Fun trivia question:

I have 6 bottles of wine for which I know three are good and three are corked. I put them into three 2-bottle lots. In the first lot, there are two corked bottles. In the second lot, there is a corked bottle and a good bottle. In the third lot, there are two good bottles. Suppose before you decide to buy, I let you pick a lot and sample a single bottle from the chosen lot. The bottle you sampled is good.
Original Post
quote:
Originally posted by Berno:
quote:
Originally posted by Rock&RollCowboy:
50/50

Wrong.


It's not wrong, missionary. You sampled a good bottle, which means it came from the lot with one or two good bottles. The first bottle being good is a given. In 1/2 of the two remaining lots, the other bottle will be good. If the question were what is the probability that two bottles are good with no given, the answer is 1 out of 3. This is elementary school stuff.
The probability is 2/3.

Disregard the set with 2 corked bottles. Only 4 bottles are left in play. 3 are good, 1 is corked.

You just opened one of the good ones. Now only 3 are left in play. That means the next one you open will be one of the 2 remaining good ones, or the 1 bad one.

Cool
Instinctively, the answer seems like it should be 1/2 so I understand why people chose it. But the answer is 2/3.

There are 3 distinct sets but 6 distinct bottles. And you are twice as likely to pick a good bottle from the good set as you are to pick a good bottle from the mixed set.

Or to put it another way, there are only 3 possibilities now-

1) you picked the good bottle from the mixed set
2) you picked good bottle sub 1 from the good set
3) you picked good bottle sub 2 from the good set.

In two of those three possibilities, you are in the good set. So the answer is 2/3.
The corked set is out of play. You know that because it's a given.

Two sets left, one mixed, one not. You now have a 50/50 chance of having picked a good set or a bad set.

Knowing one of the bottles you opened was good, and knowing it was more likely that you'd pick a good bottle to open doesn't change the fact that it is still a 50/50 proposition that you chose the good or the mixed set. All it does is take the fully corked set out of play.

You've already made the choice. How many good bottles are left doesn't matter.
The answer is 2/3 by the logic Vinyrd Skynyrd and winetarelli provided.

Alternative explanation:
There are only three good bottles, call them bottles A, B, and C. Lot three contains bottles A and B (the good lot). You just tasted a good bottle which means it is A, B, or C. What's the probability that the other bottle is good is exactly equal to what is the probability that the bottle you tasted is A or B given you've tasted either A, B, or C. That, is clearly 2/3.

Edited to provide credit for winetarelli as well. I just read your post. Nice explanation. Smile
quote:
and knowing it was more likely that you'd pick a good bottle to open [from one set] doesn't change the fact that it is still a 50/50 proposition that you chose the good or the mixed set.


No no no no...

Use an extreme example to understand.

Suppose you are dealing with two sets (the third set is just a mind-f**k set, this is a 2 set problem).

1 set has 1,000 good bottles in it. 1 set has 999 bad bottles and one good bottle. You taste 1 bottle from one set and it is good. Which set are you more likely to be tastng from?
quote:
Originally posted by WEc:
The answer is 2/3 by the logic Vinyrd Skynyrd provided.

Alternative explanation:
There are only three good bottles, call them bottles A, B, and C. Lot three contains bottles A and B (the good lot). You just tasted a good bottle which means it is A, B, or C. What's the probability that the other bottle is good is exactly equal to what is the probability that the bottle you tasted is A or B given you've tasted either A, B, or C. That, is clearly 2/3.


Ahem. Look two posts ahead of you. Ahem Wink

"Or to put it another way, there are only 3 possibilities now-

1) you picked the good bottle from the mixed set
2) you picked good bottle sub 1 from the good set
3) you picked good bottle sub 2 from the good set.

In two of those three possibilities, you are in the good set. So the answer is 2/3."

Big Grin
A,B,C = Good
D,E,F = Bad

Subset A,B = Good
Subset C,D = Mixed
Subset E,F = Bad

There is a 1/3 chance you will pick the corked set. You haven’t. You can eliminate Subset EF

There is a 50% chance you have opened one of the following two sets. We can agree on that right? How is there a higher probability that you chose the good set than the mixed set? There isn’t. You had a 2/3 chance of picking a set with at least one good bottle.
Subset A,B
Subset C,D

You have opened a good bottle A,B, or C. This is given. You have no way of knowing which subset you have opened. If picked the good subset there is a 100% probability the other bottle is good.

