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Monty Hall, Co-Creator and Host of 'Let's Make a Deal,' Dies at 96
Monty Hall, the genial host and co-creator of "Let's Make a Deal," the game show on which contestants in outlandish costumes shriek and leap at the chance to see if they will win the big prize or the booby prize behind door No. 3, died at his home in Beverly Hills, Calif., on Saturday. He was 96.
[...] "Let's Make a Deal" became such a pop-culture phenomenon that it gave birth to a well-known brain-twister in probability, called "the Monty Hall Problem." This thought experiment involves three doors, two goats and a coveted prize and leads to a counterintuitive solution.
[...] Mr. Hall had his proud moments as well. In 1973 he received a star on the Hollywood Walk of Fame. In 1988, Mr. Hall, who was born in Canada, was named to the Order of Canada by that country's government in recognition of the millions he had raised for a host of charities. In 2013 he was presented with a lifetime achievement award at the Daytime Emmys.
The Monty Hall problem:
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?
Vos Savant's response was that the contestant should switch to the other door. Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance. [...] Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong. Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating the predicted result.
Related: Get Those Brain Cells Working: The Monty Hall Problem
(Score: 4, Funny) by crafoo on Monday October 02 2017, @01:25PM (22 children)
It's only difficult for people to understand because this is the 3 door version of the question.
(Score: 2) by opinionated_science on Monday October 02 2017, @01:30PM
i mod'd funny. Did this in grad school...
(Score: 4, Interesting) by VLM on Monday October 02 2017, @01:56PM (13 children)
Modded as funny but the 3 doors is the lowest odds of winning at merely 2/3 but with more doors the odds of winning approach 100% if you switch.
For example lets play "Guess VLM's credit card number" which is a 16 digit number.
The odds of you guessing it off the top of yer head are essentially zero, or more realistically 1 in 1e16. The actual ratio is much larger because the first 4 digits are the bank and there are or used to be checksum algos for the remaining 12 digits. Also I have more than one CC. For the sake of numerical simplicity we will assume my CC number is a single 16 digit random number for the remainder of the analysis.
Monty "helps out" by providing you a list of (1e16 - 2) sixteen digit card number that are definitely not VLM's credit card number. That means there's two numbers left, the one you picked with 1 in 1e16 odds of getting it right, and the other number, which has odds of (1 - 1/1e16) odds of actually being my credit card number. Hmm, 1/1e16 or (1 - 1/1e16) which is bigger? You should always switch, only 1 in 1e16 trials will your first guess be my actual credit card number.
Or you could play the age guess or weight guess or partner count guessing game. I'll pick a number. 42. Some magical oracle crosses off every wrong number except for my first guess, 42, and, of course, the real answer. Should I switch to the remaining not crossed off number, the one that is my CC number with odds of only 99.99999999999999% of being correct?
This is a typical intelligence test fail WRT Marilyn Voss and her column etc. IQ is really good at testing raw processing horsepower, which ends up correlating well with all kinds of measures of success in life, etc. IQ is not a trivia contest, and this door puzzle is mere statistical algorithm trivia. If you know the algo its an easy puzzle, if you don't, well best of luck to ya. There are a lot of people born before mathematical algorithm innovation X, Y, or Z that were indeed IQ smart but didn't know various pieces of trivia like an algorithm invented long after their lifespan ended. Like the difference between a historian good at analysis and a historian good at memorizing trivia, and mistakenly (perhaps intentionally to muddy the waters?) claiming both have a high IQ or are "smart".
Its also a good troll topic. Trolling wasn't invented on 4chan last week. People been doing stuff like that, using stuff like the door puzzle, for millennia, most likely. Its a good troll topic because people who don't know the analysis algorithm can get all agitated by this puzzle quite easily.
(Score: 3, Funny) by VLM on Monday October 02 2017, @02:01PM
Correction, there's IQ, trivia knowledge, and blood caffeine level, and a small value for the latter leads to weird post editing mistakes.
