Quirky math.

[Indon]

In the spirit of the 2=1 thread, I present to you a perfectly logical and yet odd bit of probability.

You're on a game show and given the choice of three doors. Behind one of the doors is a big prize (be it a new car, luxury getaway, lifetime supply of xp-in-a-can, whatever) and behind the other two are hungry velociraptors (it's a japanese game show).

After picking a door, the game show host (from behind his raptorproof shield) opens up one of the _raptor_ doors, and trained handlers drag the raptor off. Now there are only two doors, and the host asks if you want to stick with your old choice, or switch to the remaining door.

The question: What impact would switching doors have on your chances to get the prize?

Answer and reference:

Spoiler

Show

It increases your chances from 33% to 66%.

Applicable link enclosed.

Post other math quirks on this thread, we'll start something that can get anyone's head scratching.

[I'm da Rogue!]

I can't remember of any right now, but I saw this thread dangerously rollong to the bottom of the page and I thought it's a really nice topic to be forgotten
Come on people, give us your best!

[Lucky]

That's wrong.

[Syka]

Spoiler

Show

Doesn't it increase your chances to 50% since you now only have two choices, the third having been eliminated? Apologies for wordiness, I'm a Classics major and have been translating Greek all day. Also apologies if I'm completely off.

Cheers,
Syka

[Lucky]

It doesn't increase anything by re-guessing. After the first door with the raptors is opened, your chances have increased to 50%. Changing will keep it at 50%. No improvement.

It's like flipping a coin twice. You flip a coin, it lands on heads. Now you flip another, what are the chances of it being tails? Would it not improve since you flipped heads last time?

No, it would not. It would still be 50%.

The probability of guessing right is equal to the number of right choices available, divided by the number of choices, time 100%.

1/2 x 100%= 50%

That's how it works.

[I'm da Rogue!]

That's wrong.

Ahhh...
OK. I'll write something. It's 1.53 in the morning here, but OK.

Riddle:
You have three vases: one vase containing two white pearls, one vase containing one white and one black pearl, and one vase containing two black pearls. From one of these vases, a pearl is taken. This pearl turns out to be white. What is the probability that the other pearl in the same vase is also white?


The answer:

Spoiler

Show

There are three pearls that can be the white pearl that was taken from the chosen vase:

* The first pearl from the vase with two white pearls: in this case, the other pearl is also white.
* The second pearl from the vase with two white pearls: in this case, the other pearl is also white.
* The white pearl from the vase with one white and one black pearl: in this case, the other pearl is black.

The probability that the other pearl in the same vase is also white, is therefore 2/3.

[The Great Skenardo]

Actually, it's right. I can sort of put together an intuitive proof for you to follow along, if you like.

Spoiler

Show

There are only three configurations the doors can be in:
( = Quesadillas n' Root Beer
= Raptors)

You can have
1__2__3



and no others, right?
Let's say you pick door #3, initially. The host reveals that Door #1 has a raptor behind it. Now, this tells you that there are two possible configurations, and it's this that gives the impression that you have 50/50 chances. Two options, so an equal chance, right?

Well, not quite. If you pick a door at random at the beginning, you have a 2/3 chance of picking a raptor, right? If you select a raptor, then the host must open the door revealing the last raptor, meaning the other door has the quesadillas and root beer behind it. If you switch doors in these two cases, you get the food!

If, however, you pick the correct door right off (a 1/3 chance), then the host reveals a raptor, but the other one is still hidden. If you switch, you will get it and dies.

Make sense?

[Skippy]

No, it doesn't... There are two doors now, but remember that those doors are part of a set of three. The possibilities now are 2/3. When you had all three doors you only had 1/3 chance to get the right door...

[Lucky]

Actually, it's right. I can sort of put together an intuitive proof for you to follow along, if you like.

Spoiler

Show

There are only three configurations the doors can be in:
( = Quesadillas n' Root Beer
= Raptors)

You can have
1__2__3



and no others, right?
Let's say you pick door #3, initially. The host reveals that Door #1 has a raptor behind it. Now, this tells you that there are two possible configurations, and it's this that gives the impression that you have 50/50 chances. Two options, so an equal chance, right?

