humber is trolling.
The Monty Hall problem
- Thread starter jazzmojo
- Start date
humber is trolling.
It is not exclusively matter of probability, and certainly not of intuition. Each game is an individual case study.
That is the same error Vos Savant makes.
If there are 100 doors, and Monty opens 98 of them to reveal only goats, then the probability that the car is behind the remaining doors is 1. Only because there are 3 doors, can it be it be inferred that the remaining door has p = 1 - 1/3, hence 2/3. ( The given 98/100 door example, differs from a 100 door game where Monty opens one door.)
It is inconsistent to talk of concurrent intentional hiding, and random behavior within the same individual. At each game, Monty can show the contestant the car or a goat. If it's the car, one may make up all sorts of trivial outcomes, but a goat always means that the contestant should swap, no matter how that goat is revealed. Monty's intent plays no part in that 2/3 case; the well known case.
Only in a very limited way. If Monty shows the contestant a goat that was behind one of the remaining doors, that informs the contestant that the best strategy is to swap. That gains the contestant 2/3 chance. If Monty shows a goat that was behind the contestant's door, the contestant still learns to swap, but with only a 50/50 chance of success with the two remaining doors. For Monty to improve upon that, additional rules must be invoked to handle the cases where he shows the contestant the car.
In terms of game theory, making that useful information available to the contestant, it the price Monty pays participating, while not wanting to freely give the car away.
Very evil Monty, Evil Monty....Very benevolent Monty.
It's the apparent 2/3 shift in probability that brings the problem any attention. That is readily explicable by logic. If it ain't there, it's here.
Probability is used to make mess of that, and cause so many to wonder.
When that is done, the straight-forward logic that produces the 2/3 result no longer applies. If Monty is seen as an opponent trying to keep the car, and the contestant trying to win it, then as that paper explains, it is an example of a two stage, two person zero sum game.
If follows that repeated iterations of that same strategy, will result in the same outcome, and so may be seen as probability.
Last edited:
Jack by the hedge
Safely Ignored
- Joined
- Oct 14, 2009
- Messages
- 23,277
Which error?
The 98/100 game is equivalent to the usual 3 door game as in both cases Monty reveals all the unchosen doors but one. In each of those games the choice is effectively the original selection versus all the other doors combined.
...except in cases like Evil Monty who only offers a swap when you chose correctly first time.
Rolfe
Adult human female
I'm inclined to think you may be right. That would explain a lot, including this part.
Nobody can be that dense by accident. But if he is trolling, A for effort, definitely.
Rolfe.
Yep, and that means the chance for the contestant is 50/50 for the remaining doors. The car can't have been exclusively in the "other" door.
Make it 1000 doors and it will be the same. Only 3 doors produces the doubling.
And how does he inform the contestant that he has done that, if he doesn't open the door to show the car? If he simply tells the contestant that the car is there, then you must take into account that the contestant has a strategy, so will stick with that door. If he says nothing, the contestant has no logical reason to swap.
That is why I made the distinction between being in the garage, and using the light switch or not. I can act against my own best interests, and mimic evil Monty.
Rolfe
Adult human female
Rolfe.
sol invictus
Philosopher
- Joined
- Oct 21, 2007
- Messages
- 8,613
That's because it is gibberish.
This is a poster that has posted tens of thousands of times - literally, no exaggeration - over the last two years or more on the topic of DDWFTTW, and still denies it's possible despite overwhelming evidence to the contrary (including a world record).
I don't know what his (?) deal is, but I advise you to ignore his posts unless they entertain you.
Slayhamlet
Master Poster
- Joined
- Apr 26, 2007
- Messages
- 2,423
In that case (if I understand you correctly) your overall chances of winning the prize can be no greater than 1/2. Let me explain:
After Monty's coin lands on the "give contestant a choice" face the chances are of course exactly the same as in the classic Monty Hall problem: 2/3 if you switch, 1/3 if you don't. But when you factor in the 50:50 probability that you might not get a choice to switch in the first place, your overall chances are reduced to 1/2 (assuming you always switch when given the choice). And obviously if you decide never to switch, your chances will remain 1/3 regardless of Monty's coin toss.
