School-Safe Puzzle Games

The best game-show math brain teaser and the controversey that surrounded the answer

This has to be my all-time favorite brain puzzle, mainly because the answer plays with your sense of intuition, but also because of the incredible controversy it created.

The question was submitted to a famous magazine about 15 years ago, and the woman with the world’s highest recorded IQ provided the solution. Shortly thereafter, dozens of math PhD’s wrote in, blasting her for giving the wrong answer. She was, however, correct.

Here it is:

• Suppose you’re on a game show, and you’re given a choice of three doors. Behind one door is a car; behind the others, goats. You pick a door — say, No. 1 — and the host, who knows what’s behind the doors, opens another door — say, No. 3 — which has a goat. He then says to you,"Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Try to give it a shot on your own, and in a day or two if no one gets it right I’ll post a link to a detailed description of the answer and the story that surrounded it.

77 Comments to “The best game-show math brain teaser and the controversey that surrounded the answer”

1. Josh | Guest

It is to your advantage to switch your choice because you had a 33% chance initialy and now have a 50% chance.

2. Lee | Guest

It is to your advantage to switch, because, since the game show host knew what the doors contained, he intentionally chose a remaining door(3) with a goat. That means it is to your advantage to pick the remaining door(2). In picking door 3, the host is basically telling you there is a chance that door number 2 has the car, when he has made no such comment on the door you picked(1).

3. Mark Erdmann | Guest

There’s no point in switching your answer. The fact that door three is a goat doesn’t have any effect on your odds of being correct. In fact, it is now more likely that you’ve already chosen the correct answer if we assume that the host’s goal is to steer you towards the wrong answer. He knows what’s behind the doors, so he wouldn’t be wasting his time showing you door three and offering you another chance if you’d already lost.

4. Michael Nunziante III | Guest

Assumptions
The host knows where the car is and intentionally picks door 3 because he knows it has a goat.
2. He cannot eliminate door 1, which you have chosen.
3. The host’s ‘job’ is to entertain the audience, NOT to make you lose, or win.

Possibilities
1. You have chosen a car. (1 in 3 chance) Both 2 and 3 have goats, and the host could have chosen either to eliminate a goat.

2. You have chosen a goat (2 in 3 chance). Host can choose either door to eliminate.

It is in your interests to change your answer, as you have a 67% chance of having chosen a goat on your first pick. The odds of door 2 having the car is now 67%, given the new information; i.e., door 3 has a goat.

5. Lee | Guest

I want to expand on my answer a bit.

The host will always pick one of the goats first, so it’s just a matter of probabilities. If you separate the goats, rather than just treating them like they are the same, you can get the proper probabilities. There are three possible outcomes:

The host shows goat A, goat B is behind door 1, and you get the car by switching.

The host shows goat B, goat A is behind door 1, and you get the car by switching.

The host shows either goat, the car is behind door 1, and you lose by switching.

Therefore, if you switch, there is a 2/3 chance of winning the car, and only a 1/3 chance of winning if you stay with door 1. This is, of course, assuming the host knows what’s behind all the doors and he’s not trying to steer you towards or away from the prize.

6. Martine | Guest

What kind of a gameshow is this? You asked for door No. 1, and the host opened No. 3!? Obviously this host is a jerk and wants the car for himself. You already got the right answer, but you’re not getting the car if this jerk’s in charge. Door No. 2 has a goat too. What a jerk.

7. Kate | Guest

this website shows that it is in your advantage to switch as explained in Josh’s response (the first one):

8. Michael Nunziante III | Guest

The key is that your first pick had a 2/3 (67%) chance of being a goat. The remaining odds are NOT 50/50 because the host eliminated a goat.
Change your pick and win a car 2 out of 3 times.

9. Bill | Guest

Kate would be right except the link has a slightly different rule, “If the host knows you have picked the wrong door, he must show another door.” Gotta be 50/50.

10. JC | Guest

My off the wall answer:
I would ask him why he did not open door number 3 like I originally asked.

11. JC | Guest

Correction to previous comment:
*…open door number 1 like I originally asked.

12. PuppetSoul | Guest

-If the contestant picks the wrong door initially, the host must reveal the remaining empty door in the second stage of the game.-
^this actually makes the odds of winning 100%. If you pick the correct door, you win, if you pick the wrong door, you just change it after the other wrong door is revealed.

This clause is not present in the question given, and thus, you have a 50% chance of winning regardless of the door choosen, because any door you’d chosen, right or wrong, the host reveals the 3rd door. You then have a 50-50 chance of picking the right door, meaning 50% of the time the car will be under the door you’d originally choosen, and 50% under the other.

The paradox is that people actually believe their chances of choosing the car door increase by changing their answer, which is obviously not true. The odds of having picked the correct door increase just the same as having not done so, because the possibilities are reduced to only two options (the third option is always removed). Even if he revealed YOUR door as being empty, you’d still have a 50-50 chance on each of the other doors, the same as if you’dve stayed and he’dve revealed the other empty door.

However, if the host follows a set pattern, say he usually opens the highest numbered empty door (such as the applet in that link does), then the odds that you will get the car are greatly increased if you choose door 1 and he opens door 2 (meaning the car is likely behind door 3), or if you choose door 2 and he opens door 1 (again behind door 3), or if you choose door 3 and he opens door 1 (door 2 this time). The odds then reduce to 50-50 for switching if he opens door 3 when you choose door 1 or 2, or door 2 when you choose door 3. Thus, if the host follows a semi-predictable pattern, your odds are greatly increased by switching, because you now have the additional clue if he opens the “tell” door.

13. RK | Profile

The question was submitted to the Ask Marilyn column in Parade Magazine back in 1990. Marilyn vos Savant is listed in the Guniness Book of World Records Hall of Fame under “Highest IQ”.