Well, not quite. If you pick a door at random at the beginning, you have a 2/3 chance of picking a raptor, right? If you select a raptor, then the host must open the door revealing the last raptor, meaning the other door has the quesadillas and root beer behind it. If you switch doors in these two cases, you get the food!

If, however, you pick the correct door right off (a 1/3 chance), then the host reveals a raptor, but the other one is still hidden. If you switch, you will get it and dies.

Make sense?

Hmm. So it seems I am wrong. Very well explained good sir! Have a cookie!

@V Simu-pwned 3 times over.

[Silkenfist]

It doesn't increase anything by re-guessing. After the first door with the raptors is opened, your chances have increased to 50%. Changing will keep it at 50%. No improvement.

It's like flipping a coin twice. You flip a coin, it lands on heads. Now you flip another, what are the chances of it being tails? Would it not improve since you flipped heads last time?

No, it would not. It would still be 50%.

The probability of guessing right is equal to the number of right choices available, divided by the number of choices, time 100%.

1/2 x 100%= 50%

That's how it works.

Uhmm...no. And I am willing to bet you large amounts of money on it.

Long explanation

Spoiler

Show

The problem is, that you are ignoring one part of the game show. It is correct, that in the final situation there are two doors with one prize and one raptor. If you would not have been given any information earlier, it would come down to the flip of a coin. But - fortunately - you have been given information. You have been shown one door that is not a raptor.
How does this affect the game? It becomes evident, if you rephrase the problem. You have one door that has been chosen earlier. Now if there are only two options, switching from one option to the other one will inverse your chances.
You have chosen your door at the beginning of the show, when there were still three options present - two raptors, one prize. Choosing correctly is - under those circumstances - 1/3. After you have chosen your door, the host opens another door and shows a raptor. Note, that this doesn't affect the chance of your current door being the correct one (tricky bit). The point is, that the game host will ALWAYS show you a raptor, since he will ALWAYS have a door with a raptor left. There is no case whatsoever, when - at this point - he will not be able to reveal a raptor.
Previously the chances were 1/3 "I have the correct door" and 2/3 "The correct door is one of the two others"
After revealing a raptor, this has not changed. The chances are still 1/3 "I have the correct door" and 2/3 "One of the others is correct"
Now we return to the final stage of the game: Two doors left. But the choice of doors left is already affected by the earlier part of the game. The door, you had chosen earlier has a chance of 1/3 to be correct. The other one has an inverse chance, read: 2/3.

Short explanation (or if you are still not believing me):

Spoiler

Show

We have three doors: A, B and C. Let's say, C has the prize and A&B have raptors. Now we start the game show and bring in one candidate. Let's say, he chooses his door randomly.
1/3 chance, he chooses door A
1/3 chance, he chooses B
1/3 chance, he chooses C

Now let's calculate his winning chances if he chooses to keep his previous option.
IF he chose A earlier, the host opens B and reveals a raptor. Our candidate will keep his guess and be ghastly devoured.
IF he chose B, the host opens A. The candidate keeps his guess and is once again eaten
IF he chose C, the host opens either A or B. Either way, the candidate keeps his guess and gets the prize.

Each of these three events has an equal probability - 1/3. In one case, the strategy "keep" wins the prize, in two cases, it loses. Thus, keeping your door has a success rate of 1/3 exactly. Choosing the other door, accordingly, has a chance of 2/3

[The Great Skenardo]

Mmm....cookie ^_^

[Gygaxphobia]

Ahhh...
OK. I'll write something. It's 1.53 in the morning here, but OK.

Riddle:
You have three vases: one vase containing two white pearls, one vase containing one white and one black pearl, and one vase containing two black pearls. From one of these vases, a pearl is taken. This pearl turns out to be white. What is the probability that the other pearl in the same vase is also white?

Surely not! If you take the white pearl out and do not replace it, you have to have a 50:50 chance.

[Pyrian]

Most of this "math" is so chock full of undeclared assumptions that no real correct answer can be given.

Let's take the gameshow example. In the scenario as presented, on no account should you switch your choice! You guessed right the first time. You know this is true because if you had guessed a raptor, they'd've simply let it out and you'd be pet food by now. The host is only trying to convince you to switch because he knows you picked right on the first try.

