Tuesday, March 16, 2010

The Wine Problem

Those who enjoy thought provoking puzzles have probably heard of the wine problem in one version or another. It offers a particularly compelling example of a counter intuitive result.

There is a bucket of pure red wine and a bucket of pure white wine. The buckets are NOT necessarily of equal size, but they can be. One dips a cup (any size) into the pure red wine and pours it into the bucket of white wine. This is then well stirred. One then dips the same cup into the mixed wine and transfers it back to the red wine.

Now both are mixed. Is there more red wine in the "white bucket" or is there more white wine in the "red bucket"?

Intuition suggests more red is in the white because the first transfer was 100% red, while the return transfer contained a weaker mixture. The correct answer is that they are identical. After some thought I came up with what I consider a rather elegant proof using only eighth grade algebra, and it holds for any size of buckets or cup. Any one care to try? We can compare.



Anonymous Anonymous said...

You're wrong. The first cup is pure 100% red. The second cup is diluted. It is physically impossible to return as much white as the amount of red that first arrived.

3/16/2010 5:53 PM  
Anonymous Anon said...

They are the same. Take r (for red) and w (for white) as constants. a is a variable.
With r=w.
We begin with:
1) ar and aw (equal amounts of wine in both)
2) (a-1)r and aw + r (take one unit out of red wine and put into the white wine)
Now we he have a-1 units of red wine in one bucket.
In the other we have a units of white wine and 1 unit of red wine: total a+1 units

After mixing, take 1 unit out of the bucket with a+1 units and put it into the bucket with a-1 units:
3) (a-1)r + (aw+r)/(a+1) and aw+r – (aw+r)/(a+1)
Write out the algebra:
4) ((a-1)r(a+1) + (aw+r))/(a+1) and ((aw+r)(a+1)-(aw+r))/(a+1)
5) ((a^2-1)r+(aw+r))/(a+1) and ((aw+r)(a+1-1))/(a+1)
6) (a^2r-r +aw+r)/(a+1) and ((aw+r)a)/(a+1)
7) (a^2r+aw)/(a+1) and (a^2w+ar)/(a+1)

As you can see in the numerator: the amount of white wine in the red wine bucket equals the amount of red wine in the white wine bucket.

3/16/2010 6:41 PM  
Anonymous Observer said...

Well, this defies my thinking. The initial transfer, like the first anon says, is pure. It has to be true that the amount of white wine going back is less than the red that went over first. I guess somehow, like the math above says, the little bit of red that goes back with the second transfer makes up for the difference.

Something tells me x4mr's proof will be shorter.

3/16/2010 7:31 PM  
Blogger Sirocco said...

Observer, the amount of red wine going back IS less than the amount of red that comes over. The trick is that SOME amount or red comes back over, so the amount of white coming back is not a full glass ... and the amount of red left behind exactly matches the amount or white coming over.

As x4mr says, the relative sizes of the red and white buckets don't matter. However, for the sake of an example, lets posit a bucket with 100 glasses of red in it, and a bucket with 9 glasses of white in it.

Remove a full glass of red and pour it into the white. We are left with:

bucket 1: 99 glasses red
bucket 2: 9 white, 1 red

Completely stir in bucket 2. Now, when you take out a glass, it will have a ratio matching the amount of red and white in it. In our example, therefore, we would take out a glass that is .9 white and .1 red. Now pour that glass into the red bucket. That gives us:

bucket 1: .9 white, 99.1 red

Meanwhile, we originally had 1 glass of red in bucket 2. However, we just removed .1 of a glass when we made the pour back into bucket 1. We also removed .9 of a glass of white. That leaves:

bucket 2: .9 red, 8.1 white

And we have as much white in the red bucket as we have red in the white bucket.

3/16/2010 8:30 PM  
Blogger TexPatriate said...


I now need about 99 cups of white wine to accept that Sirocco is right.

Egad. Man, I hate admitting that.

I'm going to bed. *humpf*

3/16/2010 8:37 PM  
Blogger The Navigator said...

I'll leave the math to you guys, but am I the only one who thinks the red/white metaphor is political?

He's really talking red/blue, and the transfers don't really help either, because we're all in the same boat?

3/16/2010 8:38 PM  
Blogger The Navigator said...

Should have said "the red/white is a political metaphor."

3/16/2010 8:40 PM  
Blogger x4mr said...

Anon (6:41) Correct but (no offense) clunky, and you assumed r = w. The superior treatment allows for different size buckets. Sirocco's remarks are also correct, and his concrete illustration helps one understand, but it is not a proof of the general case.

Thinking further, I've realized the most elegant solution requires no math, but I'll share my initial proof anyway.

I require two variables:
V = original volume white. C = volume of cup.

After the first transfer, the white bucket has C red and red concentration C/(V+C).

After the second transfer, (vol)(conc) products yield:
White provided to red bucket = C (V/(V+C)) = CV/(V+C)
Red remaining in white bucket = V (C/(V+C)) = CV/(V+C).

The truly elegant solution which indicates mastery of the problem notes that each bucket has the same volume as before. What each lost of its original color has to be in the other bucket, replaced with what came from the other bucket! QED.

Nav, close.

3/16/2010 9:37 PM  
Blogger Sirocco said...

Heh. I had just logged in to type something similar to your last paragraph.

The point is the first pour doesn't really matter for the calculation. It's the second pour, where you end up with 1 - (amount red) of white in the red bucket, and since you had 1 full red in the white bucket you are left with 1 - (amount red) of red in the white bucket.

3/17/2010 6:55 AM  
Blogger Sirocco said...

To be clear, when I reference (amount red) in the above, it's the amount of red wine contained in the second glass, the mixed glass poured back into the red bucket.

3/17/2010 6:57 AM  
Anonymous Thomas said...

I suffered through two years of algebra with little sense of what it was good for.

Thank you Thank you!! And I mean everyone. This has caused a bunch of light bulbs to flash.

For his proof, x4mr used variable AS REQUIRED only, and the (vol)(conc) equations show exactly what is happening.

Then the most elegant, well, duh!! Sirocco sees it, too.

I get "elegant." FINALLY.


3/17/2010 8:03 AM  
Anonymous Observer said...

I don't believe this, but I am actually saying that a math problem is cool. This problem is REALLY COOL.

The more I think about it, the cooler it gets. The simple physics of it. I know Sirocco and x4mr know what I mean.

Great post, x4mr.

3/17/2010 12:28 PM  

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