Originally published at: https://boingboing.net/2019/08/27/the-problem-with-the-monty-h.html

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# The problem with 'The Monty Hall Problem' was Monty Hall

That linked Let’s Make A Deal show, what in the hell? They gave a woman from Iowa a lion cub in the first game! WTF do you do with a lion cub in 1973?

My human brain still can’t deal with the Monty Hall Problem in terms of probability. I like the idea of separately blaming it on Monty Hall. Much easier.

The lion cub was presumably valuable. When it was time to pick the two contestants for the Big Deal of the Day, they would ask earlier-round winners in the order of their prizes’ worth. So such impractical, valuable prizes went to folks who were soon asked to give them up for a shot at the top prize. Of course, all the giraffe-winners I ever saw gratefully accepted.

That’s the wrong question to be asking. How about:

What

couldn’tyou do with a lion cub in 1973?

I agree it’s counter-intuitive when dealing with only three doors. The only way I was able to wrap my head around it is to think of it in terms of a million doors.

Let’s say that you’re trying to find one prize behind a million doors. You guess randomly, and the host then eliminates all but two choices, the one you picked and one other and tells you that the prize is behind one of these two. In that case, it’s intuitive that switching your guess greatly increases your odds from the literal one in a million shot you had originally.

Also, I disagree with the article because the human element, the “Monty Hall” of the problem *isn’t* a problem at all. Mathematically, if you *can* switch you are better off doing so, regardless of any goading one way or the other. That’s the math.

Thanks for the explanation.

But doesn’t the math assume random probability and blind choice? If Monty Hall were to, say, always only offer the choice of switching when switching was bad, wouldn’t switching be bad?

*Monty’s* actions are not random, because (in the case where you did not initially pick the prize door) he bases the decision about what doors to open on his prior knowledge of where the prize is and isn’t.

Why switch from door 1 to door 3 once door 2 is revealed empty? The chances are now 50/50 for either 1 or 3. There’s no reason to switch picks.

No. You had 1 chance in 3 of being right before Monty acted, and his actions don’t change that. But now you know where the entire other 2/3 chance of being right is.

Yes, this analysis assumes that Monty *always* offers you the option to switch (I don’t remember if that was actually the case). If he has a choice of whether or not to make that offer, the analysis breaks down.

I skipped around a little and saw the trade-ins but the lion cub winner didn’t get asked. I do wonder if they actually took the animals home or were offered a cash equivalent after the show.

Also, the big prize was a 55-day cruise around South America. These days we’re lucky to be able to string 2 weeks off at one time (in the US anyway), I can’t even imagine 2 months off.

I did some searching and found who seems to be the woman on the show, but sadly her obituary doesn’t mention bringing a lion to Iowa. Naomi Regina Fitzpatrick

If switching is only offered some of the time, and Monty is biased against the contestant as you say, then yes, that is a fundamentally different scenario than the one mathematicians call the Monty Haul problem.

If switching is offered every time, or just at random without bias, then the usual analysis holds and switching is the right move

The real problem with the Monty Hall problem as explained by Monty Hall is that the presented scenario in the problem was never part of the show, but rather was concocted as a statistics problem. He never made such an offer.

That makes it much harder to solve. The traditional version was that there was a car behind one door, and a goat behind each of the other two doors. Which makes it trivial to solve: I already have a car, so I just take the goat. I always wanted one of those.

The obligatory xkcd strip:

Alt text: A few minutes later, the goat behind door C drives away in the car.

Let’s then suppose that after Monty had eliminated ~1,000,000 of the choices, another contestant is brought out of a sound proof room, and told that the prize is behind one of the two doors. Are his odds different depending on which door he chooses? Do they change if he is told which door you have picked? Keep in mind the prize IS behind one of the two doors. It is NOT randomly determined after you pick. So the “probability” is merely a statement about YOUR current knowledge. If you and the other contestant have the SAME knowledge now, your odds are the same, irrespective of the fact that you USED to know less than you do at this moment.

Whoa, hold on now: there’s still the problem of choosing the best goat.

I guess I will never understand statistics, as I don’t understand why the new odds aren’t 1-in-2, regardless of which of the two one picks. I don’t understand how this is any different from starting out with two doors and being asked to pick one.

I would think that would be a 50/50 in a situation where there is no knowledge of the original guess.

I think yes. If the person is presented with the knowledge that you picked Door A, and some number of other non-winning doors was eliminated, and the winner is behind either Door A or B, the odds are better to pick B. That the heart of the MHP, in my understanding.

If it was a blind guess it would be 50/50, but the elimination of the other doors gives you new information that makes the odds of the door you picked significantly less than 50/50.

That’s how the 1 in a million scenario made it make sense for me. What are the odds that you picked the one correct door out of a million? Literally one in a million. But if you have the knowledge that either your 1 in a million guess is correct or this other door is, then you can see that probability of the other door is higher.