### Brain Teasers

# Quiz Show

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.Probability

Suppose you're participating in a TV quiz show and you make it to the final round. The quiz master shows you three closed boxes; two of them are empty, but the other one contains the super prize of $1,000,000.

He asks you to make a choice... When you make your choice he opens one of the rejected boxes and shows you it is empty.

Now he gives you the opportunity either to stick to your first choice or to pick the other box.

What should you do? Or does it not matter at all because the odds are now fifty-fifty?

He asks you to make a choice... When you make your choice he opens one of the rejected boxes and shows you it is empty.

Now he gives you the opportunity either to stick to your first choice or to pick the other box.

What should you do? Or does it not matter at all because the odds are now fifty-fifty?

### Hint

It surely DOES matter.### Answer

Leave your first choice for what it's worth and pick the other box. You only have to stick to your initial choice when that box holds the money. But the probability of your first choice being correct was only 1/3, or 33.33%.By changing your choice you raise your chances up to 2/3 (66.66%).

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## Comments

Nov 12, 2001

This is wrong there is no improvement in the odds to be gained by changing box.

the two selectins should be looked at as seperate systems and the emptyness of the rejected box does not affect the probability of the selected box having the money.

If you reselected the same box then by the logic of the answer you have just made it more likely to have the mony in it

the two selectins should be looked at as seperate systems and the emptyness of the rejected box does not affect the probability of the selected box having the money.

If you reselected the same box then by the logic of the answer you have just made it more likely to have the mony in it

Petera,it is correct.Odds are definitely improved.Suppose you play the same game thrice,and prize money in different box each time,if you will stick to your choice you will once ,but if you will change your choice you will win twice.Still couldn't get it??OK consider 100 box and you select one,out of 99 ,98 empty ones are opened,still you will stick to your original choice which has a probability of 1/100 ??

petera is correct. When you choose from 3 boxes, you have 1/3 chance of choosing the winning box. When the # of boxes decreases to 2, you have 1/2 chance of choosing the correct one. Your chances of choosing the correct box increase because there are fewer boxes that do not contain the prize.

petera is definately correct here... if you take one box away, you have new odds, which make it a 50/50 chance that you will choose the right box. It would certainly make game shows easier if this worked, but it doesn't!

Dec 11, 2001

it is correct. at first you probably picked an empty box. thus, of the other 2 left, one's probably the money and one's definitely empty. When they show you the empty one you should change to the box that's STILL probably the money.

Everybody is right if you have three boxes you have a 1/3 of a chance of getting the right box

2 boxes you a a 1/2 of a chance the only way to be sure is if you only had one box 1/1

2 boxes you a a 1/2 of a chance the only way to be sure is if you only had one box 1/1

Here, look at it this way:

Let's say the prize is behind Door Number Two. There are some different possibilities-

If you pick 1 and switch, you win.

If you pick 2 and switch, you lose.

If you pick 3 and switch, you win.

If you stay, it's lose-win-lose for Door 1, 2, and 3 respectively.

The host showing you that one of the doors you didn't pick has nothing behind it changes absolutely NOTHING. At least one of those two *must* have nothing behind it. Having the host show it to you doesn't mean your odds just improved.

Let's say the prize is behind Door Number Two. There are some different possibilities-

If you pick 1 and switch, you win.

If you pick 2 and switch, you lose.

If you pick 3 and switch, you win.

If you stay, it's lose-win-lose for Door 1, 2, and 3 respectively.

The host showing you that one of the doors you didn't pick has nothing behind it changes absolutely NOTHING. At least one of those two *must* have nothing behind it. Having the host show it to you doesn't mean your odds just improved.

This is the famous Monty Hall problem!

One is right, one is wrong that is a 50/50 chance.

Sorry to tell you, but your calculations are incorrect. Only your selection and not your probability changes

this is a case of faulty logic

I must correct my stance on this puzzle. I thought it would stay 50/50 but after experementing I found myself wrong and support the answer of this quesiton. The reason your odds increes is because you have a 2/3 chance of being wrong. So there is a 2/3 chance of the host removing the only box he can remove. So now there is a 2/3 of switching helping. If the host were to remove a box randomly, then your odds would be 50/50.

