Game Show Riddle Solved
by Bruce Hurley
GAME SHOW RIDDLE = = = = = = = = = = = = = = = = = =
This confounding riddle is my more palatable variation of a conundrum posed by one of my heroes, mathematician Martin Gardner, in 1959. Gardner is the subject of numerous interviews and articles and is one of the great minds of our times.
You are a contestant on a game show. The game is very simple. There are three doors: door #1, door #2, and door #3. Behind one door is a million dollars. The other two doors contain worthless joke prizes.
All you have to do is pick which door you want to open, and you get whatever is behind it. But you only get to open one door. By simple math, then, you obviously have a 1 in 3 chance of picking the correct door and becoming an instant millionaire.
You pick a door. As soon as you tell the game show host what door you want to open, he stops and says, "Okay, you've made your choice. Now, I'm going to do what we always do here on this show. I'm going to open one of the other two doors for you that I know has a booby prize." And he does so.
Then he asks, "Okay, now, would you like to stay with your original guess, or would you like to switch to the other door that's still closed? You only get one shot, so do you want to stay with your original choice, or switch?"
Is there any compelling reason to switch doors if you want to win the million dollars?
Explain your answer mathematically (no formulas required).
To be clear, there is no trickery, and the host is not cheating. Furthermore, the money has not moved, will not be moved, and if you open the right door, you win the cash. Money is either behind the door you first picked, or behind the remaining unopened door.
SOLUTION = = = = = = = = = = = = = = = = = = = =
Logically speaking, this seems obvious to anyone with a sharp mind: you can switch if you want to, but it makes no difference. The host has just eliminated one of your choices. Now you're down to two.
You didn't know what was behind the doors before, and by opening one of them, you still don't know what's behind the other two. Your odds are 50/50 no matter which door you choose. So, switch or don't, it makes no difference.
Perfectly sensible, right? Absolutely. But it's dead wrong.
You should always switch.
As crazy as it sounds, you are twice as likely to win if you switch than if you stay with your original pick. People will tell you I'm crazy, but they are wrong.
(Yes, I'm sure.)
This is pure mathematics and not a trick.
If you don't switch, your chance of winning is one in three. We already know that. You just pick one door out of three. No matter what the host does afterwards, your chance is still one in three if you stick with your original choice.
If you switch, however (and this is the part that's hard to believe), your chances become two in three. You double your chance of winning.
Here's the simplest way I can explain it. In essence, what the host is offering you is this: he says, "You've chosen Door #2. Would you like to switch for Doors #1 and #3?"
Because he knows where the booby prize is, he's obviously going to show you that door (let's say the booby prize is behind Door #1). But he can't show you Door #3 because the prize might be there.
So what he's essentially saying is "You picked Door #2. You can swap for #1 and #3."
If he had not shown you what was behind Door #1 and simply said, "Would you like to switch from Door #2 to Doors #1 and #3?" would you always switch? Of course, you would.
If he had instead showed you nothing and asked if you would switch to another door, your answer would probably be no, because every door has a 1 in 3 chance.
But when he shows you one of the booby prizes and you decide to switch, you have the same 1 in 3 chance PLUS the 1 in 3 chance given to you as a gift when he revealed the booby prize. That gives you your 2 in 3 chance, but only if you switch.
Another Way of Looking at It
I can tell that some of you are probably reading this and saying, "No, that's not right. There really is no reason to switch. You can if you want to, or if you think there's cheating going on, but in a pure game, it's 50/50! You've gotten rid of one of the doors is all!"
If you try it empirically, you'll find that if you stay with your original guess, you'll lose two times out of three. If you switch, you'll win two times out of three.
If you don't have time to try it empirically, here's an easier way: an actual computer simulation which can run through hundreds of guesses to show you the probability. If you choose "always reveal goat" you will see the probability always works out to about 2/3 for switching and 1/3 for sticking. If you choose "open random door" the probability will always be about 50/50. Hit STOP and then GO to run the simulation again.
By showing you a booby prize after your first choice, the host has given you information. Your original choice had only a 1 in 3 chance of being right. There was a 2 out of 3 chance that the money was behind one of the other two doors, right?
So now it's simple: Since he just showed you which of the other two doors had the booby prize, that means the remaining door has the 2 out of 3 chance.
The other two doors are comparable to a coin toss. Suppose you eliminate the possibility of heads by modifying the coin. Tails then picks up the odds of the eliminated possibility, doesn't it? It changes from a 1 in 2 chance to a 2 in 2 chance.
It's the same with the two originally-unchosen doors. When the host eliminated one of the possibilities (by exposing the booby prize), the remaining unchosen door picks up the odds of the eliminated possibility. It changes from a 1 in 3 chance to a 2 in 3 chance.
Yet Another Way of Looking at It
Here's another way of thinking about it: let's play the same game, but it's got a hundred doors, not just three. Only one has money behind it. You pick the first door, and then the host opens up 98 of the doors with booby prizes.