Again with the pebbles, if the pebble were picked randomly and turned out to be white, or if a random white pebble were picked, then yes, there's a 2/3 chance the other pebble in same vase is white. On the other hand, if a white pebble was selected from a random white-pebble-containing vase, then the odds are 50%.

[Starla]

I am a little confused by the original post. You are saying that the gameshow host had you pick a door first. Then opened one of the other doors? Then he gives you a chance to switch?

Okay the vase one is more clear. 3 vases and one white pearl was removed. Okay that excludes the vase with 2 black since it would have 0 white to remove. Now that would leave the other 2 vases and one had 2 white and the other had a white and a black. SO it is a 50% percent chance that the second pearl is also white... That is my final answer. If I am wrong give me a high school teacher's explanation, because the others were hard to follow.
I think High school trig was the last time I remember looking at ratios.

This is fun. Do another one.

[The Great Skenardo]

@Starla:
This is the famous "Monty Haul" problem, in which one door hides a fabulous prize, and the other two hide lesser prizes. The contestant picks a door. Then, the host reveals one of the unpicked doors to have the lesser prize. You then have the option of switching your choice to the other unopened door.

The Host always reveals a lesser prize.

[MeklorIlavator]

Most of this "math" is so chock full of undeclared assumptions that no real correct answer can be given.

Let's take the gameshow example. In the scenario as presented, on no account should you switch your choice! You guessed right the first time. You know this is true because if you had guessed a raptor, they'd've simply let it out and you'd be pet food by now. The host is only trying to convince you to switch because he knows you picked right on the first try.

The question actually has a real-world birthplace. It's from the gameshow Let's Make a Deal, which had rule that prevented killing the contestants, so the problem actually works.

[Silkenfist]

I am a little confused by the original post. You are saying that the gameshow host had you pick a door first. Then opened one of the other doors? Then he gives you a chance to switch?

Okay the vase one is more clear. 3 vases and one white pearl was removed. Okay that excludes the vase with 2 black since it would have 0 white to remove. Now that would leave the other 2 vases and one had 2 white and the other had a white and a black. SO it is a 50% percent chance that the second pearl is also white... That is my final answer. If I am wrong give me a high school teacher's explanation, because the others were hard to follow.
I think High school trig was the last time I remember looking at ratios.

This is fun. Do another one.

Oh Thor, please let me be not simu'd on this one....

I'll contradict once again. The chance is 2/3. How so?

Let's alter the situation slightly to make it more transparent. I am writing letters on the pearls, naming the white ones A, B and C and the black ones X, Y and Z. Then, I put on a blindfold, shuffle them around and place them in the vases as in the pattern written above.
Now I shuffle the vases around and draw one pearl from any of them. I remove the blindfold and see: It is the white pearl A. Let's have a look at the possible options now. There are three possibilities for me to distribute two white pearls in one vase:
1/3 chance of A&B being in one vase, C in the other one.
1/3 chance of A&C being in one vase, B in the other one.
1/3 chance of B&C being in one vase, A in the other one.

In have picked A, so in the first two cases I will draw another white pearl. Only in the third case, A shares a vase with a black pearl. Thus, the probability of drawing a white pearl again is 2/3

[Pyrian]

Let's alter the situation slightly to make it more transparent.

And, y'know, get the answer you're looking for.

[waffletaco]

My friends that were in statistics gave me a similar problem. They told me that by switching doors, your chances are 66% to get it right. I probably wouldn't bet money on that though. Sure you have an advantage to switch, but 33% to fail is too big for me.

[MeklorIlavator]

My friends that were in statistics gave me a similar problem. They told me that by switching doors, your chances are 66% to get it right. I probably wouldn't bet money on that though. Sure you have an advantage to switch, but 33% to fail is too big for me.

If you stay, you have a 66% chance to fail, which is greater than 33% last time I checked.

[Sir_Norbert]

And, y'know, get the answer you're looking for.

No, the point is that labelling the pearls can't change anything relevant to the original question. Therefore the probability in the original question is also 2/3. Don't try to be a smart-ass when you don't know what you're talking about.

Starla, perhaps this explanation will help.

It's all about conditional probability -- you didn't know beforehand that you would pick a white pearl, but given that you have, what is the probability that the other pearl in the same jar is white? Well, to begin with there are six equally probable outcomes (the six pearls). Three of these involve you picking a white pearl. Two of these involve you picking a white pearl from the jar with two white pearls. Therefore the probability of this happening, given that you picked a white pearl, is two out of three.