Once you pick the first box, you have 2 boxes left, one of which contains the money, so you have a 1 out of 2 chance or 50/50.

I must not get the question, because why would you stick to your first choice if you knew it was wrong?

This is a much greater question than the fun rating or difficultly rating reflect. I suppect that is because so many people were not able to "get it" so they rated it badly. Pity. The answer is correct and the reasoning given for the answer is correct. If you think it is wrong run some tests for yourself. Use three cards, one of which is an ace, turn them over and mix them up. Pick one and have a friend look at the other two and turn over one that is a non ace card. Stick with your initial card ten times and see how many times you are right (should be about 1/3 of the time). Then do another ten times only this time change your answer and see how many times you are right (should be about 2/3 of the time).

I Thought that there were editors to see that dross like this did not get through!

This is one of the least well understood concepts of probability, but the question is absolutely right -- the answer is switching gives you a 2/3 chance of winning. As has been said, the reason the probability is not 50% is because the host chooses which one to show you, but when you first picked your box, there was a 2/3 chance (which doesn't change) you left the winning box in the host's hands. This is the question that generated the most follow-up mail ever to Marilyn Vos Savant (Ask Marilyn), showing how poorly understood the concept is. Great question.

I love math problems and when i try to explain this type of problems to dumb friends what helps is exagerating. Imagine you have a million doors and you chose one, you're probability is 1/1,000,000. If the prize isn't in you're first selection, then there is a chance of 1/999,999 on the other doors .If someone opens the other 999,998 doors and only leaves one, the chance that the correct door is that one is 999,999/1,000,000. So you would obviously would change. It is calles hereditary probability. If you have n "doors" and you choose one then the "host" opens other (n-2) doors and leaves yours and other one, then by changing selections, you change you're probability from (1/n) to ((n-1)/n) or (1-(1/n))...

I pity people who voted this one boring and easy because believe me it's fun and difficult Great teaser!!!

I pity people who voted this one boring and easy because believe me it's fun and difficult Great teaser!!!

Moving this question from a game show situation to a quiz show situation does little to make the wrong answer right. Even when Marilyn vos Savant muddled around with this teaser she left out one important phrase. To arrive at the 2/3 answer, the reader needs to be told that the unchosed empty box is uncovered by the master of ceremonies ON A REGULAR BASIS, i.e. each and every game.

Without that knowlege or a clear clue as to motive, you have to guess why you are given the opportunity to change your mind. 2 better, broader possibilities follow:

1)As in the sidewalk gambling's shell game and three-card monte, the "host" wants you to lose so the game retains its assets. Or, maybe it is the host's decision, and he simply does not like you. No matter, you were only given the chance to switch, because you have already picked the prize. Your chance of winning by switching is ZERO.

2)The "host" likes you (or, at least, knows you) and wants you to win and would have revealed the prize if you had already picked it. Your chance of winning by switching is 100%.

There is also the possibility of enforced randomness. If this is in effest, the prize would be the first thing revealed in 1/3 of all games. The player would then be given, in essence, a replay where the chances are 50/50.

Without that knowlege or a clear clue as to motive, you have to guess why you are given the opportunity to change your mind. 2 better, broader possibilities follow:

1)As in the sidewalk gambling's shell game and three-card monte, the "host" wants you to lose so the game retains its assets. Or, maybe it is the host's decision, and he simply does not like you. No matter, you were only given the chance to switch, because you have already picked the prize. Your chance of winning by switching is ZERO.

2)The "host" likes you (or, at least, knows you) and wants you to win and would have revealed the prize if you had already picked it. Your chance of winning by switching is 100%.

There is also the possibility of enforced randomness. If this is in effest, the prize would be the first thing revealed in 1/3 of all games. The player would then be given, in essence, a replay where the chances are 50/50.

i am soooooooooooooooooooooooooooooooooooooo confused the question/answer needs more explaining for one thing. the box he took away was not the winning box right? and then he did not tell him that the box he was holding was right or wrong right?? then how would it affect the outcome? he has a right box and a wrong box left. switching doesn't mean he gets to have both boxes. right? i think it's 50/50

Okay, here is a clear explanation of why switching will give you a better chance.

