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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a better chance does not equal win.
@JustAGuy2: well you could imagine the only game i play is craps. i’m actually really good at it.
@Mollyg: I’m with you.
Ah okay. So it seems like the problem reduces to a 50/50 chance, which is what it would be if you just entered the game after a door opens. But there’s still information in the system from your previous choice which affects the outcome, so you’re still betting on the 2/3 game.
Monty Hall, you are clever. Me, I am not clever.
@snoop-blog: This is a common psychological element that come into play in these kinds of scenarios. People tend to act risky when the outcome has a positive benefit, but the other outcome isn’t too negative. However, when people are presented with a very negative outcome and a neutral alternative outcome, they will tend to be risk adverse and choose a strategy that is more probabilistic.
For example instead of a car and a goat. I present you with a goat, vs being tortured horribly for the rest of your life. In this scenario, you would choose the higher probability option, since the negative outcome is very undesirable.
@snoop-blog: Craps is in fact a near-even game with regard to the odds. If you play correctly, your odds are about 49% vs 51% in favor of the house.
Blackjack is the only other game where it is possible to gain an advantage over the house by counting cards. Of course casinos don’t like to lose money and kick people like this out for ‘trespassing’.
“a better chance does not equal win.”
Exactly, what it does mean is that you have a better chance of actually winning. To put it another way, if you go with your gut, you lose more often. That’s why this posting is entitled “Why You Fall For Dumb Things.” Going with your gut versus going with proven statistics in a game of chance is necessarily a “dumb thing.”
@satoru: But what if you’re into goats *and* torture? Is there a way to win both?
@Imafish: This is incorrect. Each individual instance of the game you have a 66% chance of winning if you switch. If you play more than once, the previous instances do not affect the subsequent probabilities.
This is different from say black jack where prior events have an impact on future probability.
@MrFalcon: You’ll note that in my example I did not specifically state WHICH outcome was negative. Merely that you would be more conservative in your choice because one outcome was negative
Thus if you are into goats and/or S&M you can take the probabilities any way you want
@satoru: my only point is that %33 vs %66 doesn’t mean the %33 is wrong. if you were on a game show faced with this game, how would could you be for sure it isn’t the first door, seeing how we don’t have a history to base any kind of odds on.
let me word this better, if you were betting on roulette, and you had no knowledge of what the previous spins were, it puts you at a disadvantage then the other tables that told you what the last 10 spins were.
if you witness 6-7 people before you change doors, and win, do you still go with your %66 or think the odds are considerably in favor of the %33 now. does that make any sense?
@satoru: thank you!!!!!!!!!!!!!!
@Applekid: Yeah, I just didn’t want to be the one to say it.
As a side note I have always wondered on “Deal or No Deal” if they ask people if they have heard about the Monty Hall Problem? Would these people be disqualified from being a contestant? Obviously they don’t want you to win money, just put on a good show for ratings. Thus it would theoretically be in their interest to remove rational individuals from the pool selection.
Obviously if you had say $1 million and $1 left in the game, you don’t want your contestant to go “OH I’m totally switching because I have a 29/30 chance of winning”.
Incidentally the game show continually perpetuates the incorrect probabilities on the show by stating in bold that “There is a 20% chance Contestant A has a $1 million case”. I think this is irresponsible and just further confuses people on how probability really works.
The trick to make it easier is not to work out the win states, but to work out the lose states. For some reason that makes people understand the odds better (this applies to other things as well like craps, blackjack, etc.). Switching loses 1/3 of the time. Since there is no “tie” state known, you can assume you will win the rest of the time. Seems better to switch in that light even if you don’t understand how to understand the win states.
@snoop-blog: The roulette example is in fact what is referred to as the “Gamblers’ Fallacy”. Every event in roulette is an individual probability. Thus having prior spin knowledge does in no way influence future outcomes. Though the table loudly displays the previous spins, they in fact are totally irrelevant information. It is similar to how people think that say I am flipping a coin and I come up with a run HHHHHHHHHH (10 heads). With your logic, you would think that tails is more likely in the next throw. In reality the previous 10 heads, while a very unlikely event and peculiar for sure, does not affect the next throw. Thus I can say that there is a 50% chance of getting heads on the next throw and 50% of getting tails.