Now given a first mostly random guess of which SN post is post, and a giant shitload of corrections that eliminate almost all the other posts, leaving your first guess, and one other post not marked as Fed up editing, should you switch your vote for well edited post to the other post? Yes statistically that will ALWAYS be the better edited post. Especially if your first guess was one of my "low caffeine level" posts.
(Score: 0, Disagree) by Anonymous Coward on Monday October 02 2017, @02:56PM (8 children)
This has nothing to do with guessing any value (like your credit card number), only selecting one of a limited set’s elements - in this case three doors. If there are three doors, and you can guess only one door for your final choice, then your odds are one in three.
If you can change your guess - before the car is revealed - then your odds are still one in three because you only can guess one door out of the three as your final choice. If you can change your mind three times, and there are three doors, you still only have a one in three chance because your final choice is still one door out of the three.
Your odds of guessing correctly are based on the number of doors you are allowed to choose for your final choice(s). But if you only get one final choice then your odds are still one in three.
(Score: 2) by VLM on Monday October 02 2017, @06:29PM (4 children)
Sorry sir, no.
Your first guess is 1 in 3.
Then monty releases the very important data point that one door is definitely not the right door, and you're not allowed to change to that door. There was a 2/3 chance it was one of the other two doors and by opening one that definitely is not, that means 2/3 chance collapses into one remaining door.
This relates to your final choice being 1 out of 2, not 3. You can't select the door that monty opened and showed was empty.
How bout you flip it. negative going logic instead of positive going logic... Your first choice is not 1/3 of yes but you're really saying its 2/3 of no. There's a 1/3 chance that both of the remaining doors put together are "no" and that is the same as saying a 2/3 chance that one of the other doors is yes. Monty opens one of the other doors proving its a "no". Instead of two doors sharing a 2/3 no, you've got one unopened door thats now a 1/3 no, aka a 2/3 odds of yes.
(Score: 2) by FatPhil on Tuesday October 03 2017, @07:44AM
However, sometimes the question is ambiguously worded, or leaves wiggle room for deliberate misinterpretation. The actual make a deal scenario *does* deviate from the monty hall problem, for example. The host must be disinterested, and must always offer a chance to swap (and know where the car is). Real Monty didn't always offer the swap, and was not entirely disinterested.
It's funny that the person you're responding to so nearly made the link between his premises and the correct conclusion, and in fact your entire argument:
"""
Your odds of guessing correctly are based on the number of doors you are allowed to choose for your final choice(s). But if you only get one final choice then your odds are still one in three.
"""
Because:
sticking means chosing one door for opening.
switching means chosing *every other door*, just with no surprises from the ones monty revealed.
Great minds discuss ideas; average minds discuss events; small minds discuss people; the smallest discuss themselves
(Score: 0) by Anonymous Coward on Tuesday October 03 2017, @07:51AM (2 children)
How about this analysis.
At first the door you choose is 1 out of 3.
The other two doors together are 2 out of 3.
When the single door is opened, that changes nothing, the other two doors are still 2 out of 3.
If the contestant changes the selection, they are effectively choosing two doors over one door
giving the 2/3 probability.
Note that of the "other two doors", at least one is always a looser, so a looser can always be opened.
(Score: 0) by Anonymous Coward on Tuesday October 03 2017, @12:02PM
Pretty much this.
Remove the temporal element and the choice becomes "do you pick this 1 door or these 2 doors", which makes it obvious to me why switching gives 2/3 odds of winning.
(Score: 2) by VLM on Tuesday October 03 2017, @03:55PM
Thats more eloquent and brief than my explanation AC, nice job.
(Score: 2) by maxwell demon on Tuesday October 03 2017, @07:13AM
You are wrong. The host gives you the information: "If you didn't chose the correct door right away, then the car is in the door I did not open". Now the probability that you didn't choose the correct door right away is 2/3, and therefore with that probability that the car is behind that other door. Obviously the car doesn't move between your choices.
The Tao of math: The numbers you can count are not the real numbers.
(Score: 2) by FatPhil on Tuesday October 03 2017, @07:19AM (1 child)
But VLM's credit card number *is* one from a limited set.
Looks like VLM wasted all that effort for nothing, you're entirely clue-resistent.