There are three boxes, you have a 2/3 chance of picking the incorrect box to start. This means that the prize is in one of the other two boxes.

So there are two boxes left, the host always eliminates an empty one, but never the one you picked (even if it is empty).

This means that 2/3 of the time you will pick an empty box to start, the host will then eliminate the other empty box, leaving ONLY the prize box to switch to.

The other 1/3 of the time you will pick the prize box to start, the host will then eliminate one of the empty boxes and you will switch to the other empty box.

As you can see, there are only two possibilities if you switch then 2/3 of the time you will switch to the prize, 1/3 of the time you will switch and lose.

If you do NOT switch then you simply have a 1/3 chance of picking the prize to start and a 2/3 chance of losing.

So you are better off switching all the time, if you want the odds on your side. Maybe some of you will do a little math before making fun of the guy who posted a great (and well known) teaser.

There are three boxes, you have a 2/3 chance of picking the incorrect box to start. This means that the prize is in one of the other two boxes.

So there are two boxes left, the host always eliminates an empty one, but never the one you picked (even if it is empty).

This means that 2/3 of the time you will pick an empty box to start, the host will then eliminate the other empty box, leaving ONLY the prize box to switch to.

The other 1/3 of the time you will pick the prize box to start, the host will then eliminate one of the empty boxes and you will switch to the other empty box.

As you can see, there are only two possibilities if you switch then 2/3 of the time you will switch to the prize, 1/3 of the time you will switch and lose.

If you do NOT switch then you simply have a 1/3 chance of picking the prize to start and a 2/3 chance of losing.

So you are better off switching all the time, if you want the odds on your side. Maybe some of you will do a little math before making fun of the guy who posted a great (and well known) teaser.

The math is flawed, you do not increase your chances....because randomness is not a perfect science......it's studied through the chaos theory....but nowhere close to being exact......

there are too many factors involved in this teaser to come up with an acceptable answer....

there are too many factors involved in this teaser to come up with an acceptable answer....

TehIgnored,

The math is not flawed, this is an extremely basic probability question. Please experiment with this and write back (Excel is great for creating random situations using the =RAND() function.) If you run this experiment as few as 15 times, you will see that the strategy laid out in this teaser works best.

Of course, when dealing with probability, nothing is ever for sure. You could play this game 100 times, always switch, and always lose, but that is very unlikely. In fact, you have only a 0.005% chance of losing 10 times in a row.

The math is not flawed, this is an extremely basic probability question. Please experiment with this and write back (Excel is great for creating random situations using the =RAND() function.) If you run this experiment as few as 15 times, you will see that the strategy laid out in this teaser works best.

Of course, when dealing with probability, nothing is ever for sure. You could play this game 100 times, always switch, and always lose, but that is very unlikely. In fact, you have only a 0.005% chance of losing 10 times in a row.

PatH,

It does not matter how unflawed the math is! Good math is not the answer to a bad question. Examine my previous comment, then quote that portion of the question which limits the host's behavior to anything predictable. I think you will find you have been hypn tized into believing that particulars of an instance are indicative of the general.

It does not matter how unflawed the math is! Good math is not the answer to a bad question. Examine my previous comment, then quote that portion of the question which limits the host's behavior to anything predictable. I think you will find you have been hypn tized into believing that particulars of an instance are indicative of the general.

Stil,

I believe you are looking for loopholes here. Of course, you could try to turn this into a sociology question about why the host would ask you to switch...however this is a probability question and the challenge lies therin.

There is quite simply no valid answer to this question if you cannot assume that the host would ask if you would like to switch every time. If this were a "situation" or "trick" teaser then you might have a point.

So, given that the host DOES ask you if you want to switch every time, do you agree that the answer is win 2/3 of the time?

I believe you are looking for loopholes here. Of course, you could try to turn this into a sociology question about why the host would ask you to switch...however this is a probability question and the challenge lies therin.

