Beeminder Forum

Water Into Wine [easy]

(Ok, easy is relative but this is one that doesn’t require any advanced math nor does is it just stupidly, perhaps painfully, hard, like the dog logic one. It’s still tricky and clever.)

You have a glass of wine and glass of water. You take a teaspoonful of wine and mix it into the water. Then you take a teaspoonful of the mixture and dump it in the wine.

Is there more wine in the water or more water in the wine?

(Ignore the fact that wine is already part water. This is a glass of 100% pure wine molecules.)

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Here’s the “cheating” way to solve it:

Suppose each glass has only a teaspoonful of liquid in it to start. Then you mix ALL the wine into the water. Obviously a teaspoon of this mixture will be equal parts wine and water, so after transferring it back both glasses will have equal amounts of wine and water. Since you didn’t specify how much water or wine we had to start, by the meta-rules of such puzzles the answer must be invariant under changes in the starting amounts, hence both glasses end up with equal amounts of the other substance.

And a more rigorously mathematical solution:

Suppose we have T teaspoonfuls of waTer and N of wiNe to start (it’s annoying that they both start with the letter ‘W’). After the first transfer, the glass of wine holds N-1 teaspoons and the other glass holds T teaspoons of water and 1 of wine. So one teaspoon of the second glass will contain T/(T+1) teaspoons of water and 1/(T+1) of wine. Removing it will leave 1 - 1/(T+1) = T/(T+1) teaspoons of wine in the second glass, exactly the same as the amount of water that is in the teaspoon we are about to add to the first glass.

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I love the cheating solution invoking the meta-rules of brainteasers! :grin:

But I have a twist that I think ruins both of your solutions: there’s no guarantee that the mixture is fully mixed! After dumping in the teaspoon of wine and scooping a teaspoon out, you might manage to get just water or, if you’re really talented, recover exactly the same teaspoon of wine you dumped in. Or anything in between!

Aha! I knew there must be an easier way. :smile:

When you scoop a teaspoon out of the mixture, it will contain some amount of wine and some amount of water. Since it was all wine when you dumped it in, any water in the teaspoon when you scoop it out must have displaced an exactly equal amount of wine which remains in the water. This bit of water in the teaspoon is destined to be the only water in the glass of wine; likewise, the wine it displaced is the only wine left in the glass of water.

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Do you mean by volume or by mass? If you mean by mass I think we need a lot more information (I think we need at least the molarity of the wine).

Volume! This is not a chemistry puzzle, as you might’ve guessed from my “100% pure wine molecules” :smile:

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I used Excel and I think got an applied version of Byorgey’s more rigorously mathematical solution:

Assuming 48 tsps in each glass as the starting point:
image

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Love it! That probably builds the intuition better than a rigorous proof! I think I have an even better way to make it super intuitive, possibly even for kids. I’ll post mine when this is a week old. http://dreev.commits.to/post_ww_soln_by_apr_30

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The easy algebraic solution:

Let the volume of a glass be X, and let the volume of water that ends up in the wine glass at the end of the dumping procedure be Y. Then the amount of wine that’s still in the wine glass at the end must be X - Y. (Because the volume of liquid in the glass is X - 1tsp + 1tsp (which is equal to X), and the only two liquids making up that volume are water and wine).

So, if at the end there is X - Y wine in the wine glass, and there started out being X, then an amount of wine exactly equal to Y must be somewhere else. The only place it can be is the water glass, and therefore there is exactly Y wine in the water glass, or in other words the same amount as the amount of water that’s in the wine glass.

By the way, this can be extended to any number of transfers back and forth: you can take a teaspoonful (or any other amount) of one of the liquid and put it in the other any number of times (not even necessarily the same amount each time, so long as both glasses end up with the same amount of liquid as started in them), and you’ll end up with the same result, that there will always be exactly as much water in the wine glass as wine in the water glass.

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I said I had an even better solution but @byorgey beat me to it and @zzq then nailed it as well. For completeness though, here’s my own explanation, inspired by how I convinced @faire (age 11):

If you think about how a teaspoonful of wine leaves the wine glass and the water gets slightly contaminated and then some but not (necessarily) all of the wine comes back to the wine glass it’s all very confusing. The trick is…

…to ignore all the back-and-forth and think only about how the glasses start and how they end up. You start with exactly a glass full of wine and you end with a glass full of mostly wine. But the glass is still exactly full at the end – a teaspoonful of something left and a teaspoonful of something came back.

So some fraction of a teaspoonful of wine got displaced by water and there’s only one place it could go: the water glass! That fraction of a teaspoonful could even be the fraction 0% or 100%. Say you move a teaspoonful of wine and it sinks to the bottom of the water (why not?) and you move only water back. Then you have 1 teaspoonful of wine in the water and 1 teaspoonful of water in the wine. At the other extreme, if you dump the teaspoonful of wine in and manage to scoop exactly that wine right back out then you have 0 teaspoonfuls of wine in the water and 0 of water in the wine.

For anything in between it’s the same story: Since the wine/water levels in the glasses end up the same, however much net wine [1] left the wine glass is exactly how much water must’ve replaced it.

QED

Footnote:

[1] Keeping this 11-year-old friendly, the amount of net wine that leaves the glass is the amount that left minus the amount that came back. So like the amount that really actually left, not just the amount that initially left. Like how net profit, for example, is the amount of money you really actually made, subtracting all the money you spent as part of making it. The final amount in your pocket. Or the water glass, in this case.

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The problem only makes sense if glasses have the same amount of liquids - otherwise the answer can be made to be anything you want.

So after both operations the glasses also hold the same amount of liquids.
The wine displaced by the water had nowhere to go except for the other vessel, and vice versa.
Which means, as much water was displaced by wine as wine by water. So, the same.

Not true! I think nothing in my answer depends on that! We do assume the 2 teaspoonfuls are the same.

You are correct, but it’s only obvious once you’ve solved the problem, which I did not at the beginning of my post:)

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