The New York Times has an interesting series of tests and explanations that show why and how the human brain makes errors in estimating probability—and consequently, why we get suckered even if we think we’re overall pretty smart. To start things off, play the Times’ online version of the “Let’s Make a Deal” game, where you pick one of three doors, then you can read up on how it works. The game brought in a bunch of reader responses (and arguments), so the author, John Tierney, offered a few more thought experiments you can try if you need something to keep you distracted from your job. In today’s column, Tierney talks about why so many people naturally make errors with probability and gets a plain-English explanation from a couple of marketing and psychology experts.
I’m proud to say I instinctively reasoned the Monty Hall game correctly, which is a surprise since I have a naturally ability to screw up pretty much any probability question thrown my way.
Part I: “The Monty Hall Problem” [New York Times]
“Interactive Monty Hall Game” [New York Times]
Part II: “Monty Hall’s Other Problems” [New York Times]
Part III: “The Psychology of Getting Suckered” [New York Times]
(Image: New York Times)
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@snoop-blog: so was you apology just snark or was that sincere?
A little of both. I didn’t fully understand your post when I said “A winner is you!”, so shame on me. But I’m stlll pretty sure it’s wrong.
All in all, it’s zero sum here.
@tweemo: Not a huge number of trials, but is their randomization a little off?
I had the same problem. I gave up after thirty tries or so where my odds were around 85% when I switched and 15% when I didn’t.
Somebody’ll have to click a few thousand times to verify. Any volunteers?
“attacking commenters on here is a troll move. you notice how@MrFalcon: manage to show a little class, something you’ll never have.”
“attacking commenters on here is a troll move. you notice how@MrFalcon: manage to show a little class, something you’ll never have.”
Would it be correct to call that an ironic statement, or just hypocritical?
Would it be correct to call that an ironic statement, or just hypocritical?
Oh jesus christ, apparently the preview checkbox doesn’t work for me. Sorry about the comment spam.
This is one of those things that will get people to argue forever. Unlike most things with that property, here, one side is absolutely, provably, wrong.
Regardless of your 1/3 chance of picking the correct door at the beginning, by showing you one door, you know it has the booby prize (goat or canned squid)
But now you are being asked “Do you want to keep your original choice, or go with the other door”
That is the same as showing two doors, knowing that one door had the prize, and saying “Which of these two doors do you want”
You can keep your original door, or pick the other door. two choices.
Both, at THIS point in the game, have a 50/50 chance of containing the prize.
Hey, I’m switching teams here.
I understand it now, and I must agree… You should switch. You have a 2/3 chance of picking the wrong door. Sure, now that there are two doors, there is a 50/50 chance that the car is behind either of them, but your original choice is most likely wrong, and all he did was eliminate the other “wrong” choice, so by switching, you pick the “right” choice — That is, 2 out of 3 times.
I think I just blew a gasket.
@snoop-blog:
Man I think that you should give up the ghost on this one and say you were wrong… You’re digging yourself a bigger and bigger hole by continuing to argue that it’s somehow NOT wiser to switch. it IS wiser to switch. That’s the point that’s being made, regardless of what you try to say about it still being “chance.” Yes it’s chance, it always was, but if you had the choice of picking 1 winner out of 100 or 1 winner out of 3, which would you pick? or would you say that it doesn’t matter they’re both chance? OBVIOUSLY both staying or switching are both chance occurrences, BUT switching yields statistically higher results, independent of past or future runs.
Give it up man. Eat your crow now while it’s still bite-size.
@snoop-blog: As a guy looking to hire a math major, you should give me your name so I know not to hire you.
@hypnotik_jello: Certainly failed manners class.
@snoop-blog: Yup. It’s a crapshoot, alright. But it’s a crapshoot with much better odds if you switch doors when offered the chance. Given that you’re a math major, I’m not quite sure I understand why you’re arguing against the probabilities unless it’s just to play devil’s advocate…?
@BloggyMcBlogBlog: It still does apply to Deal or No Deal, but not exactly in the same way. Because you don’t know what value is in each case, you don’t have a 100% chance of not knocking out the grand prize. But this doesn’t matter because the grand prize isn’t the only prize of value. If DoND had only one $1 million case and the rest were all for a penny, you’d be right. However, even if you knock out the $1 million case, you still have the $750,000, $500,000, etc.
Now, with that out of the way, when you play the game, the probability of you knocking out a specific value on the board is always 1/N, where “N” is the number of values on the board. So, with just your case and one other unopened case, the probability of the unopened case being the highest value left is 50% (1/2). However, the probability of your case being the highest value left is still the same as when you initially selected it: 1/26. Unless it sucks, you should take the deal.
This may sound stupid, but I’d rather have the goat than the car. Goats get better mileage than my car or my tractor.
@satoru:
Okay, let’s break this down.