There is quite simply no valid answer to this question if you cannot assume that the host would ask if you would like to switch every time. If this were a "situation" or "trick" teaser then you might have a point.

So, given that the host DOES ask you if you want to switch every time, do you agree that the answer is win 2/3 of the time?

PatH,

It is mildly annoying to be asked to agree with one's own statement. It feels more like a grab for cred than teasing.

I consider probability a science. I want to know the variables. A gaping hole in logic is more than a loophole. I merely pointed out that inferences must be made; with four valid possibilities, none is an absolute lock.

Consider what I have called enforced randomness. In the vos Savant version of this teaser, there is a "game" show host who knows what is behind curtains. In this version, we don't even have that assurance. (Many quiz show hosts don't get the answer until the contestants have answered.) Have you used Excel to model the case where the host picks the box to open without knowing which boxes are empty? Doesn't it run toward 50-50?

It is mildly annoying to be asked to agree with one's own statement. It feels more like a grab for cred than teasing.

I consider probability a science. I want to know the variables. A gaping hole in logic is more than a loophole. I merely pointed out that inferences must be made; with four valid possibilities, none is an absolute lock.

Consider what I have called enforced randomness. In the vos Savant version of this teaser, there is a "game" show host who knows what is behind curtains. In this version, we don't even have that assurance. (Many quiz show hosts don't get the answer until the contestants have answered.) Have you used Excel to model the case where the host picks the box to open without knowing which boxes are empty? Doesn't it run toward 50-50?

It does not matter if the host knew the boxes ahead of time or not. If he opens one of the two remaining boxes, and it is empty, you are better off switching to the other remaining box, this strategy will allow you to win 2/3 of the time.

If the host opens the prize box, well, then the game is over right then, because you know that two are empty and one is a prize.

Given the way this teaser is worded, you have a 2/3 chance of winning if you switch, regardless of the knowledge of the host.

If the host opens the prize box, well, then the game is over right then, because you know that two are empty and one is a prize.

Given the way this teaser is worded, you have a 2/3 chance of winning if you switch, regardless of the knowledge of the host.

PatH,

Let the upper case letter represent the prize box; let lower case represent the empty boxes. A bracket around a [letter] represents contestant choice. There are 9 equally possible circumstances.

1 A b c 2 A b c 3 A b c

4 a B c

Let the upper case letter represent the prize box; let lower case represent the empty boxes. A bracket around a [letter] represents contestant choice. There are 9 equally possible circumstances.

1 A b c 2 A b c 3 A b c

4 a B c

PatH,

Let the upper case letter represent the prize box; let lower case represent the empty boxes. A bracket around a [letter] represents contestant choice. There are 9 equally possible circumstances.

1 [A] b c 2 A [b] c 3 A b [c]

4 [a] B c 5 a [B] c 6 a B [c]

7 [a] b C 8 a [b] C 9 a b [C]

When the host is not in the know his choices are random.

The first column is the host's choice, the second cases of an instant winner, the third... instant losers, the fourth... winners by switching, the fifth... winners by sticking.

aye 1 2,3,4,7 6,8 5,9

bee 5 2,4,6,8 3,7 1,9

cee 9 3,6,7,8 2,4 1,5

Five-ninths of all games end with the first choice; the remaining four-ninths divided equally between wise to switch and wise to stick. 50-50.

Even though it is bad form to assume, let's assume he will not open the box the contestant has chosen. Half of the time the host opens the rightmost unchosen box; the other half... the leftmost. Which charts as follows: The first column is the host's choice, the second cases of an instant loser, the third... winners by switching, the fourth... winners by sticking. A hyphen indicates that just half of case instances is in the category.

.left 2',3',4' 6',7',8' 1',5',9'

right 6',7',8' 2',3',4' 1',5',9'

One-third of all games end with the first choice; the remaining two-thirds divided equally between wise to switch and wise to stick. 50-50.

When the host is determined not to open the prize box, the contestant can win by switching in every case 2,3,4,6,7, and 8, but will lose by switching in every case 1,5, and 9. That's the only way to get two-thirds.

