Many of us grew up watching Monty Hall on Let’s Make a Deal. It turns out that the show presents something of a mathematical conundrum: does switching your choice at the last minute between the two remaining rooms increase your chances of success? The video below presents the answer and it may surprise you.

The so-called Monty Hall paradox is a veridical paradox that is similar to the Three Prisoners problem and a less known Bertrand’s box paradox.

The paradox was first presents by Steve Selvin in a publication in the American Statistician in 1975. He argued that a player who stays with the initial choice wins in only one out of three times while a player who switches wins in two out of three times. Here is another explanation from a math site:

Imagine that Monty opens a door and shows that there’s only a goat behind it. Consider that the car is more likely to be behind a door other than the one you choose. Monty has just shown that one of those two doors – which together have the greater probability of concealing the car – actually conceals a goat. This means that you should definitely switch doors, because the remaining door now has a 2/3 chance of concealing the car. Why? Well, your first choice still has a 1/3 probability of being the correct door, so the additional 2/3 probability must be somewhere else. Since you know that one of the two doors that previously shared the 2/3 probability does not hide the car, you should switch to the other door, which still has a 2/3 chance of concealing the car.

The idea is so counterintuitive that hundreds of experts wrote into the magazine contesting the theory. However, this video from 2009 shows the theory in execution.

I will be trying this tonight with the kids as part of the Turley Blog Research Center.

24 thoughts on “How To Win A Car With Monty Hall Math”

This column and commentary make my head hurt.

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The crucial point in this is that when Monty opens one of the remaining doors, it is not at random. Monty is “injecting information into the system.”

The system starts out totally random. The winning door is picked at random. The contestant’s initial pick is fully random – she has zero information about what’s happening inside the system. After the first pick, Monty, who knows which door has the winning item, uses that information to alter the system – he picks a non-winning door to open. His action is not random. He inserts that information into the system which had previously been random.

We’re so used to thinking about totally random systems, like rolls of dice or dealing cards from a fully shuffled deck, that it’s hard to think about these sorts of problems where there are non-random actions taken that affect the odds of winning.

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Bon, I think martin was just saying that if you picked a losing door at first, then you will win when you switch. And since there is a 2 in 3 chance you will pick a losing door at first, there is a 2 in 3 chance that you will win if you switch.

Martin’s way of phrasing it is the one I use most often when trying to explain it to people. The key to a deeper understanding is in markj111’s post – the host ALWAYS uncovers a known losing door; it’s not a random choice. That’s why the 1/3 from that door “transfers” to the door you should switch to.

Cecil Adams (Straightdope) originally got this wrong, but in a mea culpa follow-up post he gave this other example:

“Suppose our task is to pick the ace of spades from a deck of cards. We select one card. The chance we got the right one is 1 in 52. Now the dealer takes the remaining 51 cards, looks at them, and turns over 50, none of which is the ace of spades. One card remains. Should you pick it? Of course. Why? Because (1) the chances were 51 in 52 that the ace was in the dealer’s stack, and (2) the dealer then systematically eliminated all (or most) of the wrong choices. The chances are overwhelming–51 out of 52, in fact–that the single remaining card is the ace of spades.”

I’m pretty sure increasing the number of doors to a million (or even 4) doesn’t work the same way. I’m going to work the math on it and I’ll get back to y’all. But it doesn’t feel right.

(of course neither does the original problem huh?)

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At first I was going to offer to take you all to Vegas but as the posts and opinions mounted I realized we’d never make it out of the casino’s lobby.

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DigitalDave,

Everyone knows yawning is infectious and about the phenomena of sympathy pains, but I never guessed there was such a thing as sympathy itching induced by a hypothetical until I read your example.

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Frankly I actually just increased the numbers to show that as long as the chances are less you picked the prize from the start they are still less at the end.

Digital showed it even more with his trillion doors.

This scenario actually played out in Deal or no deal where when 2 boxes were left and the million dollars was still in play and the contestant did not switch.

Guess who did not win the million dollar case!

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Frankly — “why does the entire 1/3 chance switch to the other unchosen door?”

This is the important question, and very revealing. Let’s consider a different scenario but just as important. Let’s imagine you’re sleeping in a climate where malaria is deadly. Statistically, let’s say that one in every thousand mosquitos carries the virus.

