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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: 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.

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  • (Score: 2) by VLM on Monday October 02 2017, @06:29PM (4 children)

    by VLM (445) Subscriber Badge 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) <pc-soylentNO@SPAMasdf.fi> 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) Subscriber Badge 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) 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) <pc-soylentNO@SPAMasdf.fi> 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) Subscriber Badge 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.