Let the upper case letter represent the prize box; let lower case represent the empty boxes. A bracket around a [letter] represents contestant choice. There are 9 equally possible circumstances.

1 [A] b c 2 A [b] c 3 A b [c]

4 [a] B c 5 a [B] c 6 a B [c]

7 [a] b C 8 a [b] C 9 a b [C]

When the host is not in the know his choices are random.

The first column is the host's choice, the second cases of an instant winner, the third... instant losers, the fourth... winners by switching, the fifth... winners by sticking.

aye 1 2,3,4,7 6,8 5,9

bee 5 2,4,6,8 3,7 1,9

cee 9 3,6,7,8 2,4 1,5

Five-ninths of all games end with the first choice; the remaining four-ninths divided equally between wise to switch and wise to stick. 50-50.

Even though it is bad form to assume, let's assume he will not open the box the contestant has chosen. Half of the time the host opens the rightmost unchosen box; the other half... the leftmost. Which charts as follows: The first column is the host's choice, the second cases of an instant loser, the third... winners by switching, the fourth... winners by sticking. A hyphen indicates that just half of case instances is in the category.

.left 2',3',4' 6',7',8' 1',5',9'

right 6',7',8' 2',3',4' 1',5',9'

One-third of all games end with the first choice; the remaining two-thirds divided equally between wise to switch and wise to stick. 50-50.

When the host is determined not to open the prize box, the contestant can win by switching in every case 2,3,4,6,7, and 8, but will lose by switching in every case 1,5, and 9. That's the only way to get two-thirds.

give it up u two!

This is getting far too complex. As the question is stated, for THIS game only, regardless of the host knowing or not, the player will have a 2/3 chance of winning if he switches.

My final explanation:

The player had a 2/3 chance of picking an empty box.

So there is a 2/3 chance the unpicked boxes are [prize] and [empty]

The host eliminates the unpicked empty box, leaving only the prize.

Switching will get you the prize in this case, and there is a 2/3 chance of this case occurring.

Done. Goodnight.

My final explanation:

The player had a 2/3 chance of picking an empty box.

So there is a 2/3 chance the unpicked boxes are [prize] and [empty]

The host eliminates the unpicked empty box, leaving only the prize.

Switching will get you the prize in this case, and there is a 2/3 chance of this case occurring.

Done. Goodnight.

If I could edit my last comment, I would drop the hyphens, and write that there are two equally probable subcases for each number and they are distributed as shown in the chart. (The previous chart involved three subcases.)

As always, my intent is to show this very interesting question lacks an essential element and needs editing. I hope we can all see why the other guy thinks he has THE right answer.

As always, my intent is to show this very interesting question lacks an essential element and needs editing. I hope we can all see why the other guy thinks he has THE right answer.

umm... no

it has to still be 1/3 because if he wont open the other box then your chances are still as good as the first

I had to read almost every comment b4 understanding this one

Thanks to all the smarties out there who tried to help us!

And i know many people voted badly on this one because they didn't understand, but i waited to vote until i read the comments.

Greay Job with it!

I'm sorry so many people didn't get it(including me at first)

Thanks to all the smarties out there who tried to help us!

And i know many people voted badly on this one because they didn't understand, but i waited to vote until i read the comments.

Greay Job with it!

I'm sorry so many people didn't get it(including me at first)

I was thinking about this teaser a very long time. I carefully read through each comment. My conclusion is that the teaser IS CORRECT. I think the confusion is mostly caused because people are forgetting that by switching it just IMPROVES your chances of winning, no GUARANTEEING it. It is definitely right.

This is truly a brilliant teaser, and it is such a big shame that it has such a low fun rating just because people are too thick to understand it. Definitely belongs in My Favourites.

This is truly a brilliant teaser, and it is such a big shame that it has such a low fun rating just because people are too thick to understand it. Definitely belongs in My Favourites.

I completely agree with you Psychic Master. I just saw it in a book called the curious incident of the dog in the night-time and was looking to submit it when I saw that it was already here. Into the favs it goes!

i saw this in a tv show

gr8 teaser, of course.

gr8 teaser, of course.