You sleep under a net, but there is a tiny hole. Late that night, a swarm of 1000 mosquitos attacks your house, and one of them gets in and bites you. You have just suffered a one-in-a-thousand chance of getting malaria.

Would you rather have had the other 999 mosquitos bite you?

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The scenario is easier to understand when the probability drops to zero. Try a trillion doors. You pick one, then Monty says “gosh, the car is actually either behind the door you picked or door number 65,234,102,332 — do you want to switch your choice?” Hmm, let me think…

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The key factor is that the box Monty shows you is not chosed at random. He ALWAYS picks a losing box. The initial chance that you picked the winner is one out of three. The odds do not change when are shown a box that is known to be a loser.

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The procedure works as long as you pick a goat the first time. This is 2/3 of the time. ğŸ™‚

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The probabilities don’t change, Martin. You still have a 2/3 probability regardless of what you pick the first time.

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If I may offer a video showing the math, Frankly and MacK, this might help.

MacK – but in your scenario if they opened one box at a time until there were only two left there would only be a 1/24 chance your original choice held the prize but a 23/24 chance the unchosen one held the prize if the relationship held. So basically no chance at all that you picked the right box to start with. This seems even less likely.

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Another way to look at it is from the original odds.

Let’s say we have 25 boxes, but only 1 has money in it, and the rest have nothing.

You choose 1, and then you’re, allowed to swap for the remaining 24 boxes if you want. Do you? Of course you do, because you know the chances are much greater the money is in the 24 boxes (24 to 1).

Switch it up.

You choose 1, and then we simply remove non money containing boxes until 1 box remains. Do you swap now? Remember the original choice was 24 to 1, and simply removing known non money containing boxes did not change that.

All people would choose to switch when there are 24 cases remaining. Not because of the pure math so much, but obviously the original odds were the 24 cases had the money.

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What if the contestant wants to win a goat?

I’m assuming that they never really gave the goat to the contestant.

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What makes this counterintuitive to me is why does the entire 1/3 chance switch to the other unchosen door? Why doesn’t half of it switch to the chosen door & half to the other unchosen one? there is a 2/3 chance the car is behind any combination of 2 doors not just the 2 unchosen ones.

If the 1/3 chance really did only switch to the other unchosen door than I would expect picking it every time would mean double the wins. The only demonstration of this that I saw was a limited (100) sample size but the ‘hit’ ratio for switching was huge and not switching was tiny, a much bigger disparity than I would have expected.

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I was always taught by the nuns that ou should always go with your first choice when working on questions. I should have watched Let’s Mae A Deal more often!

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You’d think that the GOP has relied on this problem solving skill for years….. We know Bush did it with TARP money….Oh what about Obama….

This column and commentary make my head hurt.

The crucial point in this is that when Monty opens one of the remaining doors, it is not at random. Monty is “injecting information into the system.”

The system starts out totally random. The winning door is picked at random. The contestant’s initial pick is fully random – she has zero information about what’s happening inside the system. After the first pick, Monty, who knows which door has the winning item, uses that information to alter the system – he picks a non-winning door to open. His action is not random. He inserts that information into the system which had previously been random.

We’re so used to thinking about totally random systems, like rolls of dice or dealing cards from a fully shuffled deck, that it’s hard to think about these sorts of problems where there are non-random actions taken that affect the odds of winning.

Bon, I think martin was just saying that if you picked a losing door at first, then you will win when you switch. And since there is a 2 in 3 chance you will pick a losing door at first, there is a 2 in 3 chance that you will win if you switch.

Martin’s way of phrasing it is the one I use most often when trying to explain it to people. The key to a deeper understanding is in markj111’s post – the host ALWAYS uncovers a known losing door; it’s not a random choice. That’s why the 1/3 from that door “transfers” to the door you should switch to.

Cecil Adams (Straightdope) originally got this wrong, but in a mea culpa follow-up post he gave this other example:

“Suppose our task is to pick the ace of spades from a deck of cards. We select one card. The chance we got the right one is 1 in 52. Now the dealer takes the remaining 51 cards, looks at them, and turns over 50, none of which is the ace of spades. One card remains. Should you pick it? Of course. Why? Because (1) the chances were 51 in 52 that the ace was in the dealer’s stack, and (2) the dealer then systematically eliminated all (or most) of the wrong choices. The chances are overwhelming–51 out of 52, in fact–that the single remaining card is the ace of spades.”