As I mentioned before, the only casino game where previous rounds affect future probability is black jack. There might be a few others, but I can’t think of any at the moment.
The Monty Hall problem surfaced on The Straight Dope years ago, and they basically hashed out all of the same arguments I see here. At the time, Marilyn Vos Savant(?) who is(was?) one of the smartest people in the world, answered this question saying you should switch, and people flipped out over it, saying she was wrong.
As I remember reading in the book of collected Straight Dope columns, there were about 2-3 columns worth that were dedicated to discussing this problem.
If I were to play that game, I would choose 2 doors, then say “open doors #1 and #3. You will now open both of the other doors, and if there is a car behind either of them I win.”
Monty Hall would say “No, you have to choose a single door” so I’d pick door #2. Then he’d open one of the other 2 doors, whichever doesn’t have a car. Then I would say “ha ha, I have tricked you Monty Hall, for I had no intention to choose door #2. I chose doors #1 and #3. You have opened one of them, knowing that there was no car. Now I shall tell you to ‘switch the door I picked’ but really I will have you open the second of the doors I chose. You see Mr. Hall, I am the one with the power, and if we’re going to play this game, we will play it by my rules, and I shall choose 2 doors for you to open.”
“This is incorrect. Each individual instance of the game you have a 66% chance of winning if you switch.”
I agree completely with that and that’s exactly what I wrote: “each and every time you have a better chance to win by switching than to staying put”
What that means is that each time you have a 66% chance to win if you switch and that chance occurs every time. So the increased probability of winning by switching occurs one time and ten times. The amount of times you play simply does not matter.
“my only point is that %33 vs %66 doesn’t mean the %33 is wrong.”
You’re right, the 33% is not wrong, it is just less likely to be right.
Did Monty actually make you keep the donkey?
I would so own a donkey.
@satoru: yeah the roulette was a bad example but what about what i wrote right underneath that. are you saying seeing 7 people in a row switch doors and win is not going to sway your choice or would? in this game, eventually the %33 choice will have a better chance (obviously at leas 1/3 of the time) of winning.
@mduser: Indeed when the problem was first proposed there was significant amount of discussion over it. At the time the problem itself wasn’t clearly defined and there was some ambiguity as to when Monty would offer the switch oand such, so people’s logic and conclusions were all over the map.
And despite being ‘solved’, whenever it appears it does generate a lot of talk and discussion. I think the extremely counter intuitive solution to the problem causes the confusion. But don’t feel bad! Many very very very smart people, who were math professors and such, got the answer wrong too.
@satoru: Maybe I’m wrong about how that show works, but isn’t it essentially random which cases get opened? I mean, isn’t it a straight 1 in 20 or whatever chance that you’ve picked the million dollar case?
Whether you take the cash or not is based on (I thought) a pretty straight mathematical calculation of what your expected average win is (perhaps modified a bit for TV excitement), so the decision isn’t so much “mathematically what should I do” as “how much risk can I tolerate.”
Or is there something I’m missing about the game?
@Michael Belisle: That is by far the best, most concise explanation I have ever heard for why you should always switch doors. Thank you Belisle – excellent work. You should teach for a living (if you don’t already).
now i’m not talking about the best odds but trying to predict multiple outcomes. obviously if you were playing for money, winning %66 percent of the time is better than damn good, but could you possible be right more than %66 of the time, by occasionally staying with the first door.
@snoop-blog:but could you possible be right more than %66 of the time, by occasionally staying with the first door.
Possibly yes, but probably no.
@Michael Belisle: so was you apology just snark or was that sincere? i’ve never been talked to so nicely on here so forgive me for my confusion.
@satoru:
Actually, Deal or No Deal does not exhibit this property, because the person choosing cases does not know the contents of the cases.
The Lets Make a Deal problem requires the person who’s opening doors to know where the real prize really is. Monty knows where that prize is, and he’s not going to show it to you. So, if it gets down to where you have the door you chose, and the door that Monty left behind, it is probable that the door left over is the winning door.
Deal or No Deal has the contestant opening cases, and because he doesn’t know where the money is, he may open the big prize in an early stage of the game. Since the contestant doesn’t know where the million dollars is, if it really does come down to two cases left, it really is 50/50 on which case contains a million.