For an interesting article on the controversy this question created (ie, the Monty Hall Problem), you can read here. http://www.willamette.edu/cla/.....arilyn.htm

I had a hard time understanding that by switching, you have a 2/3 chance of being right. At first glance, I thought the probability would be 50%.

This page at answers.com does a nice job explaining the problem, but the applet link supplied by Kate is really helpful too.

14. mario | Guest

Ok. so i am a sophmore in high school so this took me a while also. i do not like the explanations offered so heres my own. Base the problem on a scale of 100 rather than 3. So your chance of choosing the car is 1% and the goat is 99%. Well you will probably choose wrong on your guess. So say you do pick a door (probably wrong). Then the host lifts 98 doors, all containing goats and you are offered the switch. So at this point you have 98 goats out of the way, the door u picked, and the last door. Well since u PROBABLY chose wrong in the first place and 98 goats have been exposed, then that means the leftover door must be the. Unless you originally pick the car to begin with, which is highly unlikely on a scale of a 100, then switching will win you the car. You’re chances increase from 1% to 99% by making the switch. Apply that same concept to 3 doors and there you go. Dr. R.L. Kaplan, loved the problem and keep them coming.

15. DDynastiKK | Guest

OK, someone in my Statistics class presented this. Let me try and explain it the best I can.

The starting possibilities:

1. You are on a car
2. you have a goat
3. you have a goat.

Now lets look at the outcomes.

1. you are on a car. If you switch, you lose. thats 1/3 chance of loss by switching.
2. you are on a goat. Since the other goat has been shown, by switching you WIN, and have a 1/3 chance of winning by switching.
3. (repeat 2), so you have another 1/3 chance of winning by switching.

Therefore, you have a 2/3 chance of winning if you switch.

If my explanation wasn’t clear, just play out each possible situation starting with the original 3 doors. start on the possibility you started on a car, then goat 1, then goat 2.

16. DDynastiKK | Guest

O i’d also like to point out the possibility of the mathematicians being right, because regardless of what you land on, there is still 1 goat able for the host to open up, so I believe the 2 curtain switching possibility is completely unrelated to the original problem, and now it’s just a matter of are you on the goat or the car.

17. magtured | Guest

I have read and understood Marilyn’s answer, as well as DDynastiKK’s and all of the answers that say it is in your best interest to switch, but I do not agree. In my opinion, this is just a complex iteration of the “gambler’s falacy” in which gamblers believe that the result of a previous experiment affects the odds of a repeat of the experiment. An example would be flipping a coin and having it come up heads, so thinking there is a better chance that tails will come up next. The coin has no idea what the first result was. Every flip is 50%.

In this instance, every time the game show asks him to pick a door, it is a new experiment. When there are three doors, the contestant chooses, and we do not know the results of this experiment. When the host opens door number three, he asks again. This is a completely different experiment, and independent of the first. The odds or 50% that door one has the car, and 50% that door two has the car. There is no advantage to changing the contestants pick.

In case you need a weird illustration:

Suppose when the game show host asked the first time, the contestant said “the avocado is green.” The host then opens door number three, and asks the contestant again to pick a door. There are two doors and the probability for each is 50%. The only difference is the actual words uttered by the contestant. To say in the first instance that it is better for him to change after door three is opened is to say that somehow the words he uttered had some effect on the probability, which of course is impossible.

18. mario | Guest

magtured…..ur totally wrong. my theory works…..ur doesnt. its all about PROBABILITY. try it out. theres sum websites where u can simulate it online. its a 66% chance if u switch. believe me.

19. Gman | Guest

After the host narrows it down to two choices all you would have to do is just be patient and listen. The goats will make noises. The car won’t.

20. magtured | Guest

Mario: I understand it is all about probability. While not an expert, I have taken some college level prob and stat courses. That is why I can’t accept that the content of a previous conversation can affect the probability of the outcome of an experiment.

I still contend that after the game show host exposes one of the losing doors, he asks the contestant to choose again. This is a completely independent experiment, consisting of only two doors. The fact that the contestant chose previously when there are three doors is irrelevant.

Let’s try it this way. When the contestant is first asked, the probability of choosing the car (Pc) is:

Pc=1/3=.33

The game show host then exposes one of the three doors, taking it out of the equation. The contestant is asked to choose again. This experiment looks like this:

Pc=1/(3-1)=1/2=.5

Now show me mathematically, in equation form, how this is negated by the content of the conversation held before the second experiment. Or just put your solution in equation form.

As far as trying it out, the Law of Large Numbers indicates that you would have to try it many more times than is practical doing it one at a time. When blackjack players first used computers to test systems, they wrote programs to simulate 1,000,000 hands. Anecdotal evidence of trying it a few times, or 20, is not accurate. (I have not found the web sites you mentioned, but it seems it would be difficult to write code to simulate the contestant choosing thousands of times.)

21. elana | Guest

no because ytou still have a 33% chance

22. magtured | Guest

elena: I stand by my original assertion that this problem represents two entirely different and independent experiments. What happens up to the point that the host reveals a door and asks the contestant to choose again (which is what he is doing when he asks if the contestant wants to switch) is entirely irrelevant to THAT experiment.

The second experiment involves choosing between two doors. Everyone admits that, if the contestant did not choose prior to the host opening one of the doors, the probability would be 50%. So I am now expected to believe that this is changed because of the words uttered previously. Someone please explain to me how the content of a prior conversation can affect the probability of the outcome of the experiment. I don’t mean this in a confrontational way, I really want someone to explain it to me.

