Comments on: Let’s Make a Deal

By: JarvisLoop

JarvisLoop — Sat, 05 Dec 2020 03:20:35 +0000

And I still haven’t.

By: JarvisLoop

JarvisLoop — Fri, 21 Aug 2020 02:26:34 +0000

And I now wonder what are the odds that I will ever fully grasp this article.

By: Franklin

Franklin — Mon, 20 Aug 2018 19:59:08 +0000

I watch Let’s make a Deal every day and the Big Deal door is rarely #3. I would like to see statistics over time. Something seems wrong. It is almost always 1 or 2.

By: E&J

E&J — Wed, 08 Aug 2018 02:56:30 +0000

Let’s look at this a different way. We know that there are 3 doors, one has a good prize package, the other two do not. When the contestant makes his initial choice between 3 doors, he obviously has a 1 in 3 chance of guessing the door with the valuable prize. Then Monte reveals one of the unchosen doors is a loser, and offers the contestant a chance to choose again. The contestant’s odds are now 1 in 2 of choosing the winning door.

HOWEVER, are we forgetting that AFTER Monte reveals one of the unchosen doors is a loser, the contestant, at this point in time, has a 1 in 2 chance of being correct with his ORIGINAL choice. (There are only two doors left, one good, and one bad.) So whether he stands pat, or whether he changes his choice, his odds DO NOT CHANGE at this point in time. He has a 50/50 chance either way.

This problem is a classic example for me of what happens when people try to explain something, yet do not fully understand the problem and solution. Supposed experts like mathematicians from MIT are quoted, but there is NO fundamental explanation of the purported “correct” answer. (Not a simple explanation that holds water, at any rate.) It becomes an “Emperor’s New Clothes” scenario, where no one wants to admit they cannot understand the mathematician’s explanation for fear of looking stupid. So instead they say: “Oh yeah, I get it now”. Even though they don’t, because it’s wrong.

I suspect the whole premise of this is that the contestant is originally making his choice when the odds are only 1 in 3. And if he has a chance to choose after one losing door is revealed, his odds of making the correct choice are now 1 in 2. Thus is appears as if his odds of choosing correctly have improved.

But this overlooks something. Revealing he has already NOT chosen one of the losing doors tells him that his odds are NOW 1 in 2 that his original choice was correct.

If Monte wanted to do someone a favor, he should help out a contestant who unfortunately chose a losing door. At that point his odds are zero chance of winning. Letting him choose again would then give him a 1 in 2 chance of winning.

(Oh, and you folks who said they ran 20 simulations and the results seem to be true: Try running 20,000 simulations and see how your model holds up.)

Maybe this is more of a time concept problem than an odds problem :)

By: Rabid Accountant

Rabid Accountant — Thu, 02 Jun 2016 12:37:05 +0000

Seems like it’s already been covered 1000 times and yet I can’t help throwing out my grasp of the situation.

During the initial pick, there is a 100% chance that at least one of the doors you didn’t pick was wrong. To reveal that one of those doors was wrong doesn’t tell us anything we didn’t already know and as such has no effect whatsoever on the odds that one of the doors you didn’t pick was right.

By: Jay M.

Jay M. — Wed, 20 Aug 2014 16:08:17 +0000

This is, in my view, the simplest way to look at it.

Two-person card game between Bob and Sue. Three-card deck with one marked card, shuffled into random order.

STEP ONE: Sue picks one card, which she must place face-down on the table without looking at it.

STEP TWO: Bob gets the other two cards. He looks at his cards and is required to discard one blank card, face up.

STEP THREE: Prior to the cards being turned over, Sue is given the choice of keeping her card or swapping with Bob’s remaining card. Should she swap?

ANALYSIS: Sue should swap, because their original chances of having the winning card (1 in 3 for Sue, 2 in 3 for Bob) have not changed by virtue of Bob’s revealing his blank card. Whether he does or doesn’t have the winning card, he will always have a blank card to show, so this is just theater to confuse the issue.

[In the actual TV show, Monty Hall did not have to give the contestant the choice of switching, which changes the odds. I am not accounting for that here.]

By: Perry Curling-Hope

Perry Curling-Hope — Sun, 06 May 2012 10:57:01 +0000

People make this ‘problem’ hugely complicated, running simulations and introducing all manner of irrelevancies to try to ‘prove’ that the odds prevail.
They do, and always will, by simple logic, unless some metaphysical precognition is at work to confound them, but that is not the issue at hand here.

If the contestant applies the strategy of not switching, the odds are a simple 1 in 3 of picking the car, whatever the host does is irrelevant as it will have no bearing on the result.

If the contestant applies the strategy of switching, the odds are still a 1 in 3 of initially picking the car, and the contestant will lose, as the host will reveal the one goat, and the contestant will migrate to the other goat.
The odds of the contestant initially picking a goat are of course 2 in 3, and the contestant will win, as the host will reveal the other goat, and the contestant will migrate to the car.

It is only the initial selection by the contestant which has anything to do with randomness and probability, after that, it plays no part, and the outcome is determined.
The host has to know where the car is, and is presumably constrained to revealing a goat after the contestant’s choice for the ‘Monty Hall Problem’ to prevail

By: Evil1

Evil1 — Wed, 25 Jan 2012 18:01:22 +0000

If there were 1 Billion doors to choose from, and I picked one randomly that I thought may contain the prize, I would have a 1 in 1 Billion chance of picking the right door first time (very little chance of success). If the host then opened 999,999,998 doors that HE knew did not contain the prize (leaving my original picked door and only one other door), that would mean that either my door was right (on first choice) – or the only remaining door he has left unopened is right and the odds would suggest I should swap to his door. Much better odds than the three doors but the principle is the same, unless someone sees a flaw in my reasoning?

By: cornflower

cornflower — Fri, 29 Apr 2011 01:02:35 +0000

*rather than the first

By: cornflower

cornflower — Fri, 29 Apr 2011 01:01:47 +0000

I’d like to point out what I think about the coin tossing question. The question “What are the odds that…” is the key here, and is what may make the problem misleading.
It makes a difference when you say: “What are the odds that the coin will come up heads on the next (100)th flip?” and when you say: “What are the odds that you will get a series of 1oo coin flips that come up all as heads?” The illusion is in that we tend to think of the second formation of the question rather first, as our brain tends to recognize stark patterns more quickly, as the article says.