Subset A ,B
Subset C,D

Subset A, B
Subset C,D

If you picked the mixed subset there is a 100% probability the other bottle is corked.
Subset A,B
Subset C ,D

If the bottles were not grouped in this manner, I’d agree with the 2/3 possibility however the bottles are grouped and linked. Rearrange them as (A,B,C,D). The fact that you have a higher chance of opening a good bottle first doesn’t change what’s left. The question is what are the odds the remaining bottle is good.

That's my literal interpretation of the question exactly as posed.
quote:
Originally posted by winetarelli:
quote:
and knowing it was more likely that you'd pick a good bottle to open [from one set] doesn't change the fact that it is still a 50/50 proposition that you chose the good or the mixed set.


No no no no...

Use an extreme example to understand.

Suppose you are dealing with two sets (the third set is just a mind-f**k set, this is a 2 set problem).

1 set has 1,000 good bottles in it. 1 set has 999 bad bottles and one good bottle. You taste 1 bottle from one set and it is good. Which set are you more likely to be tastng from?
That's not the question posed. Given you've opened a good bottle, the odds in your example are still 50/50. The question isn't which set did you probably open, it's what are the odds in what's left over. If you have two leftover sets, one completely good and one bad, it's still 50/50. The original given good bottle isn't relevant to the question.
quote:
Originally posted by Gigond Ass:
quote:
Originally posted by winetarelli:
quote:
and knowing it was more likely that you'd pick a good bottle to open [from one set] doesn't change the fact that it is still a 50/50 proposition that you chose the good or the mixed set.


No no no no...

Use an extreme example to understand.

Suppose you are dealing with two sets (the third set is just a mind-f**k set, this is a 2 set problem).

1 set has 1,000 good bottles in it. 1 set has 999 bad bottles and one good bottle. You taste 1 bottle from one set and it is good. Which set are you more likely to be tastng from?
That's not the question posed. Given you've opened a good bottle, the odds in your example are still 50/50. The question isn't which set did you probably open, it's what are the odds in what's left over. If you have two leftover sets, one completely good and one bad, it's still 50/50. The original given good bottle isn't relevant to the question.


No. That IS the question asked... but in a much less obvious form. "Which set are you more likely to be tasting from?" That is the question. If you accept that it is twice as likely that you are tasting from the good set, then you accpet that it is twice as likey that the other bottle is good. These ideas are the SAME.
A. Probability is that I will not buy at all if you tell me you're trying to sell me corked bottles among good bottles. Razz Big Grin


Other than that I chose 2/3 because 2 of remaining 3 bottles in play are good...
But, because I've chosen of the set where one bottle is at least good, really only 2 bottles remain in play...odds are 1/2 or 50/50 that the last bottle of my set is good...
(starting to sound like Wallace Shawn as Vizzini in Princess Bride...)

I need more coffee...
quote:
Originally posted by winetarelli:
quote:
Originally posted by Gigond Ass:
quote:
Originally posted by winetarelli:
quote:
and knowing it was more likely that you'd pick a good bottle to open [from one set] doesn't change the fact that it is still a 50/50 proposition that you chose the good or the mixed set.


No no no no...

Use an extreme example to understand.

Suppose you are dealing with two sets (the third set is just a mind-f**k set, this is a 2 set problem).

1 set has 1,000 good bottles in it. 1 set has 999 bad bottles and one good bottle. You taste 1 bottle from one set and it is good. Which set are you more likely to be tastng from?
That's not the question posed. Given you've opened a good bottle, the odds in your example are still 50/50. The question isn't which set did you probably open, it's what are the odds in what's left over. If you have two leftover sets, one completely good and one bad, it's still 50/50. The original given good bottle isn't relevant to the question.


No. That IS the question asked... but in a much less obvious form. "Which set are you more likely to be tasting from?" That is the question. If you accept that it is twice as likely that you are tasting from the good set, then you accpet that it is twice as likey that the other bottle is good. These ideas are the SAME.
What is the probability that the other bottle in your chosen lot is also good? That is the question, not your interpretation.

The question isn't "What are the odds you CHOSE a set with two good bottles."

Is it that difficult for you to understand?
quote:
Originally posted by Gigond Ass:
quote:
Originally posted by winetarelli:
quote:
Originally posted by Gigond Ass:
quote:
Originally posted by winetarelli:
quote:
and knowing it was more likely that you'd pick a good bottle to open [from one set] doesn't change the fact that it is still a 50/50 proposition that you chose the good or the mixed set.