@snoop-blog: when monty opens an incorrect door, switching your decision becomes an inversion of states. there is one right door and one wrong door, so if you have a right door, switching will yield you a wrong door and vice versa.
so simplifying the switch operation to an inversion, consider your initial odds of being wrong and of being right, 2/3 and 1/3, respectively. now switch states.
@snoop-blog: Seeing 7 people win by switching doors would not sway me. I am of course assuming the game is fair and that they aren’t switching things behind the doors I cannot see. If the game is crooked then no matter what I am in a losing position. In this type of scenario it is more important to keep calm and simply play the probability. The recent movie “21″ is a bit flamboyant, but the base premise is that you stay calm, play the odds and you will win in the long run. If you are counting cards in black jack, you might not win a particular hand, but over many hands you are going to come out ahead.
Obviously in these kinds of games, there is a huge psychological element to them. In fact casinos and such take advantage of this to suck more money out of you!
One thing is that in a casino, reward is loudly proclaimed and advertised. Where as negative outcomes are silent. Thus you only hear and see the positive outcomes, and thus incorrectly assume that lots of people are winning. Another thing, the slot machines near the main walkways are actually set to give out slightly more money than machines that are not. Thus as you walk through a casino, you see people winning money. This positively reinforces the notion that since lots of people are winning, then I should win too! Then you sit down at a slot and lose all your cash
so if staying with the first door occasionally could increase your odds beyond %66 (and lets say that’s important to win more than %66 of the time) when would you stay? after how many people switched and were correct? 1 out of ____? that’s where i’m going with this.
@satoru:
Actually the Deal or No Deal (let’s call it Howie’s Dilemma) is a different case. The reason for this is that instead of Monty selecting the door, you are the one selecting the cases, and thus the probabilities are different.
Example, let’s play Deal or No Deal like Let’s Make a Deal. In that case, you select a door to remain closed, and then select another door to open.
That second door you select to reveal has a 1/3 chance of containing the car. In the case where Monty selects the door, it has a 0% chance of choosing the car.
So, if you’ve eliminated all cases except for 1$ and 1M$, you do have a 50/50 chance of each value being in your case.
WTF…NOW I get it without having to repeat it 30 times and looking at columns of wins and losses.
I’d never seen the show, and the asshole (he must have been because he didn’t get straight to the key factor in this) insisting I was wrong some years back when I was wrong…never mentioned the “ALWAYS OPENS A DOOR TO REVEAL A GOAT” part.
That makes it pretty obvious.
@thefncrow: I was thinking in a more hypothetical situation in the final end game. Let us assume I have chosen case 1, and magically burned through all the other 28 cases. Now on the board I have $1 or $1 million left as possible outcomes. While yes I chose the intervening cases, this is irrelevant. I am left now with a ‘goat’ and the ‘car’. The reality is that the probability I picked the $1 million case initially is still only 1/30. Therefore it is better for me to switch cases for the 29/30 odds.
I actually don’t know if this is how the game works in principle. Do they give you a switch option at the end? I’ve never seen it get to that point with any of the few shows I have watched.
@snoop-blog: ???
out of people who switched, 2/3 of them should be right. this says nothing of whether or not future people will switch
@snoop-blog: If you only have one shot at it, then your best chance is to take the action that has the higher probability of success. That is the whole point of probability. Of course you will still succeed or fail on luck.
@tizzed: I really don’t think there is a difference in the scenarios. Only in that obviously I might accidentally chose the $1 million case to be eliminated, at which point my odds become 0%.
But let’s assume you, in a fairly unlikely event, make it to the end with $1 and $1 million, we are still in the same scenario as the Monty Hall problem. Just as revealing the goats does not change the initial probability in the Monty Hall problem, choosing specific doors yourself doesn’t change the initial probability either. I will note that this applies only in this VERY SPECIFIC END GAME SCENARIO ONLY.
look, think of it this way. instead of trying to pick the car, try to pick the goats. you’re twice as likely to have picked a goat during your initial pick.
that’s where the 2/3 probability is. the host always reveals a goat. he won’t reveal the car. think about it. that means both goats are “revealed”. switching gets you the car.
truthfully, i didn’t really “get it” either. what a stupid stats problem. don’t gamble. go pay a bill. seriously, log into your online banking and pay a bill.