Great minds discuss ideas; average minds discuss events; small minds discuss people; the smallest discuss themselves
(Score: 2) by VLM on Tuesday October 03 2017, @04:02PM
Maybe it'll help OP to point out that my CC example is a ridiculously large set, and in that set its (obviously?) always a better idea to switch. Now if in theory with three doors its better not to switch or it doesn't matter, surely ya gotta travel a path from there to here of ever decreasing advantage to switch. You should be able to graph that curve, do calculus on it, find a first deriv zero below which its always in your benefit to not switch... so ... try to calculate that number. There isn't such a solution, so that implies ...
So numerically model it out with 1 in 1e6 lotto numbers or my mass in Kg or whatever. Eventually OP thinks there will be an inflection point where not switching starts to win and eventually wins. But it turns out if you actually run the numbers on a spreadsheet or something the advantage of switching never drops far enough even with merely three doors, and the protocol doesn't operate with less than three doors, so ....
The switching advantage is near 100% for a bazzilion doors and it drops to merely 2/3 chance with only three doors and there is no such number of doors where the advantage ever drops below 2/3 or 50/50 or somehow hits zero.
(Score: 3, Informative) by Arik on Monday October 02 2017, @03:21PM (1 child)
If laughter is the best medicine, who are the best doctors?
(Score: 5, Interesting) by bob_super on Monday October 02 2017, @06:30PM
The problem that people really faced is that they made a decision. Maybe it's good, maybe it's bad.
They are presented with new evidence, and they have to decide whether they made a wrong decision. Nobody ever wants to have made a wrong decision. Nobody ever likes to change the decision they made, even if they are told repeatedly that the odds that it turns out correct are significantly lower than the odds that they were right, but the odds that they were right are non negligible.
Change your mind? Hate that one of your two decisions is the wrong decision: 100% chance you have been wrong in this process. Not change your mind? The world tells you you should, yet you struggle to admit you could be wrong, and maybe not get rewarded anyway.
Totally unlike elections.
(Score: 2) by maxwell demon on Tuesday October 03 2017, @07:01AM
That depends on how you generalize to more than three doors. You are probably thinking of the generalization "the host opens all remaining doors except for one" which is usually used for helping intuition.
But from the original problem description, the host opens another door, that is, one door, therefore the obvious generalization if not having a specific goal in mind is that also in the generalized version the host opens just one door. In that case, with more doors your chances go down with more doors. Indeed, the probability when changing then is (n-1)/(n(n-2)), which, while still better than 1/n, goes closer to it the more doors there are, so (as one would intuitively expect) the help you get by the host opening that door diminishes for large n. Note that no matter whether you switch or not, the probability of getting the car approaches zero.
Another possible generalization is that the host opens exactly half of the remaining doors (this of course only works if the total number of doors is odd). In that case, your chance when switching is 2/n, so you double your chances by switching, just as in the original problem, but your chances to get the car go to zero for large n, not to 1.
Note that all three generalizations reduce to the original Monty Hall problem for n=3.
The Tao of math: The numbers you can count are not the real numbers.
(Score: 4, Insightful) by theluggage on Monday October 02 2017, @02:30PM (6 children)
Nah. If Erdős didn't spot the "N doors as N->infinity" case then it ain't gonna convince the great unwashed. Even if there are N doors - I pick one, Monty reveals a goat, there are N-1 doors left, so the probability of the car being behind any of the doors -including the one I picked - is 1/(N-1) - right? Actually, that is a true statement - its what you'd get if you picked again at random - its just not the correct answer to the question being asked.
The first tricky bit is realising that - unless you act on the new information Monty is giving you - the first door you choose will never have anything other than a 1/N chance of winning, however many goats Monty reveals. The second tricky bit is realising that - because the total probability must be 1 - the probability for every other door must go up every time Monty reveals a goat. The "paradox" is, how can the universe distinguish between you forgetting what has gone before and picking again at random (prob of winning: 1/(N-1) ) and "switching" doors (prob of winning: ( 1 - 1/N)/(N-2) or whatever)
The whole thing is grist to the mill of whether probabilities have any meaning beyond predicting the outcome of a significant number of trials - here the "individual" case is so abstract and meaningless that it turns the problem into some sort of Schroedinger's Goat mindfuck. The thing that clarified it for me was not the result of doing a computer simulation, but simply trying to write one and being forced to think "what happens when X people do this?" which rapidly leads to "hey, this is a stupid question because the two cases are completely different". You have to assume that either everybody switches, or nobody switches, and in the last case you have a couple of lines of dead code.