You people are missing the elephant in the living room. The teaser is ludicrous. What you're failing to see is that the mc will vary which one he opens to be sure to expose an empty one. And he's always got at least one empty to show. If you change it to the four possible results from your pick, you'll see that in two of them you win, and in two you lose. Here they are:

1)You've picked the prize box. The mc reveals the first empty box. You switch to the other empty and lose.

2)You've picked the prize box. The mc reveals the second empty box. You switch and lose.

3)You've picket the first empty box. The mc reveals the second empty box. You switch and win.

4)You've picked the second empty box. The mc reveals the first empty box. You switch and win.

The other two hypothetically possible situations will never happen, as the mc will not reveal the box with the prize in it.

Another way to explain it is that you make no selection whatsoever. Let's say that you win the opportunity to have an empty removed before you select. Now there are two left, each with equal chance of having the prize. It's a 50/50 proposition. It's because the possibility of the prize having been in the box that's opened first now changes from 1/3 to zero. The possibility of it being in the one you've selected now increases to 1/2.

1)You've picked the prize box. The mc reveals the first empty box. You switch to the other empty and lose.

2)You've picked the prize box. The mc reveals the second empty box. You switch and lose.

3)You've picket the first empty box. The mc reveals the second empty box. You switch and win.

4)You've picked the second empty box. The mc reveals the first empty box. You switch and win.

The other two hypothetically possible situations will never happen, as the mc will not reveal the box with the prize in it.

Another way to explain it is that you make no selection whatsoever. Let's say that you win the opportunity to have an empty removed before you select. Now there are two left, each with equal chance of having the prize. It's a 50/50 proposition. It's because the possibility of the prize having been in the box that's opened first now changes from 1/3 to zero. The possibility of it being in the one you've selected now increases to 1/2.

Paul -- you are correct that those are the four things that can happen, but the problem is that they are not equally likely from the start. Assigning probabilities to your matrix, the probability that you pick the winner (1/3) and he shows you empty door #1 (1/2) is 1/6. The probability that you pick the winner (1/3) and he shows you empty door #2 (1/2) is 1/6. In each of these cases (a total of 1/3), switching costs you the prize.

For your scenarios 3 and 4, the probability that you pick loser #1 (1/3) and he shows you loser #2 (1) is 1/3. The probability that you pick loser #2 (1/3) and he shows you loser #1 (1) is also 1/3. In these cases (which occur 2/3 of the time), you win by switching.

Therefore, switching gives you a win 2/3 of the time, precisely because the host will only ever show you a loser.

Hope this helps.

For your scenarios 3 and 4, the probability that you pick loser #1 (1/3) and he shows you loser #2 (1) is 1/3. The probability that you pick loser #2 (1/3) and he shows you loser #1 (1) is also 1/3. In these cases (which occur 2/3 of the time), you win by switching.

Therefore, switching gives you a win 2/3 of the time, precisely because the host will only ever show you a loser.

Hope this helps.

tsimkin-You did a nice job of explaining a narrow view error. Are there words in the question which force us into the narrow view that the host has and will always open an empty box and give the contestant a chance to switch?

For any more insight into the problem, please read a book named The Curious Incident of the Dong in the Nighttime. This will really clear things up and you'll get to read a great book!

Did you mean The Curious Incident of the Dog in the Nighttime? If so, please consider the nature of the work. A review by Maria G. Fung for the Mathematical Association of America (Find it at www.maa.org/reviews/dogincident.html) points out that a crucial subtlety is ignored.

Recognized it right away as the famous Monty Hall problem.

It was obviously intended to be the Monty Hall problem and I knew that the expected answer was probably that switching would give you a 2/3 probability.

However, like stil, when I first read the problem I noticed the flawed wording. The problem needs to make it clear that the host will ALWAYS show you an empty box after ou first selection in order to get the 2/3 chance.

Without that wording you're left to assume that the host has both the knowledge about which box contains the prize and the intent to always show you an empty box. Without having both of these assumptions spelled out the problem runs into all of the problems illustrated by the 44 comments preceeding this one.