23. alex | Guest

“Someone please explain to me how the content of a prior conversation can affect the probability of the outcome of the experiment”

It doesn’t change the probability. The probability that he chose the correct door in the first place is 1/3. It always will be, regardless of what happens next. It doesn’t change to 1/2 simply because one of the goat doors was revealed. YOU are trying to change the probability based on the prior conversation.

SUppose YES is the correct answer 1/3 of the time, NO 1/3 of the time, and MAYBE 1/3 of the time. You choose to answer YES. Now, suppose after choosing YES, all MAYBES are changed to be considered equal to NO. Now, NOs are the correct answer 2/3 of the time. Now, if someone gives you the opportunity to change your answer to NO, would you not do it?

Do you see the analogy? YES is door 1, NO is door 2, and MAYBE is door 3.

24. alex | Guest

Eh, perhaps that wasn’t such a good analogy, in retrospect.

This is hard. On the one hand, the host is giving you doors 2 and 3 by giving you the opportunity to change AND reveal one of the doors. However, if you stick with your original choice, it would seem that you ALSO have a 2/3 chance because you are given door 1 and 3.

I can see the logic on both sides, I just can’t decide which is faulty.

25. james | Guest

to magtured…you are forgetting one important factor that doesn’t make the second part of the experiment completely blind to the first part. When you are asked initially to choose a door, you are choosing a door and not saying avocados are green. The event in the second experiment depends on what doors you did not pick, therefore giving you an advantage by almost allowing you to choose again. The host will choose the wrong answers of the remaining choices that you didn’t choose. the host will never reveal your 1st door before asking you to switch.

if were given two doors and a picture of a goat next to the doors then it would be 50-50 and you can be asked if apples are red and bananas are brown after some time, and then choose a door.

26. james | Guest

this is a really cool problem. here is my summation from the links i’ve read.
*=car

WIN LOSE WIN (switch)
[ ] [*] [ ]
1 2 3
LOSE WIN LOSE (don’t switch)

if i choose 1 or 3 i win; 2 i lose (assuming i switch) – 2/3 chance
if i choose 1 or 3 i lose; 2 i win (if i dont switch) – 1/3 chance

choose 1, host shows 3 switch to 2…WIN
choose 2, host shows either 1 or 3 i choose the opposite…LOSE
choose 3, host shows 1 switch to 2…WIN

To WIN if you switch, you have to choose a wrong one initally.
To WIN if you dont switch, you have to choose THE right one initally.

since there are more wrongs than right its better to switch and hope to initially choose a wrong one.

27. magtured | Guest

James: Your explanation details exactly the problem I have with your solution. You say the first experiment is relevant precisely because of what the contestant says. I do not accept that the content of the conversation can affect the probability of the outcome of the experiment.

You say the host “…almost allow(s) you to choose again.” It appears to me that he most definitely allows you to choose again, and you choose between two doors.

The probability in question is the probability that the goat is behind door number one, not the probability that the correct door was chosen. The probability remains the same whether anyone chooses a door or not.

28. Seanwaa | Guest

its got nothing to do with the second choice. It’s all about that there is a 2/3 chance initially that you are wrong. This means that there is a greater chance to be incorrect, even when the third door is removed because the choice was made initially based on there being 3 doors. Now, if you have a greater chance to be wrong, and a wrong answer is removed, you still have a 2/3 chance to be wrong, based on the 3 original doors, the difference now is you know where one of the wrong answers is, but there are still 2 wrong answers behind the doors. Now that you can see one of the Goats, and the host knew that the door he revealed was wrong, you have a 2/3 chance that your door is wrong, meaning there is a 1/3 chance that the door that is left is wrong, so it makes sense logically to switch. You may not win, but you have double the chance to.

29. e-motionsickness | Guest

Maybe I’m thinking simply, without all the mathematics. The first thing I thought when I processed the question was no, it doesnt matter. Of course I would like for the host to open door number two, cause since he already opened one, any next door opened would be door number two….in theory.

30. e-motionsickness | Guest

and to add on, of course if this were true, like said above, I of course would like to choose my own door rather than the game show host playin’ with my free car!

31. Sebastian Lee | Guest

The fact that the host reveals one of the wrong answers AFTER you choose your initial answer does something really cool!

I think the easiest way to think of it is this:
1.) In order to win by switching your answer, all you have to do is choose a WRONG answer initially.
2.) You have a 2/3 chance of choosing a wrong answer initially!
3.) You have a 2/3 chance of choosing the right door by switching!

Another way of looking at it is this:
1.) The door you initially choose has a 1/3 chance of being correct. Each of the other two doors still have a 1/3 chance of being correct at this point.
2.) However, the host REVEALS one of these two other doors as wrong, meaning that door now has a 0/3 chance of being correct!
3.) So now there is 2/3 of a chance needing to be assigned to a door still… The last remaining door that you would switch to is the door that now has that 2/3 chance of being correct!

Still not satisfied? I created interactive proof that you can play with. Check it out:

32. RLP | Profile

Excellent simulation, Sebastian. Nice work!

33. JT | Guest

O.K. Hopefully this should clear things up for everyone, (or at least the mathematicians). And remember there has already been a extensive debate over this and they did choose that you should always change. But that is irrelevant considering we are trying to prove it for ourselves here.

Let the doors be called X, Y, and Z.

Let CsubX(Cx we will say) be the event that the car is behind door X and so on (X, Y, and Z).

Let HsubX(Hx we will say) be the event that the host opens door X and so on (X, Y, and Z).

Supposing that you choose door X, the possibility that you win a car if you then switch your choice is given by the following formula:

P(Hz ^ Cy) + p(Hy ^Cz)
= P(Cy)*P(Hz|Cy) + P(Cz)*P(Hy |Cz)
= [(1/3 )*1] + [(1/3)*1] = 2/3

Not my work, I take no credit.
-RAmen

34. atlas3u | Guest

Suppose there were 10 doors and you choose door 1. The host opens doors 3 to 10 revealing 8 goats. Does that meam you have 90% chance of success if you switch. I don’t think so. It is still 50% when there are only two possibilities left.