Do you stay with your original, or do you switch? The odds of your first pick being right are 99-to-1 against, and that didn't change just because he opened up 98 empty doors. Right?
In the case of the hundred doors, switching to the other unopened door would actually be the better choice 99 times out of 100. It would have to be, wouldn't it? Because your original choice is just 1 out of 100, the opened doors showing the booby prizes are, of course, 0 out of 100, and somehow it's all got to add up to 100 out of 100.
After seeing 98 booby prizes, we still don't know where the money is, but does that magically make our original guess 50-50? Was our original guess actually 50-50 all along? Certainly not. It was 1 out of 100 then, and it is 1 out of 100 now that we have seen 98 booby prizes. That means the one remaining (the one you can switch to) must be 99 out of 100.
Still don't believe me? I don't blame you. Versions of this puzzle popped up all over the online universe, and even made its way into magazines, newspapers, and newsletters. Wherever it went, most people gave the wrong answer. The Mensa monthly journal, for example, published an article claiming that the odds were only 50/50. A month later they printed a retraction.
Mathematicians and others wrote to these publications swearing up and down that the odds were 50/50. It was quite something to watch, if you're enough of a geek to care about this sort of thing.
Three Shell Game
For those of you who want a different perspective, imagine you run into a game operator.
Operator: Step right up, folks. I'll put a pea under a shell out of your sight and you see if you can guess which shell the pea is under. Double your money if you win.
After playing the game a while, you conclude that you can't win more than one out of three times.
Operator: Don't leave, Mac. I'll give you a break. After you pick your shell, I'll turn over an empty one. Then you know the pea has to be under one of the other two, so your chances of winning go way up.
Fall for that and you'll go broke fast. Can you see how turning over an empty shell has no effect on your chances?
The problem is actually the same but looked at from a different perspective. Since you have already made your choice, no Operator's action can change your chances. So, to me at least, the Shell Game makes it pretty obvious that unless you switch in the Game Show Riddle (i.e. if you play the Shell Game), you chances remain 1 to 3. However, if you switch, you select one door out of two.
I also had several experiences where I related the puzzle to friends and they refused to believe you should switch. At several points, I tried placing wagers on it. Challenging people to put money on the line usually made them wake up and try it empirically before risking their cash. But even then, I found that some people--smart people, quite often--stubbornly refused to take the bet. They merely insisted they "knew" they were right, and that was that.
It was when I encountered such people that I began to recognize a very human impulse. Sometimes, people will refuse to change their minds once they've decided what to think. You can be friendly about it, joke about it, be irate or impatient about it, or try to present it any way you want, but some people will not be convinced and will even actively avoid looking at proof that they are wrong.
I also decided I didn't want to be one of those people.
(By the way, if you still don't believe I'm right, contact me, and we can set up the same wager. I'm so certain I'm right, I'm willing to gamble $1,000 to anyone who can prove me wrong. Make that $10,000. In fact, I'll mortgage the freakin' house and go as high as you want.)
Besides giving me insights into human behavior, this experience emphasized something else to me: the world is not always a matter of subjective opinion. You can't worm out of empirical truths. 2+2 never equals 5.
More simply put: the world does not always conform itself to what we want it to be, or think it should be, or even what we are certain it must be.
Some things are a matter of opinion or conjecture, but some things are stubborn, immutable facts. And cold, hard, ruthless facts contain the keys to finding whatever provable truths exist in the universe.
Nothing in the universe is more perfect than a single, absolute answer. It is wondrous to contemplate and rapturous to unravel.
Contact me if you have any questions.
Martin Gardner's version of this puzzle, published in October 1959, involved three condemned prisoners, one of whom will be pardoned at random. One prisoner cons the warden into naming one of the other prisoners (other than the prisoner who is asking this of the warden) who will not be pardoned. Do this prisoner's (the one talking to the warden) chances of being pardoned then go up to 50%?
This is identical to the game show trap, and this prisoner's chances are still 1/3, but the probability that the third prisoner will be pardoned have gone up to 2/3. Mr. Gardner got a flood of mail about this.
Marilyn Vos Savant's column on this subject was published in Parade magazine, on September 9, 1990. Subsequent readers' comment appeared on Dec. 2, 1990, Feb. 17, 1991, Jul. 7, 1991, Sep. 8, 1991, Oct. 13, 1991, Jan. 5, 1992, and Jan. 26, 1992. Also see The New York Times of July 21, 1991 (front page) and August 11, 1991 about the furor. Several articles in mathematical journals were also devoted to this.
Here are some links to some other sites about this riddle:
Yahoo's directory. - A list of sites maintained by Yahoo.
The WWW Tackles The Monty Hall Problem - another substantial list.
The Infamous Monty Hall Problem
Monty Hall (Let's Make a Deal) Problem
University of Chicago - somewhat dry and mathematical.
Monty Hall Dilemma