It's easy to work out visually:
G = goat
P = prize
[no choice]-[choose to stay]/[choose to switch]
goat door 1:
G-G/P
goat door 2:
G-G/P
prize door:
P-P/G
G-G G-G P-P (without switching) G=4 P=2
G-P G-P P-G (with switching) G=3 P=3
P = prize
[no choice]-[choose to stay]/[choose to switch]
goat door 1:
G-G/P
goat door 2:
G-G/P
prize door:
P-P/G
G-G G-G P-P (without switching) G=4 P=2
G-P G-P P-G (with switching) G=3 P=3
Rolfe
Adult human female
I thank you. I hadn't encountered Humber before, so that provides useful context.
I think I have my brain round the Monty Hall thing, but every time this conversation happens I think I grok it slightly better. I'm always open to the possibility there could be a new angle on it, so I initially wondered if Humber was saying something useful.
It seems not, though.
Rolfe.
Jack by the hedge
Safely Ignored
- Joined
- Oct 14, 2009
- Messages
- 23,277
No. That means the chance is 1/N or (N-1)/N for the two choices with N doors. One of the unopened doors hides the car. The contestant is choosing between their original choice with its original probability or all the other doors.
I have no idea what you intend that to mean.
True. (N-1)/N is double 1/N only for N = 3. For larger values of N the ratio is more than double.
Evil Monty doesn't inform the contestant he's done anything. He just accepts their first choice if they guess wrong, and only offers a swap if they guess right. Contestant's best strategy is not to swap.
If contestants spot Evil Monty's strategy they'll all stop swapping. In that case he might switch to also offering a swap 50% of the time they guess wrong. That might sucker some contestants into switching again. Again the contestant's best strategy is not to swap.
Rolfe
Adult human female
I see your point, but that wasn't really the angle I was looking at. Monty may vary the frequency with which he offers the switch as much as he likes - the only stipulation is that he can't base his decision on whether or not the switch is offered to any given contestant on whether or not the contestant already picked the prize door.
So, if you aren't offered the switch, your chance of winning is 1/3rd. However, if you are offered the switch, and you switch, your chance of winning is 2/3rds.
Overall, going into the game, the chance that you will win depends on how often Monty is deciding to offer the switch.
Rolfe.
Litotes is a from of irony that may lead to trouble. You have fallen for the hyperbole if intuition and "getting" the problem and so inherited Vos Savant's errors.
The only variation of interest is the well known version, where there is an apparent doubling of the contestant's chance of winning the car. The problem has also been called the "Monty Hall Paradox", but there is nothing to the problem that is in the least surprising, paradoxical or intuitive.
The problem has taken on many forms, but it is Vos Savant's version that has produced so much internet chat, and that apparent need for intuition and talk of probability. If his her interpretation of the problem that is responsible for that. I will get back to that later.
I have no dispute with any of the probability calculations. They have been proven by arithmetic means, Bayes and calculus. That is to be expected, because they are all the product of the same mechanism. Information and its employment. Essentially, game theory.
You do not seem to understand what I say, so I will post only part of the way I see the problem, and then from another source, even though I think the latter is more complicated than necessary.
The TV game and goats are frippery, so I will use garages and Jeeves.
The Setup
1) There are three garages and one car.
2) My aim is to find the car using logic, and only resorting to chance as a last resort.
3) I have my Monty, in the form of Jeeves.
All about Jeeves
He knows where the car is, but I don't have access to that information. His "motive" is irrelevant, but is involvement means:
a) He will not inform me of where the car is.
b) As a direct consequence of that, he may be useful to me.
First attempt to find the car:
At first, I try to find the car without the Jeeves. Before me, are the closed garages, and within one of them lies the car. I could say " I have a one in three chance" of finding the car, but I want to minimize what is left to chance, and the process to be algorithmic, so say "I don't know where the car is, other than it is in one of the 3 garages" That is the extent of my knowledge. Can I improve upon my current situation?
Not without more information, so I recruit Jeeves.
Second attempt to find the car:
From (a), I know that Jeeves won't tell me where the car is, but that means I can be certain that he will tell me where it isn't. He acts accordingly, so opens a garage that is empty.
That leaves me with two garages. I know it is within one of them, but no reason to think I should select one over the other. That's an improvement, but can I do better? No, because that is Jeeve's best strategy for not letting me find the car, other than not being involved at all.