35. atlas3u | Guest

On further thought, the host has to leave the car plus your answer. Therefore 90% is probably right.

36. Zubeida | Guest

The contestant had already picked door number one. This means that at that point he had a 33.333..% chance of getting it right. The fact that the show host opens door number three increases his chances of winning to 66 %. So he should stick to his choice

37. The missing square: Where does it come from? - Smartkit Brain Enhancement News | Guest

[…] If you’d like to try the most controversial brain teaser of all time (seemingly easy but very counter-intuitive) that pitted ‘the world’s smartest woman’ Marilyn vos Savant vs. Math PhDs, click on the link. […]

38. milo zachary | Guest

The Exploratorium in San Francisco had an exhibit demonstrating this a while back; it is probably still there.

As I recall, there were 4 doors and one person selected a door after the other person standing behind the doors had put a token behind one. The person behind the exhibit then revealed 2 of the incorrect non-choices. The instructions were to then change your selection to the door that had not been revealed that you had not initially selected. Between my son and myself, this resulted in ending up with the correct door 19 out of 20 times.

39. ashwin | Guest

To switch to another choice is about instincts and not advantage.The host might have opened door No-3 to generate interest and nothing else.

40. Dan | Guest

It is in your advantage to switch. If you reverse the situation, and pretend the game show host must pick two consecutive goats, then work the probabilites of that, it is in your favor.

likelihood of picking 1 goat on the first pick is .666, likelihood that the host’s second pick is also a goat is (.66666 X .666666 or .44). So the likelihood the second pick is a goat is only 44%, so you now have a 56% chance that #2 is the car.

41. JDR | Guest

WRONG, WRONG, WRONG!!! The answer is 50%, not 2/3

This problem has always annoyed me because the people who say it’s a 2/3 shot if you switch use slight of hand to get there by treating the two goats as one. So for the moment, forget the fact that we have two goats. Let’s assume instead that there is a car, a goat, and a bucket instead of a second goat. Your initial setups would be:

1 car/goat/bucket
2 car/bucket/goat
3 goat/car/bucket
4 goat/bucket/car
5 bucket/car/goat
6 bucket/goat/car

If the host shows you a goat behind door #3, configurations 1, 3, 4, and 6 are eliminated. That means the only initial configurations that could have been behind your door were 2 and 5. Which means there is only a fifty-fifty shot that you would win by changing your answer. Let’s do it using goats, this time, but labeling them as specific animals, rather than the generic “goat”. Our initial configurations could be:

1 car/goat A/goat B
2 car/goat B/goat A
3 goat A/car/goat B
4 goat A/goat B/car
5 goat B/car/goat A
6 goat B/goat A/car

If the host shows you a goat behind door three, then the configurations for 4 and 6 are are eliminated. More important, however, YOU DON’T KNOW WHICH GOAT HE SHOWED YOU. Of the four choices you have left, two (1&2) have a car behind the initial door you picked, two (3&5) have a goat. Once again, you have a fifty/fifty shot of winning if you change.

The 2/3 advocates conveniently ignore the fact that the goats are distinct. They ARE NOT interchangeable. As a result, they propose only three initial configurations:

1. car/goat/goat
2. goat/car/goat
3. goat/goat/car

By eliminating the MANDATORY other three choices as noted above, they manipulate their answer through false logic and therefore come up with the wrong answer.

Even from a statistical point of view the result doesn’t change. The people who advocate the “switching gives you a 2/3 chance of being correct” have fallen into the classic “coin flipping” problem. Person #1 says: the odds that I can flip a coin ten times in a row and have it land heads each time are ridiculously large and against me.

Incredibly, however, he in fact flips heads nine times in a row.
Just before Person #1 flips the coin for a tenth time, along comes person #2, who says, “the probability that you are going to flip heads this time is 50%.

How can this be if the person #1 is correct (and he is) that the odds of flipping 10 heads in a row are insanely high against him?

Simple: the events are examined at separate times. Huh? Person #1’s chances have to be calculated as if he flipped 10 pennies at 1 time, not one penny 10 times. Person #2 always calculates his odds based on each individual coin.

Flipping coins consecutively, rather than simultaneously, has no impact on the probability that the next throw gives you a fifty-fifty chance of throwing heads. If you flip a coin for an infinite period (not counting the effects from one side of a coin being slightly heavier than the other) you will have a fifty-fifty chance of flipping heads on your next flip regardless of whether you flipped heads 1,000,000 heads prior to that. Over time it all evens out.

It’s no different with the goats. If you have a 1/3 chance of picking the car at the beginning, that 1/3 odds is only valid if you are not given any additional chances to pick a door. It doesn’t matter which door the host opens first because your odds were fixed at 1 in 3 when you were given the opportunity to choose.

*But* this did not happen in this case. Why? Because the host gave you a new chance to pick a door. Just as with the pennies, each separate choice has to be viewed on its own. As soon as the host gives you a second opportunity to choose, you are no longer dealing with a 1 in 3 situation, but a 1 in 2 choice. By permitting calculations to take into account anything that came before (just like with the pennies) creates a logical fallacy because those prior conditions have absolutely no bearing on the the choice now before you.

Suppose the rules were stated slightly differently: Contestant, choose a door. After you choose I will reveal a goat. It might be your goat or it might be one of the goats behind a door you did not pick. What he is really saying is after you pick your door, I am going to take one of the goats away. This means that only one of two possibilities exists: you have a goat or you have a car.