No no no no...

Use an extreme example to understand.

Suppose you are dealing with two sets (the third set is just a mind-f**k set, this is a 2 set problem).

1 set has 1,000 good bottles in it. 1 set has 999 bad bottles and one good bottle. You taste 1 bottle from one set and it is good. Which set are you more likely to be tastng from?
That's not the question posed. Given you've opened a good bottle, the odds in your example are still 50/50. The question isn't which set did you probably open, it's what are the odds in what's left over. If you have two leftover sets, one completely good and one bad, it's still 50/50. The original given good bottle isn't relevant to the question.


No. That IS the question asked... but in a much less obvious form. "Which set are you more likely to be tasting from?" That is the question. If you accept that it is twice as likely that you are tasting from the good set, then you accpet that it is twice as likey that the other bottle is good. These ideas are the SAME.
What is the probability that the other bottle in your chosen lot is also good? That is the question, not your interpretation.

The question isn't "What are the odds you CHOSE a set with two good bottles."

Is it that difficult for you to understand?


Yes. That is incredibly difficult for me to understand. Let me try to explain why...

IF I chose the set with two good bottles, then the other bottle is also good. You and I are in agreement that there is a 2/3 chance I chose the set with 2 good bottles.

IF I chose the set with one good bottle, then the other bottle is bad. You and I are in agreement that there is a 1/3 chance this is the set I chose.

SO...

2 out of 3 times, you and I would agree I chose the set with 2 good bottles.

SO...

2 out of 3 times, the second bottle is good.
If that is the question asked, I'd agree.

However that isn't the way the question is phrased. If that was the intent of the question you would be correct.

Take the question literally as asked and you are dead wrong. It's pretty simple.

You will have to excuse me but literal interpretation is something I do for a living. Good luck in court with "I know what it says, but this is what it means".
quote:
What is the probability that the other bottle in your chosen lot is also good? That is the question, not your interpretation.

The question isn't "What are the odds you CHOSE a set with two good bottles."

Is it that difficult for you to understand?


Perfect explanation - and I agree as it is asked, the answer is 1/2
I will try this one more time, and then stop. And by that I mean, I won't look at this for a while, and then I will re-post.

"What is the probability that the other bottle in your chosen lot is also good?" = "what is the probability that you chose the lot with 2 good bottles"

Because... there is a 100% chance if you chose the lot with 2 good bottles, the other bottle is good. And there is a 0% chance if you chose the lot with 1 good bottle, the other bottle is good.
quote:
Originally posted by winetarelli:
I will try this one more time, and then stop. And by that I mean, I won't look at this for a while, and then I will re-post.

"What is the probability that the other bottle in your chosen lot is also good?" = "what is the probability that you chose the lot with 2 good bottles"

Because... there is a 100% chance if you chose the lot with 2 good bottles, the other bottle is good. And there is a 0% chance if you chose the lot with 1 good bottle, the other bottle is good.
You are extrapolating a question that isn't asked. The 2/3 crowd is not understanding that the question as phrased is really a 1/2 probability.
quote:
Originally posted by Ozarks21:
I think we all agree that the other two corked bottles don't need to be considered at all. Of the three remaining bottles, two are good. 2/3.


What everybody doesn't seem to agree with is that one of the two remaining good bottles also doesn't need to be considered either...since it has to be in the other set, therefore only 2 bottles are in play - (1) that is good and (1) that is corked.
Exactly Chile!

I think the 2/3 crowd is either reading the question incorrectly as GA noted or forgetting the very important fact that each set has to be considered independently of the other.

It's irrelevant that there are 2 good bottles left and 1 corked bottle left. The problem with that thinking is that it combines the sets. You can't do that.

All that matters is what is in the "other bottle" of the chosen set -- it's either corked or it's good.

The answer is 1/2.
I hate to say this but there is only one answer and it is 2/3. Bottles A B and C are good. D E and F are bad.
Good Lot = A and B
Mixed Lot = C and D
Bad Lot = E and F
You've picked a lot and consequently a bottle to sample from it and it was good. This necessarily implies that you just tasted either bottle A, B, or C. The Question was:
What is the probability that the other bottle in your chosen lot is also good?

If the bottle you tasted was A, then the answer is 1. If the bottle you tasted was B, then the answer is 1. If the bottle you tasted was C, then the answer is 0. There is a 2/3 chance that you got the good lot given you already tasted a good bottle.