But, I have a strategy too, and can improve that to 2/3. I could go on to show how the 2/3 result evolves, but you can see that I can change the problem from Monty being an unwilling game show host, to me recruiting him to my benefit. And his best game, results in 50/50 for me.
Anyway, here is another view: ( my bolding, underlining)
The Holy Grail of MHP studies was for me, for a long time, to find a stupendously elementary proof of the fact that, in the case of a possibly biased host, there is no strategy for the player with a better overall success chance than 2/3. Hence the player might just as well ignore all specific door numbers and just switch.
Well, here is one, which I learned from a Wikipedia editor. I use rather a lot of words below: first to introduce the problem, and then to solve it. It could all be said in much fewer, but I hope this way there can be no misunderstanding.
A.3.1 The problem
Let's suppose the car is hidden behind one of the three doors by a fair randomization. The contestant chooses Door 1. Monty Hall, for reasons best known to himself, opens Door 3 revealing a goat. It can be shown using Bayes' theorem (or better still, Bayes' rule) that whatever probability mechanism is used by Monty for this purpose, the conditional probability that switching will give the car is at least 1/2. We know that the unconditional probability (i.e. not conditioning on the door chosen by the contestant, nor the door opened by Monty) is 2/3. Using the law of total probability
and the fact that all conditional probabilities of winning by switching are at least 1/2, proves that 2/3 overall win-chance can't be beaten.
Do we need Bayes to come to this conclusion? Surely, nobody in their right mind could imagine that there could exist some mixed strategy (sometimes staying, sometimes switching, perhaps with the help of some randomization device, and all depending on which doors were chosen and opened) which would give you a better overall (i.e. unconditional) chance than 2/3 of getting the car. Is there an elementary proof? A short proof using words and ideas, no computations? Yes there is, and I learned it from a wikipedia editor.
A.3.2 The solution
Obviously the player only needs to consider deterministic strategies for himself. Now suppose Monty Hall makes his choice of door to open, when he does it at all, by tossing a possibly biased coin (a possibly different coin for each door). He might just as well toss his three coins in advance and just "look up" the action which is needed, if and when an action is needed. Now suppose the player also gets to see the results of the three coin tosses in advance. He now knows even more, so he cannot do worse (provided he uses all the available information as best as he can).
But now we are effectively in the so-called "Monty crawl" situation: this is the problem where the door which Monty would open in each of the three situations where he does have a choice (because the player is standing at the door hiding the car) is actually fixed in advance and known to the player.
We want to show that for the Monty crawl problem, there still is no strategy with an overall win-chance of more than 2/3. There are just two cases to consider now. Suppose the coin says that Monty would open door 3, if he had a choice between 2 and 3. Then, whether the car is behind door 1 or door 2, Monty is certain to open door 3.
His action tells us nothing so we may as well switch. Suppose the coin says that Monty would open door 2, if he had a choice between 2 and 3. Then the fact he opens door 3 shows us that the car must be behind door 2, so we must switch.
Either way, we might as well switch. If we switch anyway, our overall win chance is 2/3. So, this is the best overall win chance which is available for us for the Monty crawl problem. Hence the best in general.
A.3.3 Conclusion
To sum up: you can't do better than 2/3 overall because you can't do better than 2/3 in the situation that would be most favourable to you; Monty crawl.
And therefore, because you can't do better than 2/3 overall, the chance of winning by switching must be at least 1/2 in each separate situation which you can distinguish (reductio ad absurdam and law of total probability).
We are in fact using Bayes in the Bayes' rule form, but only for the situation when the evidence we are given is certain under both hypotheses, and for the situation when it is certain under one, impossible under the other.
We are solving Monty Hall by use of the more simple problem Monty crawl. It reduces the problem just to an enumeration of two possible cases.
We are using the insight of all game theorists that one can always reduce everything to the extreme (deterministic) case. We are actually using game theory to prove an inequality about conditional probabilities!
Since 2/3 overall is the best you can do, and you can achieve that by always switching, it's a waste of time to look at the specific door numbers and a waste of time to figure out conditional probabilities with Bayes' theorem or whatever.
Back to my words.
Vos Savant
She is not the originator of the problem, nor did she pose it in Parade magazine, but answered a question put to her, but re-worded it. Subsequent replies and remarks, show that she does not understand the problem, and used naive intuition about probability.