What are the chances that you will get down to 2 cases : 1$ and 1M$? This happens in 2 cases
1) You select the 1M$ case, and then select the other cases, leaving the 1$ case unopened. The chances of that happening are 1/30 * 28/29 * 27/28 * 26/27….* 1/2
That’s 1/30 (chance of picking the 1M$ case) times 28/29 (choosing a case other than 1$…so on and so on)
Note the numerators and denominators cancel each other our, and you’re left with 1/(30*29)
Scenario 2 : You choose the 1$ case, and avoid the 1M$ case.
The chances of this are 1/30*28/29*27/28…*1/2 = 1/(30*29)
Exactly the same, so given that there are 2 cases left, and 1 M$ in one case and 1$ in another case, there is a 50/50 chance of it being in your case.
In other words, one is unlikely to pick the 1M$ case, but once it is chosen, it is very unlikely (i.e. nil) that you will reveal it upon opening cases. However, It is likely that you will not choose the 1M$ case, but it is unlikely that you will not reveal it when opening up other cases. These likely/unlikely scenarios have the exact same chance of occurring.
See how this is different from Monty’s dilemma because there is a chance you will never get to the endgame choice because the big prize is revealed, whereas in Monty’s dilemma, it is never revealed.
I’ve never understood why people have such a hard time understanding this puzzle.
Look – instead of three doors, imagine it’s a deck of cards. The ace of spades is the only winner. You pick one at random. Before you look at it, I reveal all but one of the remaining cards to be losers. Do you really think your odds are 50/50? They’re not.
Look at it this way. When you pick a card, it is 1/52 that you got the ace. It is 51/52 that the winner is in the other pile. No matter what I do to that second pile – wtiting on them with sharpies, tearing them in half, or turning most of them them over to show their faces, the odds remain 51/52 that the winner is in that pile.
So now you have two piles – your pile, with a 1/52 winning chance, and the other pile with a 51/52 chance of having the winner and only one single card still unrevealed. The choice is obvious.
When you pick a door, you have a 1 in 3 chance of it being the car. nothing that happens afterward can change that fact. So no matter what happens afterward, you sill have a 1/3 chance of having picked the winner. Since there is only one other door at the moment of choice. It must have a 2/3 chance of winning.
In the traditional Monty Hall puzzle – 3 doors, 1 car, he always reveals 1 loser after you pick – switching is ALWAYS the better strategy.
GAH!
Sorry about the wall of text there. Here it is again properly formatted:
I’ve never understood why people have such a hard time understanding this puzzle.
Instead of three doors, imagine it’s a deck of cards. The ace of spades is the only winner. You pick one at random. Before you look at it, I reveal all but one of the remaining cards to be losers. Do you really think your odds are 50/50? They’re not.
Look at it this way. When you pick a card, it is 1/52 that you got the ace. It is 51/52 that the winner is in the other pile.
No matter what I do to that second pile – writing on them with sharpies, tearing them in half, or turning most of them them over to show their faces, the odds remain 51/52 that the winner is in that pile. So now you have two piles – your pile, with a 1/52 winning chance, and the other pile with a 51/52 chance of having the winner and only one single card still unrevealed. The choice is obvious.
When you pick a door, you have a 1 in 3 chance of it being the car. nothing that happens afterward can change that fact. So no matter what happens afterward, you sill have a 1/3 chance of having picked the winner. Since there is only one other door at the moment of choice. It must have a 2/3 chance of winning.
In the traditional Monty Hall puzzle – 3 doors, 1 car, he always reveals 1 loser after you pick – switching is ALWAYS the better strategy
OK, clearly ewither the commenting is broken, or my brain is, so I’m done for this morning. Sorry about the double post, I really did format it correctly the second time, I thought.
@tweemo: “Not a huge number of trials, but is their randomization a little off?”
No, the randomization is fine, but your sample size is off. 13 trials is insufficient. Run it for a few hundred times with each strategy and you’ll get results more in line with the expected returns.
I made the mistake of assuming that it was random. In most computer-controlled games, you’d end up with a 50/50 chance on the last one.
So really this is just a simulation of a rigged contest. Boring.
@Maged: Your numbers are off… At the end of the game, there can’t be a 50% chance the other case holds the million but only a 1-in-26 (~4%) chance that yours does. You haven’t accounted for the other 46%.
Deal or No Deal doesn’t follow the Monty Haul paradigm at all because the person opening cases doesn’t have the knowledge required to pick the “goats.” After every choice the contestant makes, all the probabilities in the system re-set.
If Howie somehow knew where the million was, and had to choose cases that DIDN’T hold the million dollars, it would be just like the Let’s Make a Deal problem. It’s not, though.
At the beginning of the game, you have a 1-in-26 chance of holding the million dollars. As cases are eliminated (as long as the million isn’t revealed), your odds increase until, at the end, if there are 2 cases left and $1m still in play, you have a 50% chance.
satoru: You’re right, theoretically. A possibility remains that there’s an underlying reason why you flipped 10 heads in a row that makes the outcome actually not what it would be theoretically, like having a weighted trick coin or something.
Basically all this thing is saying is, “Do you want to pick two of the doors or just one?” Because if you switch you’ve essentially picked two of the doors. If you stay, you’ve picked one. For those that can’t understand this, I feel sorry for you.