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The other of my two favorite brain teasers is the "wine in the water" problem. What makes this such a goodie, of course, is that the correct answer is very counter-intuitive; and also that there is a solution so much simpler, more elegant and more general than the one the puzzlers themselves give.
I've only ever seen one person get this right on his first shot, and that's only because he was afraid to give the "obvious" answer and went off with paper and pencil before committing himself.
The problem is this: There's a bucket of wine and a bucket of water and you transfer a cup of wine to the water bucket, and then a cup of the mixture back to the wine bucket. Is there more wine in the water or water in the wine?
Since you're web surfing and can't possibly have a moment to stop and actually think about something like this, and since you'd be on your guard about giving the obvious, intuitive answer, anyhow, I will state it for you. "Well, since you're transferring pure wine to the water, and just a mixture of the two back to the wine bucket, there has to be more wine in the water, right?"
Let's not be too hasty. Here's the operation in pictures.
1. Start with a bucket of wine, a bucket of water and an empty cup. Wine is depicted by "o"; water by ".".
Bucket of wine cup Bucket of water ! ! ! ! !oooooooooooo! !............! !oooooooooooo! | |-\ !............! !oooooooooooo! | |_/ !............! !oooooooooooo! | | !............! !oooooooooooo! ====== !............! !oooooooooooo! !............! !------------! !------------!
2. Scoop out a cup of wine.
! ! ! ! ! ! !............! !oooooooooooo! |oooo|-\ !............! !oooooooooooo! --> |oooo|_/ --> !............! !oooooooooooo! |oooo| !............! !oooooooooooo! ====== !............! !oooooooooooo! !............! !------------! !------------!
3. Dump it into the water bucket.
! ! !.oo.........! ! ! !.......o.o..! !oooooooooooo! | |-\ !....o.......! !oooooooooooo! | |_/ !.......oo...! !oooooooooooo! | | !..o.........! !oooooooooooo! ====== !...o.....o..! !oooooooooooo! !o......o....! !------------! !------------!
4. Scoop out a cup of the mixture.
! ! ! ! ! ! !........o..o! !oooooooooooo! |...o|-\ !.....o......! !oooooooooooo! <-- |o...|_/ <-- !......o.....! !oooooooooooo! |oo..| !..o.........! !oooooooooooo! ====== !...o......o.! !oooooooooooo! !.......o....! !------------! !------------!
5. Dump it into the wine bucket.
! ! ! ! !ooooo.oooooo! !...........o! !oo.ooooo.ooo! | |-\ !....oo......! !oooooooooo.o! | |_/ !............! !ooo.oo.ooooo! | | !.o....o.....! !oo.ooooooooo! ====== !...o........! !ooooo.oooooo! !.......o..o.! !------------! !------------!
Those who are inclined to solve this problem algebraically always seem to choose some special case. For instance, they may say the volume of the cup is 1 cup, and the volume of each bucket is 10 cups. Clunking our way through this (and you don't really need to pay attention here), we see that after the first transfer the proportion of wine to total mixture in the water bucket is 1/11. Thus, on the return trip, the proportion of water to total mixture in the cup is 10/11. After dumping it in the wine bucket we get back to the original volume of 10 cups. The proportion of water to total mixture is
volume of water in wine bucket 10/11 1 -------------------------------------- = ----- = -- volume of total mixture in wine bucket 10 11
Well, what do you know. This is the same proportion as wine to total mixture in the water bucket. Since both buckets are back to the same, original volumes, the amount of wine in the water bucket is exactly the as the amount of water in the wine bucket.
That wasn't so bad, but it's not too satisfying solving it for one special case. Try doing it for the general case, using variables for all of the volumes involved, and it gets real hairy real fast! (See Appendix at bottom.)
But let us throw all of this algebra in the trash can. Step aside. Here's how to solve it with a VERY SIMPLE VISUALIZATION - no numbers or variables in sight.
Take another look at the cup on its return trip to the wine bucket.
|...o|-\ |o...|_/ |oo..| ======
It has a mixture of wine and water. Imagine that the wine and water separate out. It doesn't need to in reality, but we force it to in our minds. (This is what Einstein called a Gedankenexperiment.) Here we show the wine settled on the bottom.
|....|-\ |....|_/ |oooo| ======
Now look. And think. That cup was full of wine on its first trip to the water bucket. Now it is partially full of wine on the return trip. The amount of water in the cup has to be exactly equal to the amount of wine left behind in the water bucket. After dumping it in the wine bucket, there will be exactly the same amount of water in the wine as wine in the water.