The 1/2 answer does not give credit to the possibility that the good lot has 2 possible configurations even given you've tasted a good bottle. Namely, it could have been (A and B) or (B and A). (Think of the first bottle as the bottle you tasted.) Whereas, the configuration of the mixed lot can only be (C and D) given the bottle you tasted was a good bottle.

This problem is exactly the same as the following scenario. Suppose I have two kids and I tell you at least one of them is a boy. What is the probability that the other is a boy? (Assume it is equally likely to have boys or girls.)

Answer: 1/3 not 1/2. Because there are four possible configurations, (BB, BG, GB, GG). The information I told you just stipulates that it can't be GG. So of the three remaining possibilities, only one will result in the other being a boy. The 1/2 answer is correct if you assumed the first of my two kids was a boy in which case the comparison is only between BB and BG. The order matters!
WEc,

I understand your reasoning, but I still disagree.

Let's use your A,B,C,D,E,F wines.

You are answering the question of what is the probability that a *specific bottle of wine* left in the set is good.

In other words, you are asnwering the question of what specific bottle of wine of the 3 possibilities (A, B, or D)is left in the set, not whether its good or bad.

I am answering the question of the probability that the bottle left is good or bad.

The question says that I got to choose a bottle. I picked a good bottle.

Let's physically remove that wine from the set.

I am looking at one other bottle of wine in my set. Either it's (1) A or B (it can't be both) or (2) it's D.


I think your boy/girl analogy fails for the same reason. You configure BG and GB as different answers. That configuration is completely irrelevant to the question of whether the other child is a boy or a girl. In other words, it adds a "false choice" to the problem (configuration).

Similarly, your wine example adds a false choice to the problem, i.e., specificity. The actual bottle, A, B, or D is irrelevant to whether the other wine is good or bad.
quote:
Originally posted by Instant Access:
The question says that I got to choose a bottle. I picked a good bottle.

Let's physically remove that wine from the set.

I am looking at one other bottle of wine in my set. Either it's (1) A or B (it can't be both) or (2) it's D.


That's precisely where the problem lies, you don't know if it is A or B if it came from the good set where as you KNOW it is C if it came from the bad set.

As for the B/G question think of this way, I have a one year old and a two year old. They are two distinct individuals. Let XY denote the pair of sexes for the two kids where X is the sex of the one year old and Y is the sex of the two year old. There are 4 possibilities (BB, BG, GB, GG). You cannot argue that BG and GB are the same because in the former that means my one year old is a boy and the latter the one year old is a girl. (I don't have kids but if I did I wouldn't want ones that can change sex! Eek) So when I tell you that at least one of them is a boy, you cannot disregard that all you know is (BB, BG, GB) are the possibilities. Again, BG is not equal to GB.

This is the same with the wine example. Let's say that XY is the lot you chose and X is the bottle you sampled. Now let's literally write the letters ABCDEF on the bottles and then brown bag them. Now you taste a good wine, this means you have tasted either A, B, or C. Now the problem is, if you want the answer to be 1/2 you are thinking the following. If it is A or B I am tasting, I will remove the brown bag. If it is C I won't. So without removing the brown bag (and we don't in the original problem), you could be holding AB, BA, or CD. I agree that AB is the same as BA, but only if you knew whether you tasted A or B (which is not what we assumed since the brown bags are still on).
quote:
Originally posted by Instant Access:
Let's physically remove that wine from the set.


Actually, a even simpler way of thinking about it is this way.

Would you agree that if you knew you tasted A then the other bottle is good? - Yes
Would you agree that if you knew you tasted B then the other bottle is good? - Yes
Would you agree that if you knew you tasted C then the other bottle is bad? - Yes

Which bottle did you just remove?
quote:
Originally posted by WEc:
quote:
Originally posted by Instant Access:
Let's physically remove that wine from the set.


Actually, a even simpler way of thinking about it is this way.

Would you agree that if you knew you tasted A then the other bottle is good? - Yes
Would you agree that if you knew you tasted B then the other bottle is good? - Yes
Would you agree that if you knew you tasted C then the other bottle is bad? - Yes

Which bottle did you just remove?


No from your description of the problem that's an incorrect assumption.

- 3 sets of bottles
- we only have to look at 2 of the sets because you said the bottle we picked up tasted good.
- question was what was the probability the other bottle is good.

from those assumptions, what are the chances out of 2 lots that one is good and one is bad.

answer is 1/2

your answer is right if you still had to consider the first lot where both bottles were bad...