42. Toon | Guest

On ‘Let’s Make a Deal’ the game had three steps:

A) the contestant picks but can not yet see one of three prizes, let’s say \$5, \$10 and \$1000.
B) Monty Hall reveals one of the remaining prize, but he is not allowed to reveal the \$1000 prize,
C) the contestant is offered the chance to switch prizes.

The key is that MH is not allowed to reveal the \$1000 prize.
If step B were a simple random elimination then you would have twelve possible outcomes. You could:

1)Initially chose \$5, MH shows \$10, you switch to \$1000 and WIN
2)Initially chose \$5, MH shows \$10, you stay with \$5 and LOSE
3)Initially chose \$5, MH shows \$1000, you switch to \$10 and LOSE
4)Initially chose \$5, MH shows \$1000, you stay with \$5 and LOSE

5)Initially chose \$10, MH shows \$5, you switch to \$1000 and WIN
6)Initially chose \$10, MH shows \$5, you stay with \$10 and LOSE
7)Initially chose \$10, MH shows \$1000, you switch to \$5 and LOSE
8)Initially chose \$10, MH shows \$1000, you stay with \$10 and LOSE

9)Initially chose \$1000, MH shows \$5, you switch to \$10 and LOSE
10)Initially chose \$1000, MH shows \$5, you stay with \$1000 and WIN
11)Initially chose \$1000, MH shows \$10, you switch to \$5 and LOSE
12)Initially chose \$1000, MH shows \$10, you stay with \$1000 and WIN

At step A you have a 1 in 3 chance of winning. When step B is random then at step C you still have 4 in 12 or 1 in 3 chance of winning. 2 of 6 times when you stay you win and 2 of 6 times when you switch you win. This is simple statistics. What gets tricky is how step B changes those statistics. Intuitively one wants to just delete possibilities 3, 4, 7, and 8; but that would falsely exaggerate the probability of initially choosing \$1000.

So adding step B changes some of MH’s actions and some of the outcomes. You can:

1)Initially chose \$5, MH can show \$10, you switch to \$1000 and WIN
2)Initially chose \$5, MH can show \$10, you stay with \$5 and LOSE
3)Initially chose \$5, MH must show \$10, you switch to \$1000 and WIN
4)Initially chose \$5, MH must show \$10, you stay with \$5 and LOSE

5)Initially chose \$10, MH can show \$5, you switch to \$1000 and WIN
6)Initially chose \$10, MH can show \$5, you stay with \$10 and LOSE
7)Initially chose \$10, MH must show \$5, you switch to \$1000 and WIN
8)Initially chose \$10, MH must show \$5, you stay with \$10 and LOSE

9)Initially chose \$1000, MH can show \$5, you switch to \$10 and LOSE
10)Initially chose \$1000, MH can show \$5, you stay with \$1000 and WIN
11)Initially chose \$1000, MH can show \$10, you switch to \$5 and LOSE
12)Initially chose \$1000, MH can show \$10, you stay with \$1000 and WIN

Step A has not changed, you start with a 1 in 3 chance of winning. Add in step B, and at step C you have 6 in 12 or 1 in 2 chance of winning. But if you stay you still have a 2 in 6 chance of winning but if you switch you now have a 4 in 6 chance of winning.
What happened? At step A the three choices are divided into two groups. One contains your initial choice with a one-third chance of winning and the other with the two remaining choices with a two-thirds chance of winning.
In step B MH must reduces the group with two choices to one choice without eliminating its two-thirds chance of winning, that is, if the winning choice was in the group of two it will remain in that group when MH shows and eliminates one of the two choices. So in step C you are better off switching prizes.
Still confused? MH cannot show (and thus eliminate) the prize in step B. This changes his actions in outcomes 3,4 7, and 8. In numbers 3 and 7 he is forced to eliminate a losing choice and you gain two more chances to win by switching. In 4 and 8 the change in MH’s actions do not change the outcome.

43. Lath | Guest

Ah, I love the classic Monty Hall problem. I see it’s an old thread, but not everyone is convinced, so I can’t resist throwing in my two cents.

The assumptions that Michael Nunziante gave in post number 4 are absolutely critical.

If the host has no knowledge of what lies behind the doors, or doesn’t base his action on his knowledge, then the people claiming 50-50 are correct. In this case, the problem is simply reduced to two doors. In this case your choice of a door and what lies behind the door chosen by the host are independent events. Therefore the remaining possibilities are equally likely. Since you would win 1 out of 2 of them, it’s 50%

However, given that the scenario is a game show, I think the most reasonable assumptions are that the host knows what lies behind the doors and will always choose to reveal a goat and give you the option to switch. In other words, the host bases his choice on your choice, therefore they are not independent random events. This means that the two remaining possibilities are not necessarily equally likely. It can not be safely concluded that just because 1 out of 2 possibilities results in a win means there is a 50% chance. Given these assumptions, as others have stated before, using the strategy to switch wins the car whenever your initial choice was not the car – which is 2/3 of the time.

Also, it makes no difference whether you consider the goats separately or recognize that by symmetry that can be treated in the same.

If you are still not convinced, consider the following games – or better yet find a friend to play both games with 3 playing cards instead. Suppose you use an Ace for the car and 2 Jacks for the goats.

Scenario A – Your friend has no knowledge
– Shuffle the three cards, and deal them face down.
– Choose a card.
– Let your friend reveal a card
– 1/3 of the time your friend will reveal the Ace at which point the game is over
– 2/3 of the time your friend will reveal a Jack
– If your friend has not revealed the Ace, then check to see if switching would win.
– Repeat the trial (keeping track of wins and losses) as long as you need to convince yourself that you will win half the time (that your friend did not reveal the Ace).

