I was a little confused by the mixing problem because of the phrase, “any sequence of transfers ….” Are the transfers always A-to-B followed by B-to-A, or can we use ANY sequence of transfers, such as A-to-B, then A-to-B again, then B-to-A, …? I think the former is what was intended (?); the latter is a more general question.

I think I have a proof that even in this more general case, it impossible to ever get a 50-50 mixture. First note that if at any point one bottle has a 50-50 mixture, then the other bottle does also. (If A has 2 oz wine and 2 oz water, then B has 8 oz wine and 8 oz water.) So if one bottle is not 50-50, neither is the other.

Second, note that if, say, we’re transferring from A to B, the ratio of water to wine in A after the transfer will be the same as the ratio before the transfer. (For example, if A is 63% wine, and I scoop out 3 ounces, 63% of what’s left will be wine.) The ratio in the bottle I’m scooping from can’t be changed by the transfer. (The ratio in to bottle I’m transferring to can be changed.) So if the “from” bottle was not 50-50 before the transfer, it can’t be afterward. And if the “from” bottle is not 50-50 after the transfer, then neither is the “to” bottle.

The only exception to the above argument is if I scoop out ALL of the liquid from one bottle: then I can’t say that if A was 63% wine before the transfer, it is still 63% wine afterward. If I manage to scoop out all of the liquid from A, then B will be a 50-50 mixture. But in this problem we have 10-ounce bottles and a 3-ounce scoop, so I can never empty a bottle. (If it were 12-ounce bottles and a 3-ounce scoop, then I could.)

So no sequnece of scoops can result in a 50-50 mixture.

(I think Harry Kaplan’s proof could be modified to cover this more general problem as well.)

Evan Romer

Windsor NY

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Let W(0) be the amount of wine in the wine bottle before the dual 3 oz. transfers, and let w(0) be the amount of wine in the water bottle at the same time. Similarly, let W(1) and w(1) be the amount of wine in those two bottles after the two transfers are complete. If the first transfer is from wine bottle to water bottle, it’s easy to derive:

10/13 W(0) + 3/13 w(0) = W(1)

10/13 w(0) + 3/13 W(0) = w(1)

The problem is to determine whether we can ever end up with W(1) = w(1) = 5 after any series of dual transfers. Assuming we can, let’s consider the final pair of transfers that created the desired outcome. We can arbitrarily claim that the first of the transfers is out of the wine bottle. The difficulty is that if we set W(1) = w(1) = 5, our equations tell us that W(0) = w(0) = 5 as well. Therefore our assumption is wrong, and we can only end up with two identical bottles post-transfers if we started with two identical bottles. Yes or no?

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[spoiler] If we have, say, 1,000,000 molecules at one step, well, 1,000,000 = 2^6 x some junk; at the next step, whether or not we multiply by 1/2 or 3/2, we will have one less factor of 2, namely 2^5 * some other junk. So when we’re thinking about the physical problem, we’re limited by how many “factors of two we can get into a glass of water”

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[Edit: Now, wait a sec. These were what I meant. It’s not “close” I guess, but it’s “closer”: 3^31/2^49 is about 1.097, which is about as close as you can get up to that size. Byon’s right of course, that you can get a lot closer going about twice as long…]

]]>Also, 2^49 is not very close to 3^31. Are those the numbers you meant? Thanks!

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