Give credit where credit is due. Marlyn Vos Savant was posed this question in 1990 and answered it correctly despite much disagreement with mathmeticians.

I’m pretty sure increasing the number of doors to a million (or even 4) doesn’t work the same way. I’m going to work the math on it and I’ll get back to y’all. But it doesn’t feel right.

(of course neither does the original problem huh?)

At first I was going to offer to take you all to Vegas but as the posts and opinions mounted I realized we’d never make it out of the casino’s lobby.

DigitalDave,

Everyone knows yawning is infectious and about the phenomena of sympathy pains, but I never guessed there was such a thing as sympathy itching induced by a hypothetical until I read your example.

Frankly I actually just increased the numbers to show that as long as the chances are less you picked the prize from the start they are still less at the end.

Digital showed it even more with his trillion doors.

This scenario actually played out in Deal or no deal where when 2 boxes were left and the million dollars was still in play and the contestant did not switch.

Guess who did not win the million dollar case!

Frankly — “why does the entire 1/3 chance switch to the other unchosen door?”

This is the important question, and very revealing. Let’s consider a different scenario but just as important. Let’s imagine you’re sleeping in a climate where malaria is deadly. Statistically, let’s say that one in every thousand mosquitos carries the virus.

You sleep under a net, but there is a tiny hole. Late that night, a swarm of 1000 mosquitos attacks your house, and one of them gets in and bites you. You have just suffered a one-in-a-thousand chance of getting malaria.

Would you rather have had the other 999 mosquitos bite you?

The scenario is easier to understand when the probability drops to zero. Try a trillion doors. You pick one, then Monty says “gosh, the car is actually either behind the door you picked or door number 65,234,102,332 — do you want to switch your choice?” Hmm, let me think…

The key factor is that the box Monty shows you is not chosed at random. He ALWAYS picks a losing box. The initial chance that you picked the winner is one out of three. The odds do not change when are shown a box that is known to be a loser.

The procedure works as long as you pick a goat the first time. This is 2/3 of the time. ğŸ™‚

The probabilities don’t change, Martin. You still have a 2/3 probability regardless of what you pick the first time.

If I may offer a video showing the math, Frankly and MacK, this might help.

http://mathfour.com/monty-hall

And Nal, you’re funny!

MacK – but in your scenario if they opened one box at a time until there were only two left there would only be a 1/24 chance your original choice held the prize but a 23/24 chance the unchosen one held the prize if the relationship held. So basically no chance at all that you picked the right box to start with. This seems even less likely.

Another way to look at it is from the original odds.

Let’s say we have 25 boxes, but only 1 has money in it, and the rest have nothing.

You choose 1, and then you’re, allowed to swap for the remaining 24 boxes if you want. Do you? Of course you do, because you know the chances are much greater the money is in the 24 boxes (24 to 1).

Switch it up.

You choose 1, and then we simply remove non money containing boxes until 1 box remains. Do you swap now? Remember the original choice was 24 to 1, and simply removing known non money containing boxes did not change that.

All people would choose to switch when there are 24 cases remaining. Not because of the pure math so much, but obviously the original odds were the 24 cases had the money.

What if the contestant wants to win a goat?

I’m assuming that they never really gave the goat to the contestant.

What makes this counterintuitive to me is why does the entire 1/3 chance switch to the other unchosen door? Why doesn’t half of it switch to the chosen door & half to the other unchosen one? there is a 2/3 chance the car is behind any combination of 2 doors not just the 2 unchosen ones.

If the 1/3 chance really did only switch to the other unchosen door than I would expect picking it every time would mean double the wins. The only demonstration of this that I saw was a limited (100) sample size but the ‘hit’ ratio for switching was huge and not switching was tiny, a much bigger disparity than I would have expected.

I was always taught by the nuns that ou should always go with your first choice when working on questions. I should have watched Let’s Mae A Deal more often!

You’d think that the GOP has relied on this problem solving skill for years….. We know Bush did it with TARP money….Oh what about Obama….