[LCR]

It doesn't increase anything by re-guessing. After the first door with the raptors is opened, your chances have increased to 50%. Changing will keep it at 50%. No improvement.

It's like flipping a coin twice. You flip a coin, it lands on heads. Now you flip another, what are the chances of it being tails? Would it not improve since you flipped heads last time?

No, it would not. It would still be 50%.

The probability of guessing right is equal to the number of right choices available, divided by the number of choices, time 100%.

1/2 x 100%= 50%

That's how it works.

I can't present you any numbers, but you are wrong. This is one of the few things I remembered from high school stochastic, because it stroke me as weird.

[KoDT69]

Actually, it's right. I can sort of put together an intuitive proof for you to follow along, if you like.

There are only three configurations the doors can be in:
( = Quesadillas n' Root Beer
= Raptors)

You can have
1__2__3



and no others, right?
Let's say you pick door #3, initially. The host reveals that Door #1 has a raptor behind it. Now, this tells you that there are two possible configurations, and it's this that gives the impression that you have 50/50 chances. Two options, so an equal chance, right?

Well, not quite. If you pick a door at random at the beginning, you have a 2/3 chance of picking a raptor, right? If you select a raptor, then the host must open the door revealing the last raptor, meaning the other door has the quesadillas and root beer behind it. If you switch doors in these two cases, you get the food!

If, however, you pick the correct door right off (a 1/3 chance), then the host reveals a raptor, but the other one is still hidden. If you switch, you will get it and dies.

Make sense?

This only works when you factor in the starting number into your statistics. The problem is that either door now has a 66% chance to be the winner in that respect. Sure say all you want "at the beginning you picked 1 of 3 doors so that's 33%" but door #3 gets eliminated and now you're down to 2/3 doors. It doesn't matter now because the factor of the problem has decremented from 3 to 2 giving a 50/50 chance. But if you count the eliminated door #3, then each has a 66% chance in that sense, but that's not realistic. Why is an eliminated door still being calculated? The problem has changed. Here look at this from Skenardo's example:

33% chance




By door #3 being eliminated, it removes the 3rd row down AND the 3rd column to the right leaving a 50% chance regardless if you switch:



If you still leave in a configuration other than that, it's not realistic. Configuration 3 was clearly not right altogether being eliminated, as was door number 3 eliminated from the first 2 configurations. Can anybody really justify the chance to pick a door that's already been revealed a loser?

[Azrael]

This is the famous "Monty Haul" problem.

No, the Monty Haul problem involves tearing through dungeons and not having sufficient carrying capacity for all the loot, raising the issue of which items to leave behind before you've properly identified them.

This is the Monty Hall Problem.

[KoDT69]

Ahhh...
Riddle:
You have three vases: one vase containing two white pearls, one vase containing one white and one black pearl, and one vase containing two black pearls. From one of these vases, a pearl is taken. This pearl turns out to be white. What is the probability that the other pearl in the same vase is also white?

The answer:There are three pearls that can be the white pearl that was taken from the chosen vase:

* The first pearl from the vase with two white pearls: in this case, the other pearl is also white.
* The second pearl from the vase with two white pearls: in this case, the other pearl is also white.
* The white pearl from the vase with one white and one black pearl: in this case, the other pearl is black.

The probability that the other pearl in the same vase is also white, is therefore 2/3.

This quirky answer is also incorrect for a good reason. Now that vase #3 is gone, so are 2 of the pearls. That leaves 4 total in which one white has been pulled so there are 2 more white and a black pearl between the 2 remaining vases. The chance would be 2/3 ONLY IF you chose another pearl from either vase. If you chose from the same vase you have a 50% chance to get the black one. Remember, you are pulling the second pearl from the SAME vase, which also eliminates another vase with its 2 pearls.

[The Great Skenardo]

@KODT
I think it's a still a relevant part of the problem, because there's a 2/3 chance (taken altogether) that the door revealed will be the other raptor door, meaning that the unopened door has the behind it.