However, like stil, when I first read the problem I noticed the flawed wording. The problem needs to make it clear that the host will ALWAYS show you an empty box after ou first selection in order to get the 2/3 chance.

Without that wording you're left to assume that the host has both the knowledge about which box contains the prize and the intent to always show you an empty box. Without having both of these assumptions spelled out the problem runs into all of the problems illustrated by the 44 comments preceeding this one.

Like everybody here said and explained, given:

1. You want to win (take the RIGHT box)

2. You choose AT RANDOM at first (1/3 RIGHT probability, 2/3 WRONG)

3. A WRONG box is always SELECTIVELY removed after your first choice

You should switch to increase winning chances.

As to those who object to the logic of "switching increases winning probability", in fact, you're right, in a way. Switching RANDOMLY after your first choice (e.g: you choose box 1 then change to 3 WITHOUT knowing what's in any of the boxes 1, 2 or 3) would NEITHER INCREASE NOR DECREASE your chances. You chose a random and substituted it by another random. You had 2/3 chances of being wrong, then by changing your choice you now have 2/3 chances of being wrong. Nothing changed.

In our case here, however, a WRONG choice is SELECTIVELY removed. Your 2/3 chances of being WRONG are THE SAME if you don't switch. Losing odds are decreased to 1/3 by switching, since now a WRONG option has been omitted. You CAN'T now switch to this WRONG choice if you switched, hence the change in losing odds. That's it.

1. You want to win (take the RIGHT box)

2. You choose AT RANDOM at first (1/3 RIGHT probability, 2/3 WRONG)

3. A WRONG box is always SELECTIVELY removed after your first choice

You should switch to increase winning chances.

As to those who object to the logic of "switching increases winning probability", in fact, you're right, in a way. Switching RANDOMLY after your first choice (e.g: you choose box 1 then change to 3 WITHOUT knowing what's in any of the boxes 1, 2 or 3) would NEITHER INCREASE NOR DECREASE your chances. You chose a random and substituted it by another random. You had 2/3 chances of being wrong, then by changing your choice you now have 2/3 chances of being wrong. Nothing changed.

In our case here, however, a WRONG choice is SELECTIVELY removed. Your 2/3 chances of being WRONG are THE SAME if you don't switch. Losing odds are decreased to 1/3 by switching, since now a WRONG option has been omitted. You CAN'T now switch to this WRONG choice if you switched, hence the change in losing odds. That's it.

READ:"When you make your choice he opens one of the rejected boxes and shows you it is empty." What says this is more than the simple reportage of a single instance it seems to be? The only only way to be certain of the so-called answer is to asssssume the 'quiz master'

ALWAYS opens an unchosen empty box as an act of volition. That assumption is imagination, not math.

ALWAYS opens an unchosen empty box as an act of volition. That assumption is imagination, not math.

Dec 29, 2009

I'll make it real simple. There's 3 doors...one with a car, one with a goat (goat A), and another with a goat (goat B). You have a 1/3 chance of initially selecting the car.

Scenario 1: you choose the door w/ the car. The host then can open either of the other doors, showing you goat A or B. In this scenario, switching doors will lose.

Scenario 2: you choose the door w/ goat A. The host now can only open the door w/ goat B. In this scenario, switching doors will win.

Scenario 3: you choose the door w/ goat B. The host now must open the door w/ goat A. In this scenario, switching doors will win.

In 2 of the 3 scenarios, switching doors will win.

Now the same examples, but without switching doors.

1: you pick the door w/ the car. The host can then open either of the other doors revealing one of the goats. If you stick w/ your initial pick, you win.

2: you pick the door w/ goat A. The host will open the door revealing goat B. You stick to your initial pick, you lose.

3: you pick the door w/ goat B. The host will open the door revealing goat A. You stick w/ your initial pick, you lose.

So in the second example, sticking to your initial pick will win 1/3 of the time.

Scenario 1: you choose the door w/ the car. The host then can open either of the other doors, showing you goat A or B. In this scenario, switching doors will lose.

Scenario 2: you choose the door w/ goat A. The host now can only open the door w/ goat B. In this scenario, switching doors will win.

