Water into wine and back

[bonanova]

A familiar problem concerns two glasses, one containing water, the other wine [edit: in equal amounts]. A certain amount of water is transferred to the wine, and an equal amount of the mixture is transferred back to the water. Is there now more wine in the water glass than there is water in the wine glass? A moment's thought shows the two amounts must be the same.

Is it possible, by thus transferring the same amount of fluid, any number of times, to reach a state where the percentage of wine in each mixture is the same?

[Morningstar]

Are we assuming that the mixtures are perfectly mixed and that all samples of the mixtures will have the same percentage of each liquid as the entire glass?

if we transferred half the water, and then managed to transfer back pure wine by chance, both glasses would have an equal amount.

[Barcallica]

Lets say there were x amount of water and wine. First we transfer y amount of water to wine. Now wine occupies x/(x+y) percent of mixture and water y/(x+y). When transfer back Y amount of mixture there goes xy/(x+y) amount of wine with it (and y^2/(x+y) water as well). And y-y^2/(x+y)=yx/(y+x) water stays in the second glass meaning amount of wine in the first glas is equal to water in the second bottle.

Another approach: if we put infinitesimal amount of water to wine and bring back equal amount of mixture obviously first glass would have infinitesimal amount of wine and second glass would have infintesimal amount of water. As opposite if we transfer all the water to second bottle and return half of the mixture of course there would be as much wine in the first glass as water in the second glass. And as we increase amount of transition; amount of water and wine in not original glasses will increase as well meaning this is continuous function (above mentioned regulatority will be present no matter how much liquid we transfer).

Another thing is there can never be more water in second bottle than wine and vice versa. So after every pair of transfer there is equal or more than 50% wine in the second glass and 50% or less wine in first glass. And if and if only we put all the liquid to second glass and transfer back the half we can have same amount of wine in both glass.

Edited May 8, 2012 by Barcallica

[phil1882]

hmmm.

let's start off assuming we have 12 liters of water and 12 liters of wine.

we want a mixture that's 6 liters water/wine in both glasses.

it seems to me the quickest way to achieve this is to pour all 12 liters of wine in the water. we now have a 50/50 mix.

then transfer 12 liters back. done.

[bonanova]

Are we assuming that the mixtures are perfectly mixed and that all samples of the mixtures will have the same percentage of each liquid as the entire glass?

if we transferred half the water, and then managed to transfer back pure wine by chance, both glasses would have an equal amount.

Yes. Assume complete mixing.

[bonanova]

Lets say there were x amount of water and wine. First we transfer y amount of water to wine. Now wine occupies x/(x+y) percent of mixture and water y/(x+y). When transfer back Y amount of mixture there goes xy/(x+y) amount of wine with it (and y^2/(x+y) water as well). And y-y^2/(x+y)=yx/(y+x) water stays in the second glass meaning amount of wine in the first glas is equal to water in the second bottle.

Another approach: if we put infinitesimal amount of water to wine and bring back equal amount of mixture obviously first glass would have infinitesimal amount of wine and second glass would have infintesimal amount of water. As opposite if we transfer all the water to second bottle and return half of the mixture of course there would be as much wine in the first glass as water in the second glass. And as we increase amount of transition; amount of water and wine in not original glasses will increase as well meaning this is continuous function (above mentioned regulatority will be present no matter how much liquid we transfer).

Another thing is there can never be more water in second bottle than wine and vice versa. So after every pair of transfer there is equal or more than 50% wine in the second glass and 50% or less wine in first glass. And if and if only we put all the liquid to second glass and transfer back the half we can have same amount of wine in both glass.

Yeah, you have it. Nice.

[bonanova]

hmmm.

let's start off assuming we have 12 liters of water and 12 liters of wine.

we want a mixture that's 6 liters water/wine in both glasses.

it seems to me the quickest way to achieve this is to pour all 12 liters of wine in the water. we now have a 50/50 mix.

then transfer 12 liters back. done.

Precisely. Well done.

The answer when only a fraction of the fluid is transferred each time depends on whether one takes into account the discrete nature of matter. If one counts the number of their molecules, then the answer is reachable with non-zero probability. But if one measures the liquids in liters or ounces, using real numbers capable of infinitesimal change, then the state of equality is an asymptote that is never reached.

[Peekay]

From the first part of the problem it is clear that the only time you can have the same percentage of wine in both vessels is if the percentage is 50%. The only way this can be done is to transfer the entire wine into the water and then transfer back half of the mixture (same volume as step 1).

[crazy]

The question doesnt say that both glasses contain equal volume of liquid.

[bonanova]

The question doesnt say that both glasses contain equal volume of liquid.

It should. I've made that edit.

Thanks for making the point.