I played 20 times, 10 of each. I won 50% of the tme I switched, but only 30% of the time I stayed.
(And I wanted to try the new image commenting thingee.)
When Casinos put in the boards showing the previous results, they saw their roulette income rise dramatically, because people saw a string of black and thought either “bet on black, its hot” or “bet on red, its overdue”. While the odds of 20 Black in a row is 1 : 1,048,576, if you see 19 Blacks in a row, a 20th black is 50/50. (does not count 0 and 00 for easier math, remember 1 out of 19 times the house wins all red/black bets)
All Casinos know that past results have no affect on future outcomes. In the monty hall problem, just because losing by switching is 1:3, it never becomes ‘due’. If you did the game 99 times, you would expect 66 wins and 33 losses, but if you won 98 by switching, you should still switch the 99th time, you’ll win 66% of the time.
@satoru: The Monty Hall Problem does not extend to Deal or No Deal because the cases are opened at random and the host does not know what’s in them. Deal or No Deal is much more of an “Expected Value” game. Expected Value = Total amount of dollars in all of the cases divided by the number of cases left. If the deal they offer you is higher than the expected value, you should take the deal. In the earlier rounds, the deal that the banker offers you is usually only 50-80% of the expected value. This discourages people from taking the deal and extending the game to make it more exitciting. The deal approaches expected value as the game progresses (an exception is when a big amount gets knocked out and the banker usually offers way less than expected value.)
@monkeyboy13: Interesting I did not know that the casinos saw a revenue increase when they put those roulette boards up. I suppose this is logical, since everything in a casino is purposefully placed to maximize the amount of money the suck out from people. Playing into player’s instinctive urge to follow the Gambler’s Fallacy is a good way to increase revenues.
@snoop-blog: Actually knowledge of previous spins in roulette does not help at all since each spin is statistically independent ([en.wikipedia.org]). In fact, psychologically speaking, knowledge of previous spins would probably be a disadvantage since it may bias you to (incorrectly) favor some bet when there is no mathematically justifiable reason to do so (basically what this article is trying to say).
@jeff303: if you read all of my comments (which there are a lot) then you will see i admitted the roulette was a bad example.
Just watch the movie “21″, Spacey explains it.
I already knew about it from High School Physics anyway, so it’s fun to see everyone cry about how it doesn’t work when they don’t have a clue!
@snoop-blog: Yes, there is still a possibility that if you stay, you will win. Which means, out of 100 chooses, it is possible to win %100 of them all if so many of your choices are to stay.
But, mathematically, there’s no way to know which of the times you should stay and which times you should switch. You would need some extra information about the situation that isn’t given. In this problem, everything is assumed to be random, so there is no extra information. On a game show, you could try to gleam some psychological information like you’re proposing, but you’d be better of if you just assume they randomized things as well, because it would be to their benefit to randomize it once they’ve fixed the rules, even if the same outcome arises ten times in a row.
Therefore, all you have to go on is the chance the prize is behind your first choice or behind the other door. That’s a simple %33 and %66, respectively, so your only choice should ever be to switch, because there’s a higher chance of winning that way.
Um. Firts I played the game about six times. I switched for half and stayed for half, and I got a goat every single time.
Then I read the explanation. Then I played seven more times, switched every time, and got a car every time.
Not a huge number of trials, but is their randomization a little off?
@chrisjames:
said: But, mathematically, there’s no way to know which of the times you should stay and which times you should switch. You would need some extra information about the situation that isn’t given. In this problem, everything is assumed to be random, so there is no extra information. On a game show, you could try to gleam some psychological information like you’re proposing, but you’d be better of if you just assume they randomized things as well, because it would be to their benefit to randomize it once they’ve fixed the rules, even if the same outcome arises ten times in a row.
you win. you just proved my point. there’s no way of telling when to stay and when to not. as it is not probable for your stay to be correct say 100 times in a row, it is still possible. so really, it’s a crapshoot.
@tweemo: It’s possible that it’s rigged (like this game: [www.guessyournumber.com] ) but why would they do that?
You should always switch; No matter what. You win 66% of the time that way. It’s not worth it to try to be right the other 1/3 of the time. Case closed.