Scenario B – Your friend knows the cards and always reveals a Jack
– Let your friend deal the cards
– Choose a card.
– Have your friend reveal a Jack
– 1/3 of the time (when you chose the Ace) your friend may choose either remaining card
– 2/3 of the time (when you chose a Jack) your friend will have no choice which card he must show (the only remaining Jack)
– Check to see if switching would win.
– Repeat the trial (keeping track of wins and losses) as long as you need to convince yourself that you will win 2/3 of the time by switching.

44. Tommy | Guest

I think that the main reason that most of the letters that she received disagreed with her was because the people that knew the answer had no reason to write. There is also the possibility that it was the cause that probability has not been stressed much in math competitions until recently.

45. charlotte_elle | Profile

curious incident of the dog in the night time anyone? [: you should have read that anyway on principle, but if you havent, the book proves you have more of an advantage by switching, in both a basic explanation and an advanced one for you high-fliers =p

46. wyrdo | Profile

Well, from what I’ve heard, your best play is to randomly choose between the two non-goat doors by flipping a coin. In other words–in this situation, intuition is bunk, and it’s best to assume that there’s an equally likely chance that the prize could be behind either of the doors that are still closed, never mind your guess. Your original guess was made when there were three closed doors (~33% chance). If you flip a coin now, you’ve got 50% odds.

Furry cows moo and decompress.

47. ivydog | Profile

You will win twice as often if you switch doors.

That is a fact jack.

48. george1994 | Profile

you are twice as likey to win if you switch, i like this cuz you really have to think about it to understand it.
and yes the curious incident of the dog in the nightime is an amazing book

49. Anastasia11 | Profile

I got it eventually. After trying it with eggcups.
If you pick a door, whether you open it before or after the host opens his, its probability of being the car is 1/3. Its probability can’t change because of what the host does.
Before the host opens the door the probability that one of them contains the goat, is 2/3. Once one of the doors has been opened its probability is 0/3 so the probability that the remaining door contains the car must be 2/3 for the probabilities to add up to 1.
I thought this problem was really cool. Its really weird when you get it because the answer is the opposite of what you would expect.

50. Kevinpkennedy | Profile

The easiest explanation is a simple spreadsheet. The goats are irrelevant – its all about your initial choice & where the car is. There are 9 possibilities:

You Car Monty
Chose Behind Shows Stay Switch
1 1 2 or 3 W L
1 2 3 L W
1 3 2 L W
2 1 3 L W
2 2 1 or 3 W L
2 3 1 L W
3 1 2 L W
3 2 1 L W
3 3 1 or 2 W L

51. Kevinpkennedy | Profile

I’ll try this again, with the spreadsheet turned around.

The best way to see it is a simple spreadsheet. The goats are unimportant – it is all about your initial choice & where the car is. There are 9 possibilities:

You choose 1 1 1 2 2 2 3 3 3.
Car behind 1 2 3 1 2 3 1 2 3.
Monty show 2or3 3 2 3 1or3 1 2 1 1or2.
Ifyou stay W L L L W L L L W.
Ifu switch L W W W L W W W L.

52. wyrdo | Profile

I really don’t think it work that way. Almost every single person on this board believes that’s how it works, but I still disagree.

I think the moment one of the doors gets opened and you find out what’s behind it, you have to redo all the probability stuff. The open door with a known object (goat) will *now* have that *known* quantity (goat) 100% of the time.

To figure the probability on the other two doors, you merely ignore the goat you can see. This leaves you with only *two* doors. You’ve got a 50% chance of picking the car, whether you stick with your original choice, or whether you switch makes no difference.

See, the reason why I feel sure this is the correct answer, is that if the *other* 2/3 answer is correct, then this seems to imply that simple probability has weird properties similar to quantum physics (which it doesn’t).

Furry cows moo and decompress.

53. noxsmarts | Profile

I say no, it’s no advantage in it. No.1 is the choice made and you should stick with your choice believing strongly you made the right choice first.
Noxsmrt

54. seananderson | Profile

The way I think it works is you start off with a 1/3 chance. no matter what you choose the host is always going to choose another door with a goat. meaning that you may think you have a better chance but you are still relying on that 1/3 chance you took before. this means that the first box you chose has a 66% chance of being wrong while the other one has a 50% chance because you know it was not chosen by the host

55. rishirocks | Profile

see first of all the question had be answered once only and the host asking the second time was wrong as the game would be unfair. there was no clause in the game rules to change the answer. and the quastion had to be answered between 3 doors not 2, so the particpant shd not change the answer andgo for another rule also because luck favours the brave and choosing another door would be unfair coz if she won the car others might argue why the host ask to change the door and if loose he might ardue why the host asks so.

56. eskimo | Profile

If you still don’t believe that switching increases your chances of winning from 33% to 66%, even with all the previous, well-explained posts on here, the only way for you to be convinced is to gather data for yourself. Yes, you will be doing the Monty Hall “game” at least 30 times, preferably more.
To conduct this experiment, get a deck of cards and a patient friend or family member. Pull out 2 black cards and 1 red card. I use the two black queens (goats) and the ace of hearts (car). Mix them behind your back thoroughly so the ace is in a random position. Bring them back in front of you and glance at the cards, noting the position of the two queens, without revealing any of the cards to your friend. Lay the cards face down and allow your friend to touch the back of one card. Now, eliminate one of the queens from the game as long as it’s not the card she chose. To gather data on the outcome of switching, always have her reveal the card she didn’t chose after one of the queens has been eliminated. Record the face of the card, mix, and repeat. In my experiment of 100 trials, always switching, 69 resulted in winning. Most results will hover around 66%. You don’t have to do this 100 times, but for all you non-believers, at least try this experiment until you are convinced that there is no way in hell the odds are 1:1. Switching DOES make a difference. If you don’t believe me, see for yourself…

57. heinscott | Profile

After reading all these responses (especially the one about trying it yourself with a deck of cards), I thought I would try to write a test case in C#. Here is code that will show the 2/3 positive results of always switching. Obviously, the more trials you run, the closer the results will be to 66.666667.

