 3 cases of win, 6 cases of loss if you don't switch.
How does monty revealing the car 1/3 of the time change the odds that you chose wrong initially the other two times?
How can it?
What if you don't know what monty knows, only that this time, he didn't reveal the car? Should you switch then?
(Note: I realize that the answer to the ignorant monty version has been published at 50%, i'm being hardheaded on purpose, until I hear it in a way that makes me click)
I host a podcast about movies.
No, you'll get two where Ignorant Monty reveals the car, 2 where he reveals Goat A, and 2 where he reveals Goat B.
No, you do not know this. If you pick a goat initially, there's a 50% chance Ignorant Monty reveals the car, meaning that there is a 33% chance that you can pick a goat and make it to the second choice, just as there is a 33% chance that you can pick a car. You can't just throw out those possibilities.
Imagine that 50% of the time you chose a goat initially, Monty said "That one doesn't count" and you lost. This would result in you having less chance of choosing a goat and getting to make the second choice. If there's less chance of having a goat, switching becomes less viable.
It's the exact same situation. Monty revealing the car 50% of the time you choose a goat does exactly the same thing as the above situation.
You need to assume one way or the other if you want to follow a strategy.
The only times that Monty reveals the car and you lose without a second choice are when you picked a goat first. So half the time you pick the goat first you lose without a second choice, the other half you want to switch.
Also, I don't think there are nine cases. There are only 2 cases for when you lose outright: when you pick goat A then he goes car, and when you pick goat B and then he goes car. 2 cases for when you want to switch: You pick goat A and he reveals goat B, you pick goat B and he reveals goat A. 2 cases for when you pick the car first: you pick the car he reveals goat A, you pick the car he reveals goat B.
Only 6 permutations.
You could do it with six trials or with 3 or with 33 I guess.
But how does the quality of what monty's revealing too you change the chances you chose right back when you were just picking 1 of 3 choices?
Your odds then were 33%
If he reveals a goat and KNOWS he's revealing a goat your odds of winning by switching are 66%
If he reveals a goat but wasn't sure what he was revealing your odds are now 50/50?
He gave you the same information. It should be the same decision
It shouldn't matter what you know about what he knows.
I mean, forget doing it 1000 times and counting the results: How can it be 50 percent likely one way, and 66% likely the other way, when the only difference is that monty knows it's a goat in one case, and in the other case it turns out to be a goat?
You made the same initial choices, you were presented with the same information, you're manipulating the same physical objects and face the same end game in both cases.
How can the odds be different?
Once again, I'm looking for an answer that makes some intuitive sense. I'm aware that it does come out 50/50, but it doesn't make intuitive sense to me.
I host a podcast about movies.
If you can only have a goat by picking a goat and then enduring a coinflip, how can you have a 66% chance of having a goat?
And I must apologise to ASimPerson, Smasher, et al.
I think I was making the same mistake as most make in the original problem. Changing conditions doesn't change the probability of the original choices that are made - so just like it can't reduce to 50/50 in the initial problem it can't rise in the case of Ignorant Monty.
As soon as I drew the probability trees for Monty and Ignorant Monty it was clear.
That may seem counter-intuitive, but it's true. With normal Monty he will never reveal the car, so the fact that he doesn't reveal it doesn't change the odds that you picked it. With ignorant Monty on the other hand, all of the cases where he reveals the car are cases where you didn't pick the car. Therefore there are the same number of cases out of a given number where you pick the car and Ignorant Monty doesn't reveal it, but fewer cases where you don't pick the car and Ignorant Monty doesn't reveal it. That means your chances of having picked the car given that he revealed a goat are greater than those chances if he hasn't revealed anything yet (or than if you had normal Monty).
t apo: no worries. It's amazing how counter-intuitive such a seemingly simple problem can be; I was starting to doubt my logic for a while too.
I have here a standard 52 card deck of playing cards. I spread them out and ask you to take one. Got one? Okay, now I'm going to look at the deck of cards for a certain one... Ah, here it is! I toss the rest of the 50 cards.
Now, here's the deal. If you have the Ace of Spades, I'll give you $50 bucks. If I have the Ace of Spades, you give me $50 bucks. Want to play? :P
No? Why not! You're no fun.
Okay, fine, how about this game. We do the same thing, you pick a card randomly, and I'll randomly toss 50 cards and keep the last one. Same deal, if you have the Ace of Spades I give you $50, if I have it you give me $50. If neither of us have it, nothing happens.
(Depending on how loose you are with money, this isn't that bad of a game :P)
See the difference? In the first game it's practically a guarantee that I will have the AoS. But in the second game, we have the same chance. Why? Because in the first game I deliberately got rid of the losing cards, in the second game, by some miracle I might have gotten rid of the losing cards and left the AoS for myself, but the odds of that happening are the same as the odds of you picking the AoS from the start.
I've been approaching the problem the whole time from the angle of you, the player, are just presented with this choice, and it was infuriating me that the odds of the choice could vary. Knowing monty is basically cheating for you by actively killing a torpedo. Ignorant monty is spreading a "fair" game.
Techboy gets a cookie for using the right metaphor for me.
I was treating the problem like a hand of poker - if you don't know if a card is live or dead, you assume its live, in an effort to find the best way to play the hand in lieu of information. But Knowing Monty is going to take some of my opponent's outs out of the deck for me. Ignorant monty is just going to deal. The problem was I was assuming neither monty would tell you his dealing strategy.
I just illustrated the reason poker players on TV have different odds in their minds at any given time then the actual odds in the little box on the television...
I host a podcast about movies.
It may help if you change the numbers, ignoring the practicalities. If Monty reveals the car 99% of the time you choose a goat, and in a particular case he does not reveal the car, would you bet that you had a goat or the car?
So now I repeat my followup question:
If you don't know which monty you're dealing with, and he shows you a duck, what do you do?
I host a podcast about movies.
Deja-vu... this seems somehow very familiar but I cant put my finger on it... is it like a joke or line from something else?
If it's just between normal and ignorant Monty, switching won't hurt and it might help.
I just had a thought about this, and I don't think this offers any advantage.
You choose an incorrect answer at the start (3/4 chance), use your 50:50, and there is a 1/3 chance that the answer you chose is not eliminated. You then switch to the correct answer. 3/4 * 1/3 = 1/4.
If your answer is eliminated, you pick one of the two remaining choices at random: 3/4 * 2/3 * 1/2 = 1/4. There's no other way to win, so the strategy offers a 1/4 + 1/4 = 50% chance.
Using your 50:50 and then picking one at random gives you, obviously, a 50% chance.
This situation is fundamentally different from the Monty Hall problem in that it's quite possible for your original choice to be eliminated by the 50:50. Because of this, your first choice really has no impact at all.
Edit: This doesn't take into account an asshole host who asks for your best guess, then ensures that guess is not eliminated, whether it is correct or not.