Scenario 3: you choose the door w/ goat B. The host now must open the door w/ goat A. In this scenario, switching doors will win.

In 2 of the 3 scenarios, switching doors will win.

Now the same examples, but without switching doors.

1: you pick the door w/ the car. The host can then open either of the other doors revealing one of the goats. If you stick w/ your initial pick, you win.

2: you pick the door w/ goat A. The host will open the door revealing goat B. You stick to your initial pick, you lose.

3: you pick the door w/ goat B. The host will open the door revealing goat A. You stick w/ your initial pick, you lose.

So in the second example, sticking to your initial pick will win 1/3 of the time.

Futile###-

In your scenarios, you turn the clock back to the moment before the first goat is revealed. You also ASSUME that the MC's only goal is to reveal a goat first. This shifts the odds. Honest would oblige you to consider other possibilities.

Suppose the car's cost comes out of the MC's pay. He would want you to lose, and because there is no proof he couldn't have, he would have revealed the car if you have not picked it. Change and you lose 100% of the time.

and vice versa.

In your scenarios, you turn the clock back to the moment before the first goat is revealed. You also ASSUME that the MC's only goal is to reveal a goat first. This shifts the odds. Honest would oblige you to consider other possibilities.

Suppose the car's cost comes out of the MC's pay. He would want you to lose, and because there is no proof he couldn't have, he would have revealed the car if you have not picked it. Change and you lose 100% of the time.

and vice versa.

It's not 1/2. It actually is 1/3. I understand this more than the other ones! I hope this'll help you understand... (If you actually read it)

Door 1-?

Door 2-?

Door 3-?

You have only 1/3 chance of getting the car-however, you can change after the host shows you a goat.

(Let's use A as a goat and B as a car, and they'll be shown just so you know which one is real.)

Door 1-A-Your choice

Door 2-B

Door 3-A

The host revealed that door 3 is a goat. Now, it might seem like it's 1/2, but you have to consider your first choice- 2/3 will be goat, but if you switch, 2/3 will be car!

Door 1-A

Door 2-B-Your choice

Door 3-A-Shown

Another situation:

Door 1-A

Door 2-B

Door 3-A-Your choice

End result:

Door 1-A-Shown

Door 2-B-Your choice

Door 3-A

Both of them show you winning the car-which is 2 out of 3.

However:

Door 1-A

Door 2-B-Your choice

Door 3-A

Door 1-A-Shown

Door 2-B

Door 3-A-Your choice

As you can see, you lose. But that's only 1 of 3! Switching improves your chances of winning! (Now, how can you go against THIS?)

Door 1-?

Door 2-?

Door 3-?

You have only 1/3 chance of getting the car-however, you can change after the host shows you a goat.

(Let's use A as a goat and B as a car, and they'll be shown just so you know which one is real.)

Door 1-A-Your choice

Door 2-B

Door 3-A

The host revealed that door 3 is a goat. Now, it might seem like it's 1/2, but you have to consider your first choice- 2/3 will be goat, but if you switch, 2/3 will be car!

Door 1-A

Door 2-B-Your choice

Door 3-A-Shown

Another situation:

Door 1-A

Door 2-B

Door 3-A-Your choice

End result:

Door 1-A-Shown

Door 2-B-Your choice

Door 3-A

Both of them show you winning the car-which is 2 out of 3.

However:

Door 1-A

Door 2-B-Your choice

Door 3-A

Door 1-A-Shown

Door 2-B

Door 3-A-Your choice

As you can see, you lose. But that's only 1 of 3! Switching improves your chances of winning! (Now, how can you go against THIS?)

racoonieboy is just the latest to "SEE" more than he has read.

Before you can answer this puzzler accurately, you must know the answers to two questions which will effect how those fractional goats and car are distributed. Does the host know what is behind the doors? Is he required to reveal a goat behind 'remaining' doors? My past comments detail how.

Before you can answer this puzzler accurately, you must know the answers to two questions which will effect how those fractional goats and car are distributed. Does the host know what is behind the doors? Is he required to reveal a goat behind 'remaining' doors? My past comments detail how.

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