Scott

int trials = 1000000;
int count = 0;
bool[] door = new bool[3];
Random ran = new Random();
for (int x = 0; x < trials; x++)
{
door[0] = (ran.Next(0, 3) == 0);
//selects a random number between 0 and 2. If 0 set to true;
door[1] = (ran.Next(0, 2) == 0 && !door[0]);
//Same as above (0-1), but, set to false if first is true;
door[2] = (!door[0] && !door[1]);
//third door only true if first two are false;
int pic = ran.Next(0, 3);
//select a random “guess” for contestant
//Switch door to opposite (always end result of switching)
//If true, increment the counter.
}
Response.Write((double)count/(double)trials * 100);

The hardest part to really get your head around, I think, is realizing that “switching” always means that you will have the opposite that you started out with, due to the fact that the host will always show you a door with a goat. Therefore, your 1/3 chance changes to 2/3.

58. kaxxito | Profile

sorry for my english. suppose in game are 2 players. David and John. David choosed door number 1, and John choosed door number 2. both of them has 33% chance to win. then host opened door number 3 and there was nothing. after that David’s and John’s chances will be equal. David and John are in equal positions, everything effects on them identically, so if one’s winning probability increases, seconds will increase too. my answer is: it has no matter to change answer

59. smartguy | Profile

Here is another way of looking at it – sorry if anyone has already mentioned it. Suppose you have a deck of playing cards with 51 Aces and 1 King. All the cards are faced down. You are asked to select a card that you think is the King. I then turn over 50 cards, showing 50 Aces. Now surely if you are given the opportunity of changing your card, you would do so – as the chance of you selecting the king at the start was 1 in 52. It would be very unlikely that you chose the King at the beginning. The same principle applies here.

60. drshacket | Profile

Stay with the same door. Your odds were 33%. Now with door number 3 revealed (also 33%), your odds of having selecting the right door increased by 33%. Therefore, your odds of winning is now cumulative, to 66%, buy staying with door number 1.

61. experiment | Profile

ok, really all you PHD’s, professors, MIT guys, highest IQ lady, etc, why is this so hard to understand and explain, apparently most cannot do what science always dictates, run the experiment. To better understand, make it ten doors, one car, nine goats. Contestant picks a door, host removes 8 goats doors. One of two things just happened:
1.contestant picked the one car door out of ten doors-10% chance
2.contestant picked one of the nine goats doors-90% chance
That means the other door the host left is one of two things:
1.a goat-10%chance
2.a car-90% chance
GET TEN PIECES OF PAPER, ONE CAR, NINE GOATS,PICK ONE.AFTER THE HOST REMOVES THE EIGHT GOAT DOORS IT BECOMES BLATANTLY, GLARINGLY, RIDICULOUSLY OBVIOUS WHERE THE CAR IS(90% OF THE TIME ON AVERAGE). JUST DO IT.

My G.E.D. came in real handy here, it helped me understand this immediately, some people are just thinking too much instead of running the experiment.
This is actually my favorite brainteaser of all time now

62. experiment | Profile

it replaces:
HOW FAR CAN A DOG WALK INTO THE WOODS?

63. experiment | Profile

Ok, 2 more ways to look at this.
1. The key is that the host KNOWINGLY exposes a goat. He exposes a goat no matter what the contestant chose. If the host DID NOT know where the car is and the exposed door could be either car or goat then it would be 50/50 that the contestant had the car door IF a goat was exposed.

2.Doors 2 and 3 must be one of 3 combinations
goat goat
goat car
car goat
The host automatically eliminates a goat from whichever combination exists leaving:
goat
car
car
2 OUT OF 3 chance that a car is left.

end of story

64. wyrdo | Profile

There are many, many answers that indicate that you will have a 2/3 (66%) chance of getting the car if you switch. This is one of those times when people let themselves fail to “see the forest for the trees”. I mean–if you think about it for a moment, how could your choosing door one over door 2 possibly have any impact on the odds that it will have a car or a goat? All the “monte hall” simulations in the world aren’t going to help you work out the odds because, in this instance, you’re basing them on a false premise.

What really happens is this: as soon as the host reveals there is a goat behind one of the doors, the original problem:

( 33+1/3% car, 33+1/3% goat, 33+1/3% goat )

goes away. Supposing the host reveals a goat behind the 3rd door, we now have:

( 50% car, 50% goat, goat *for sure* )

The goat we can see doesn’t figure into our probability calculations. And which door we choose has no affect on the probability that we get a car or a goat. How could it?

In an instance such as this, game theory tells us the best play is to make the decision truly random: “let the chips fall where they may” as they say. So you should do a heads pick door 1, tails pick door 2 type of thing.
—————————————–
I have to say, the frequency with which people get this wrong is very depressing. But on the plus side, I’m starting to understand why casinos succeed in pulling in so much cash. Various people that think they are smart are unable to detect a run-of-the-mill trick question, and they think their choice can somehow affect the odds. That’s insane. I want to open my casino now please.

Furry cows moo and decompress.

65. Benson | Profile

It is obvious that we have to switch as door 2 now has the probability doors 2 and 3 combined.

Despite the GED, why do some people find it so hard to understand? This has nothing to do with dogs walking far into the woods. But you still can do what science always dictates, run the experiment. To better understand: make it a cocker spaniel and a medium sized forest. Explain to the dog that its grandmother is waiting for it at the other side of the woods. Make sure you give it plenty of food and water. We would not want some lame dog excuses like no food or water ruining the experiment. Then equip it with a tracking device to find out how far a dog can run into the woods.