Original question:
"I've worked out two different situations based on whether or not Monty
knows what's behind the doors. In one situation it is to your advantage to
switch, in the other there is no advantage to switch. What do you think? "
Re-written by Savant:
"Suppose you're on a game show, and you're given the choice of three
doors: behind one door is a car; behind the others, goats. You pick a door,
say No. 1, and the host, who knows what's behind the doors, opens another
door, say No. 3, which has a goat. He then says to you, Do you want to
pick door No. 2?" Is it to your advantage to switch your choice? ".
Why the 100 door remark is wrong:
By changing one aspect of the problem, this way of understanding why the contestant indeed should switch may become even more compelling to the reader.
Consider the 100-door problem: 99 goats and one car. The player chooses one of the 100 doors. Let's say that he chooses door number 1. The host, who knows the location of the car, one by one opens all 99 of the other doors but one; let's say that he skips door number 38. Would you switch?
The simple solutions often implicitly used frequentist picture of probability: probability refers to relative frequency in many repetitions. They also do not address the issue of whether the specific door opened by the host is relevant: if the player has initially chosen door 1, could it be that the decision to switch should depend on whether the host opens door 2 or door 3? Intuition says no, but we already saw that naive intuition can be misleading.
The switch in probability is apparent, and the result of sloppy thinking: Erdos is right.
And BTW, Sol thinks that a runner on an exercise treadmill is exactly like a runner running downwind at the speed of the wind, and does not know that air pressure is a result of gravity.
And we have met before over at Homeopathy.
Last edited:
Jack by the hedge
Safely Ignored
- Joined
- Oct 14, 2009
- Messages
- 23,277
You appear to be considering a range of different versions of the game here but, so far as I can tell, you always conclude switching is the best strategy. However that's obviously not true in the specific case of Evil Monty who does not always offer a switch and biases his offers toward players who guessed right.
Also, I cannot understand your attempted explanation of why "the 100 door remark is wrong". Oh, and who, by the way, is Erdos?
PS A runner on a treadmill is like a runner running downwind at windspeed.
Also, I cannot understand your attempted explanation of why "the 100 door remark is wrong". Oh, and who, by the way, is Erdos?
PS A runner on a treadmill is like a runner running downwind at windspeed.
No, and would you care what the result would be for a following competitor?
Get over it. It is not about frequency.
Jeeves' best strategy results in 50/50 for me, even if he does his worst.
The conditional probability of winning by switching is never less than 1/2.
There is no strategy giving more than 2/3, so you may as well swap no matter what.
It is best to remember that the possibility that the chosen door contains the car remains at 1/3, and there is still a 1/3 chance that the "2/3" door won't.
Makes false assumptions. That is clear if you have the "right stuff".
Q.E.D.
It was not just he who objected, but many others. However, shouty blockheads and their "intuition" forced the other case.
A great deal has been written about this, but mostly esoteric papers. But, their authors don't bother with forums. They know what to expect.
Jack by the hedge
Safely Ignored
- Joined
- Oct 14, 2009
- Messages
- 23,277
You might find it to your advantage to observe what befell previous competitors.
Probability is not about frequency?
Jeeves is playing a different variant.
The probability of winning by switching with Evil Jeeves is zero, since Evil Jeeves only opens a door if you originally guess right.
All variants make assumptions. The 100 door version only makes the same assumptions as the familiar 3 door version with the 1/3 vs 2/3 probabilities. i.e. after you choose, Jeeves will open all but one of the remaining doors and will not reveal the prize.
Intuition helps people grasp this version as intuition tells them a) their first guess is almost certainly wrong and b) there's probably something special about the door Jeeves doesn't open. And 99 times out of 100 their intuition will be right.
Jack by the hedge
Safely Ignored
- Joined
- Oct 14, 2009
- Messages
- 23,277
Thanks. I realise now that I knew of him by reputation, but the name hadn't stuck.
If Monty always allows a swap when you guess right, and allows a swap 50% of the time you guess wrong, it doesn't matter whether you swap when offered the chance: the probability is 1/2 that your original choice is right.
Dave Rogers
Bandaged ice that stampedes inexpensively through
Only for a simple first-order incarnation of Evil Jeeves, though. A second-order Evil Jeeves would reason that you will never switch, and will therefore bluff you by only opening a door if you guess wrong.
Dave