It doesn't matter the order of the bottles because you said there were only two bottles in a lot and that one you pulled out was ALREADY good.
quote:
Originally posted by WEc:
quote:
Originally posted by Instant Access:
Let's physically remove that wine from the set.


Actually, a even simpler way of thinking about it is this way.

Would you agree that if you knew you tasted A then the other bottle is good? - Yes
Would you agree that if you knew you tasted B then the other bottle is good? - Yes
Would you agree that if you knew you tasted C then the other bottle is bad? - Yes

Which bottle did you just remove?


I just removed "AB" or "C". Thus, I know that the other bottle is either "BA" (good) or "D" (bad).

If I only care about the probability of the other bottle being good, there is no need to draw a false distinction between A and B as they are exactly the same.

Perhaps we can look at it this way. We have 3 "G"s (good) and one "B" (bad) in the 2 sets.

I choose a "G" from one set. I know the other bottle has to be either the other "G" from the "all good" set or the "B" from the "mixed set."
quote:
Originally posted by GlennK:
The way I first read it without reading everyone else’s responses was…..You have tasted 1 (good bottle), you have 3 left to choose from and you know 2 are good. The “probability” of picking a good bottle is 2 out of 3.


Glenn,

I read it as...you tasted a good one. What is the probability of the one sitting next to it being good?

Either it is good b/c you picked the "all good" set, or it is corked b/c you picked the "mixed" set.

Thus, what I don't understand is why it matters whether I picked "A" or "B".
Ok ok ok ok....

IF you really want to see for yourself, and you definitely will, do the following (I'm serious, if you still don't believe the "2/3" crowd after my explanation then obviously I'm not doing a good job explaining and you owe it to yourself to do this. In fact, you will/can ONLY understand if you do this).

Take a piece of paper and divide it into three equal sizes/shapes. On the first piece, write G on one side and G on the other. On the second piece, write G on one side and B on the other. On the third piece write B on both sides. (G means good, B means bad).

Now shuffle the pieces of paper under your desk and pull one out, if it shows G ask yourself, what is the probability that you are holding the GG paper? If it shows B ask yourself, what is the probability that you are holding the BB paper? If you do this enough times, you will figure it out.
wow, didnt expect this to get this crazy...

quote:
Originally posted by WEc:
Ok ok ok ok....

IF you really want to see for yourself, and you definitely will, do the following (I'm serious, if you still don't believe the "2/3" crowd after my explanation then obviously I'm not doing a good job explaining and you owe it to yourself to do this. In fact, you will/can ONLY understand if you do this).

Take a piece of paper and divide it into three equal sizes/shapes. On the first piece, write G on one side and G on the other. On the second piece, write G on one side and B on the other. On the third piece write B on both sides. (G means good, B means bad).

Now shuffle the pieces of paper under your desk and pull one out, if it shows G ask yourself, what is the probability that you are holding the GG paper? If it shows B ask yourself, what is the probability that you are holding the BB paper? If you do this enough times, you will figure it out.


WEc,

problem with the above scenario is that you already eliminated the B, B paper by the 1st bottle being good... do your paper trick with only the G, G paper & the G, B paper with the G's side up, now what's the probability the one you pick is a G on the other side, 50/50, 1 out of 2...

or

Lot 1 = G, G
Lot 2 = G, B
Lot 3 = B, B

Lot 3 eliminated by proof of 1st bottle being good (G)

Remaining possibilities

Lot 1 = G, G
Lot 2 = G, B

What is the probability that the remaining bottle in YOUR lot is good as well… Either it is good, or it is bad, 1 out of 2
WEc,

My head is hurting and my productivity at work is hurting today. Smile

Using your example... if it shows a G, isn't the proper question, "What is the probability that there is a G on the other side?

If that is the proper question, either there is a G on the other side or there is a B.
That is the proper question.

"If it shows a G what is the probability that there is a G on the other side?"

And the answer you provide is right, either there is a G on the other side or there is a B, problem is they don't happen with the same probability. Eek

Hint: You are more likely to see a G with the GG piece than the GB piece.

You have to see it to believe it... Wink

Fortunately for me, this kind of thing falls into my "work".

Kumazam: Problem is which side of the GG side do I put up? That matters as much as which side of the BG piece I put up.

Now I have to go and be productive.

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