Don’t bother telling anybody the results as I doubt that anybody would be interested in them, that is unless they are craving for a good laugh.

end of story

66. experiment | Profile

hmm, so its seems you agree with me about the doors while showing no proof or reason for your thinking but do not know how far a dog can walk into the woods. OK, brainiac. Maybe you should stick to the word search puzzles.

67. experiment | Profile

and wyrdo, you are embarrassing yourself. The challenge is not to prove that the wrong answer is correct, the challenge is to understand why switching doors gives one a 2/3 chance of winning. There are many great explanations here. good luck.

68. Benson | Profile

experiment, I have summed up the proof in a very simple sentence. Not everybody is expected to grasp it, so there is no need to be on the defensive if you didn’t. Derogatory terms won’t help your case either.

As for the dog walking far INTO the woods, maybe you’d like to share the details of your experiment. A good laugh won’t hurt us.

69. XSLT brain teaser « Tech Team | Guest

[…] address values “n/a”, “na” and “Dr”. This reminds me of the game show brain teaser where by stepping through the scenarios in a logical fashion gives the correct answer; trying to […]

70. fanfan | Profile

Let us start by assuming that we cannot second guess the motive of the show host in changing the game rules:
a. By offering a 2nd choice, the game rules have been changed, from a 1 in 3 choice, to a 50:50 probability choice, having no relationship at all to the earlier 1 in 3 starter (it is as if this did not exist at all). In this case, it remains a 50:50 probability, and there is no point in changing the original choice.
b. If the game host has a motive, it would, by the very nature of these programs, to prolong delivery of the reward, and not make it too easy for the contestants. This motive would have a higher probability, looking at the very nature of man. In that case, in case the contestant had initially made a wrong choice with the 1 in 3 option, the probability should be less than 50% that, instead of promptly declaring the contestant failed, and extending the program, that the host would try to give a second chance with a new 50:50 option. In this case, the host is attempting misdirection, to create self-doubt in the contestant. So, in this scenario too, contestant should stick to original choice.

71. wyrdo | Profile

I agree for the most part with fanfan. It’s nice to know that my opinion (which I fully believe to be the mathematically correct one) is shared by at least one other person on this site. :-)

fanfan could very well be correct about the psychology angle, but my understanding of how these game shows operate, at least in the USA where I am, is that because of a controversy back in the 50s over the show “Twenty-One” (see wikipedia), no one on the show including the show host has access to info that’s supposed to be hidden. That is to say: in order to prevent even the illusion of the possibility of rigging the show, the things that are supposed to be hidden/random/unknown really *are* hidden/random/unknown.

And, from what little game theory I can remember, in a situation like that, you’re best game strategy is to be truly random. So: don’t *always* change your original guess and don’t *always* stick with your original guess. Rather: completely ignore you’re original guess (since it was based on different odds) and make a new random choice (50%/50%) using a coin or something.

I don’t know if would fanfan agree with me on that part, but it’s cool either way. I don’t want to start a dispute with fanfan over that point.

I’m just relieved that there’s at least one other person out there that understands that once one door has been opened and you’re given the opportunity to change your guess, the old 1/3, 1/3, 1/3 odds no longer applies and we have a brand new probability problem with 50%/50%. :-)

Furry cows moo and decompress.

72. LydiaFre | Profile

I’d say yes because door 1, which you picked first had a 1 out of 3 (33.3%) chance of having the car. Now, with only 2 doors to chose from, the other door has a 1 out of 2 (50%) chance of being correct. The odds are in your favor if you switch your choice.

73. Mev | Profile

The answer is to switch and you have a 2 in 3 chance. People assume it’s 50/50 because there’s 2 doors left.

The easiest way I can explain this is to think of it as you against the host, you have one door and he as two. He opens one of his which he knows wrong, then offers to switch his with yours. In fact he is offering to swap your 1 in 3 chance with his 2 in 3 chance.

Always swap and you’ll double your odds.

74. Kevinpkennedy | Profile

There are 9 possible combinations of the door you choose and the door the car is actually behind. This is easy to see with a spreadsheet, but I cannot get it to format here. So:

1. You choose Door 1, car is behind Door 1, Monty shows 2 or 3 – if stay, you win; if switch, you lose.

2. You choose Door 1, car is behind Door 2, Monty shows 3 – if stay, you lose; if switch, you win.

3. You choose Door 1, car is behind Door 3, Monty shows 2 – if stay, you lose; if switch, you win.

4. You choose Door 2, car is behind Door 1, Monty shows 3 – if stay, you lose; if switch, you win.

5. You choose Door 2, car is behind Door 2, Monty shows 1 or 3 – if stay, you win; if switch, you lose.

6. You choose Door 2, car is behind Door 3, Monty shows 1 – if stay, you lose; if switch, you win.

7. You choose Door 3, car is behind Door 1, Monty shows 2 – if stay, you lose; if switch, you win.

8. You choose Door 3, car is behind Door 2, Monty shows 1 – if stay, you lose; if switch, you win.

9. You choose Door 3, car is behind Door 3, Monty shows 1 or 2 – if stay, you win; if switch, you lose.

If you switch, you win 6 times. If you stay, you win 3 times.

75. Jasha | Profile

The host knows whats behind the door so he open door 3 revealing the goats. So it is to his advantage thatby switching to door 2 is the car since the host could not make that choice at the first place.

76. brainache | Profile

On average,player will have goats 2/3 times and car 1/3 times.If both player and host have goats 2/3 times then for those 2/3 times the car must be behind the other(closed) door.Choose to swap and player gets contents of closed door which is 2/3 times the car and 1/3 times a goat.Choose not to swap player gets 1/3 times the car and 2/3 times the goat.Swapping therefore doubles your chances.

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