You don't have any way of knowing what will be behind the door you choose at the outset. If you only knew that each door may have either or behind it, without knowing how many of each exist, then yes, the problem comes down to a 50/50 chance. But since you know that there are two raptor doors and only one door, then the revelation of one raptor door has a very large impact on the problem. It means that you've either chosen the raptor door to begin with, or that you've chosen the reward door.

But the chances of these two events happening is not equal. You're twice as likely to pick a raptor door at the outset as you are to pick the reward door. Therefore, your best chance comes from switching doors.

No, the Monty Haul problem involves tearing through dungeons and not having sufficient carrying capacity for all the loot, raising the issue of which items to leave behind before you've properly identified them.

This is the Monty Hall Problem.

Indeed. I exist corrected.

[Attilargh]

You have a one in three chance to pick the right one from the get-go. Then one of the wrong doors is opened.

Now, you've either picked the right door and should not switch, OR picked the wrong door #1 and should switch, OR picked the wrong door #2 and should switch.

Yes, you either have the wrong door or you don't, but you're more likely to have the wrong door to begin with.

[Ikkitosen]

I'll just add my voice the the "yet another way to explain why you should swap doors" thing...

You pick one door from three. There is a 1/3 chance you have the right door. So there is a 2/3 chance that the correct door is in the other doors, which we'll group into a set of 2.

Now, if the quizmaster offered you your door or both of the other 2 - and all you have to do is find the prize, not avoid any "raptors" - then you'd take the other 2 since there's a 2/3 chance the prize is in there. And with the quizmaster automatically eliminating one wrong door from the set of 2, they effectively become one choice with a 2/3 chance of being correct.

The acid test: Do it. Get 3 pieces of paper and mark one on the back. Do the quiz; pick one, have a friend look and eliminate one blank one from the other 2 and then look to see where the winner is. Within 10 tries you'll notice a 2/3 to 1/3 win ration of swap to don't swap. I have done it, I know

[KoDT69]

But it still comes down to a 50/50 chance because door #3 is a loser, so if you pick #1 to start with it HAD a 33% chance before door #3 was eliminated. If you count that eliminated door your win chance becomes 66% as well. All doors have an equal chance to be a winner. Why would it matter if you changed the choice, there are only 2 doors left. I just really don't understand why the 3rd door is calculated with a changed door choice, but not the option to stay?

Let's scale it up double to make a dice reference. Instead of 1-2-3 let's make it 1-2, 3-4, and 5-6 and start with a d6 dice. You roll and get a 1 or 2. The game host now says since you rolled low, he will decrease the dice size to a d4. You have to roll it again. He gives you the choice of rolling for a target of 1-2 or 3-4 on the d4. You have a 50% chance either way regardless if you pick the 1-2 that you got on the first round.

The real issue is that regardless of what you picked to start, the division factor changes the equation when one is eliminated. Your 33% increases to 50%. If you cut a pie into 3 slices and Skenardo eats slice #3, you indeed have no chance of getting the other 2 slices when you only chose 1 of the 3 to begin with.

[Ikkitosen]

But it still comes down to a 50/50 chance because door #3 is a loser, so if you pick #1 to start with it HAD a 33% chance before door #3 was eliminated. If you count that eliminated door your win chance becomes 66% as well. All doors have an equal chance to be a winner. Why would it matter if you changed the choice, there are only 2 doors left. I just really don't understand why the 3rd door is calculated with a changed door choice, but not the option to stay?

Let's scale it up double to make a dice reference. Instead of 1-2-3 let's make it 1-2, 3-4, and 5-6 and start with a d6 dice. You roll and get a 1 or 2. The game host now says since you rolled low, he will decrease the dice size to a d4. You have to roll it again. He gives you the choice of rolling for a target of 1-2 or 3-4 on the d4. You have a 50% chance either way regardless if you pick the 1-2 that you got on the first round.

Sorry mate, not the same problem at all. I have met people before that couldn't or wouldn't believe/understand the answer of 1/3:2/3, so I suggest you go and try it, accept that it is true, and then try to understand it without the niggling worry that it's the wrong answer.

EDIT2: Note that the chance would be 50:50 if the quizmaster took away one random box from teh two you didn't choose, but since they take away a guaranteed loser you have the situation where they're effectively saying "if the winner is in these 2, I guarantee you'll get it if you swap since I'll remove the losing one", and that's where the 2/3 chance comes from.