Pretty neat, eh? But this is isn't the end of the problem; it's where the fun begins. Armed with this understanding, we see that the answer is independent of almost everything. It doesn't matter how big the wine and water buckets are. It doesn't matter whether or not the buckets are the same size. It doesn't matter how filled the buckets are (although there must enough wine in the wine bucket to completely fill the cup.)
It also doesn't matter how well the wine is mixed into the water in the water bucket! It's fun to think about the extreme cases. Suppose there is no mixing and you scoop out pure water to transfer back to the wine bucket. Ok, there will be 1 cup of water in the wine and 1 cup of wine in the water. Check.
Or suppose that you somehow manage to scoop the cup's worth of wine back out of the water and transfer it back to the wine bucket. There would be exactly no water in the wine, and no wine in the water. Roger.
Think of what a nightmare the general problem would be to solve algebraically. You would need variables for the volumes of the 3 containers. How one could work in partial mixing, I have no idea.
I thank my college friend Ronald Takemoto who, upon hearing the "wine in the water" problem read aloud, instantly saw the Elegant Solution (No. 1) in his mind. Durn, I wish it'd been me. I've carried it around with me for the last few decades and have had some fun with it, but now I have to confess, to my slight embarrassment, there's an even better way of thinking about it. You know, the web really connects you with a bunch of really smart people.
This is so simple and elegant that, when it clicks, you can't imagine why anybody would bother asking such a trivially easy brain teaser.
Chris Reiss was first. He had this to say: "The problem really doesn't require any visualization. Each bucket ends at the same level it started. Liquid is conserved, so wine and water must have been equally traded."
Later (Nov 2002), Brian Hinton wrote: "The easiest way to explain this problem is, that if the volumes of liquid return to the exact same amounts, then, after the transfer of the wine and water, the wine that was transfered to the water has to be the same amount of water that was transfered to the wine. Or you could think about it this way: the amount of wine transfered to the water has to be the same amount of water transfered to the wine in order to keep both volumes of water the same in the end."
Since it's my web page, let me state it a third time in my own words: After both transfers, imagine all the water in the wine bucket floats together in a "blob". Well, that blob of water has to have replaced an identical blob of wine that is now in the water bucket.
This is the same thought process we used in Elegant Solution (No. 1), except now we look at the "water blob" after being dumped in the wine, rather than just before. I guess I didn't think of it because Elegant Solution (No. 1) seemed so darn good. The brain teaser turns out to be even better than I thought - the transfer cup itself is a red herring!
There could be any number of transfers back and forth with any number of cups of different sizes, or for that matter, a connecting hose pumping the mixtures back and forth, but as long as the contents of the buckets are returned to their original levels, the final "water blob" in the wine bucket has to have "forced", so to speak, an identical "wine blob" over to the water bucket.
Another point comes to mind here. Notice that when you are aware of how independent the answer is of everything - bucket shapes and sizes, how filled they are to start with, the number and manner of transfers back and forth, and thoroughness of mixing - you might be inclined to say this in the statement of the problem, or let it slip when people press you for more information.
If you do this, you've changed the problem! If a person is told that the answer is independent of bucket sizes, solving the problem then becomes a matter of choosing a convenient special case - even one involving vessels that would have little in common with what we think of as buckets and cups. A simple special case would be starting with 1-cup "buckets' worth" of both wine and water. After transferring, thorough mixing and returning, there would be half wine and half water in each bucket.
Likewise, if you let it out that it doesn't matter how much mixing takes place, one could simply examine the special case of returning all water, or all wine, back to the wine bucket.
(This brings to mind another problem. Suppose you have a metal band wrapped snugly around the Earth's equator. You snip it, insert an extra 10-foot length of metal band, and then adjust the band so that it is a uniform height above the equator all the way around. How high above the equator will the band be? The answer is very astonishing. See hint below.)
And beating the water and wine problem into the ground (actually, you're all dismissed now), let's work in real-world chemistry and physics. The wine is made up of "wine molecules", so to speak, and the water of water molecules. Since the volumes of a wine and water molecule are certainly not related by a rational number ratio, it would be impossible to transfer the exact same volume of mixture back to the wine bucket.
If we are adament that the same volume is transfered both ways, the only possibility is that all the wine has to come back (in which case our answer is still correct, but not very interesting). If we do transfer a mixture back to the wine bucket, it will certainly not be the exact same volume as the original wine transfer. A fraction of a molecule will either stick up above or dip down below the rim of the cup. And then there will be a different amount of water in the wine than wine in the water.
