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posted by martyb on Monday October 02 2017, @01:02PM   Printer-friendly
from the deal++ dept.

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


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  • (Score: 4, Interesting) by VLM on Monday October 02 2017, @01:56PM (13 children)

    by VLM (445) on Monday October 02 2017, @01:56PM (#575891)

    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.

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  • (Score: 3, Funny) by VLM on Monday October 02 2017, @02:01PM

    by VLM (445) on Monday October 02 2017, @02:01PM (#575893)

    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)

    by Anonymous Coward on Monday October 02 2017, @02:56PM (#575915)

    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)

      by VLM (445) on Monday October 02 2017, @06:29PM (#576053)

      If you can change your guess - before the car is revealed - then your odds are still one in three

      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

        by FatPhil (863) <reversethis-{if.fdsa} {ta} {tnelyos-cp}> on Tuesday October 03 2017, @07:44AM (#576483) Homepage
        My favourite use of monty hall is not for distinguishing smartypants who know the answer from those who don't, but for distinguishing those who can understand a sound logical explanation from those that can't. There's no shame in getting the answer wrong when initially posed the question, it's not immediately obvious, but there is something wrong without being able to understand an explanatio such as yours.

        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)

        by Anonymous Coward on Tuesday October 03 2017, @07:51AM (#576486)

        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

          by Anonymous Coward on Tuesday October 03 2017, @12:02PM (#576536)

          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

          by VLM (445) on Tuesday October 03 2017, @03:55PM (#576627)

          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.

          Thats more eloquent and brief than my explanation AC, nice job.

    • (Score: 2) by maxwell demon on Tuesday October 03 2017, @07:13AM

      by maxwell demon (1608) Subscriber Badge on Tuesday October 03 2017, @07:13AM (#576473) Journal

      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)

      by FatPhil (863) <reversethis-{if.fdsa} {ta} {tnelyos-cp}> on Tuesday October 03 2017, @07:19AM (#576476) Homepage
      > This has nothing to do with guessing any value (like your credit card number), only selecting one of a limited set’s elements

      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

        by VLM (445) on Tuesday October 03 2017, @04:02PM (#576635)

        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)

    by Arik (4543) on Monday October 02 2017, @03:21PM (#575924) Journal
    After reading the article I see why I misunderstood this all these years. The key is in the assumptions which are usually left unsaid, but need to be. The assumption that the host will always open a door, for instance. That *completely* changes the problem. Kind of funny that mathematicians call this the 'Monty Hall' problem when it's an entirely different problem to the one players on his show actually faced.
    --
    If laughter is the best medicine, who are the best doctors?
    • (Score: 5, Interesting) by bob_super on Monday October 02 2017, @06:30PM

      by bob_super (1357) on Monday October 02 2017, @06:30PM (#576055)

      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

    by maxwell demon (1608) Subscriber Badge on Tuesday October 03 2017, @07:01AM (#576467) Journal

    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.

    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.