Oh, PS, Obligatory XKCD [xkcd.com]
(Score: 0) by Anonymous Coward on Monday October 02 2017, @03:29PM (1 child)
Is it so complicated? When you first pick there is p = 1/3 chance it was correct (and hence q = 2/3 incorrect).
If p happened, there is 100% chance the host will offer you the wrong door to switch to. If q happened, there is 100% chance the host will offer you the correct door. So the probability of winning if you switch is just q*1 = 2/3.
(Score: 1, Interesting) by Anonymous Coward on Monday October 02 2017, @05:59PM
But remember, there are four cases for the rules:
1) The host only shows you a goat when you have picked the car.
2) The host only shows you a goat when you haven't picked the car.
3) The host always shows you a goat.
4) The host is capricious about showing you a goat (the real Monty Hall).
What trips people up is that they don't stick to a single case while they do their analysis. The problem statement makes it easy to fall into that trap.
(Score: 2) by fritsd on Monday October 02 2017, @05:46PM
Nicely explained!
And that, dear people, is why it's so bloody difficult to get your head around Bayesian statistics :-)
It just feels un-natural.
(I think)
But as Monty adds information, the prior changes. The probability density function changes from a flat uniform 1/N line to a series of taller block wave functions with small goats inbetween.
(Score: 2) by VLM on Tuesday October 03 2017, @04:14PM (2 children)
The car can't move on its own, and that darn Monty keeps shrinking the territory the car could be hiding in. Its not the universe messing with the odds, its Monty. Monty is a cheater and he knows where the car is and he's telling you where it isn't, which is almost the same as telling you where it is, once he's leaked enough info.
Maybe a hunting / fishing analogy. Ya got 100 ice fishing ponds and exactly one fish. The odds of you ice fishing on your first guess and catching the fish are really low. That darn Monty tosses dynamite into 98 of what he knows are empty ponds and nothing floats... "obviously" the fish isn't in your first guess pond, and it must be in the one remaining pond Monty didn't dynamite, right? The numbers scale all the way down to precisely three ponds. Or some similar weird analogy with deer hunting.
(Score: 2) by theluggage on Tuesday October 03 2017, @09:55PM (1 child)
Sure, but the slippery bit (that has fooled the great and the good) is getting why that hasn't changed the odds of the car being behind your door, but has changed the odds for the remaining door. The "total probability must be 1" thing proves it, but the trouble is that proving something like that is not the same as explaining something. What explained it for me was the experience of writing code to simulate it - which forces you to "depersonalise" the whole thing and think in terms of how repeated trials might work.
(Score: 2) by VLM on Wednesday October 04 2017, @07:43PM
Hmm well flip positive and negative thinking like I alluded to in another post.
I'm not saying I think there's a 1/3 chance the car is behind my door, I'm really saying there's a 2/3 chance its behind the other doors. And that pesky Monty leaked information such that "the other doors" has collapsed down from a buncha doors, down to precisely one remaining unopened unselected door. All the 2/3 is sitting on that one remaining door.
The 2/3 chance of it being behind other doors has remained constant from the first step. Its just the set of "other doors" has collapsed thanks to Monty from a buncha doors to precisely one door. That one remaining unopened unselected door owns the whole 2/3 chance subspace now, it doesn't share with multiple doors.
Its an English Language puzzle. Instead of all this "where is the car" odds, it seems clearer if we talk about "where isn't the car" odds. I've seen attempts at artificial languages from esperanto to lojban and it would be interesting to see mathematicians write a general purpose language. Well, interesting might not be the correct word. But it would be something, thats for sure.