On top of that, we have the complicating factor of the tendency of alcohol molecules to hide, to some extent, between water molecules. In other words, if you mix a cup of water and a cup of alcohol, you get slightly less than 2 cups of mixture. Pooey on molecules.
In closing, if you are inclined to try out this cool brain teaser on your friends, make sure they stew over it before you haul out our beautiful solution. The impact derives from how intuition leads to the wrong answer. "There's more wine in the water" seems obvious to almost everyone. But people are wary and have a tendency to remain close-lipped for fear of being tricked and subject to public humiliation. I say, no guesses, no solution.
Hint on the metal band problem: You don't need to know the size of the earth; the answer is independent of it. And now that I've told you that, you can solve it for a sphere of any size. Try one with a radius of 0.
Thanks to my 7th-grade math teacher Mr. Lieske (Johnnycake Junior High School) for this really neat problem about the metal band girdling the earth. The answer, by the way, is 10/(2*pi) which equals about 1.6 feet - you could crawl under it!
Afterword: Since writing the above, I came across the wine and water problem in a 1946 book, Fun With Puzzles by Joseph Leeming. It's on page 25 in the Brain Twisters section. Let's see how he handles it.
31. WINE AND WATER
There are two glasses of the same size and shape, one containing water and the other containing wine. A man takes a spoon and transfers a spoonful of water from the water glass to the wine glass. Then he takes a spoonful of the mixture from the wine glass and puts it into the water glass.
Now, did he transfer more water to the wine glass, or more wine to the water glass?
The answer (page 95) is:
31. WINE AND WATER
It is hard to believe, but equal amounts of water and wine were transferred. Assume that each glass contained 100 units of liquid and that the spoon held 10 units. The spoon first removes 10 units of water, so the water glass contains 90 units of water, and the wine glass contains 100 units of wine and 10 units of water.
With 110 units in the wine glass, the spoon will remove 1/11 of each liquid in that glass. Thus it will transfer to the water glass 9 1/11 units of wine and 10/11 units of water. The water glass will then contain 90 10/11 units of water and 9 1/11 units of wine, and the wine glass will contain 90 10/11 units of wine and 9 1/11 units of water.
You can unglaze your eyes now.
Rainbow Sky a.k.a. Mike Mann made a nice catch regarding Leeming's statement of the problem just above. He wrote (Nov 2002):
I hope you can indulge this minor observation: I was reading the wine and water problem and it occurred to me that Joseph Leeming answers a different question than the one asked in "Fun With Puzzles". He asks "Now, did he transfer more water to the wine glass, or more wine to the water glass?" Of course, a whole teaspoon of water was transferred, as opposed to the diluted teaspoon of wine. Okay, maybe I do have a grudge against misworded puzzles. *grins*
Added April 2009.
More than 10 years after putting this page up, Franklin Garson sent an algebraic solution for the general case of vessels of three different sizes. If you follow it through you will see that it has a certain beauty of its own, although I doubt many people could see their way through it in their heads in a party situation!
H = volume of water (H2O) bucket
G = volume of wine (Grape juice) bucket
C = volume of Cup
Let us work in milliliters (ml). It could be any unit of volume, of course, but choosing one puts us on more familiar ground than the fuzzy, generic term "units".
Add C ml wine to the water.
This gives total mixture = (H+C) ml
with proportion of water = H/(H+C)
and proportion of wine = C/(H+C)
Mix thoroughly, then remove C ml of mixture:
Cup contains C * H/(H+C) ml water [Expression 1]
and C * C/(H+C) ml wine.
We are left with C - (C*C/(H+C)) ml of wine in the water bucket.
Manipulating this expression:
C - (C*C/(H+C))
= (C(H+C) - C*C)/(H+C)
= (C*H + C*C - C*C)/(H+C)
= C * H/(H+C) ml of wine in the water [Expression 2]
Stopping to compare Expression 2 with Expression 1 we see that the wine that is left in the the water bucket is the same as the water in the cup being transferred to the wine bucket. Voila!
Notice how the size of the wine bucket (G) never entered into the analysis. I'm guessing that decades ago, when I attempted an algebraic solution, I got to this point and did not notice I was finished, but tried to plow on, algebraically mixing the return cup into the wine bucket and from there trying to determine how much water is in the wine bucket.
Thanks again, Franklin!
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