- 30 Oct '11 07:00It is fairly well known that if you have two sacks of beans, red and white, with the same number of beans in each, and take a cupful of red beans from the red to the white, mix perfectly, and then move as many beans from the mix of mainly white and some red into the red sack, the number of red beans in one sack equals the number of white beans in the other.

What if there are three sacks and three kinds of beans?

From red to white, and mix

From mainly white to black, and mix

From mainly black to red

Are the proportions of different kinds of beans the same in each sack, though of course with colors changed? - 31 Oct '11 12:07

Even that is only a guarantee if you can mix perfectly. Contrast the case of two bags, where it's a guarantee no matter how badly you mix, as long as the two amounts are the same.*Originally posted by talzamir***so.. if someone moves a bucketful from black to red, and you can only move to black to red to white to black.. is the only way to get the proportions to be the same in a finite time to put all the beans together and then dividing by three?**

Richard - 31 Oct '11 13:49Of course. Entirely possible to have two sacks with a million beans in each, take a bucketful of beans from white bean sack to black bean sack, mix, and take a bucketful back - and by a stroke of luck not only is the number of beans in both bucketfuls the same even though beans are of different sizes and take more or less space depending on how they arrange themselves - but that all the white beans are in the original sack again and no black bean ever moves between sacks. Just extremely unlikely. For all I know, the solution on various problems can be affected by precision. It is also true that very few things are equal. And who knows, maybe the gravity of the moon messes up some problems too.

I would hope that this of all places - a puzzle section inside a virtual chess club - is an environment for people who simply enjoy the mental challenge of puzzles. Apparently not. Sadly, some of the other ways take the fun out of this for me. - 31 Oct '11 19:04

Don't give up hope, I enjoy them!*Originally posted by talzamir*

I would hope that this of all places - a puzzle section inside a virtual chess club - is an environment for people who simply enjoy the mental challenge of puzzles. Apparently not. Sadly, some of the other ways take the fun out of this for me. - 01 Nov '11 09:01

I haven't done the math but so long as there is a probability you could end up with a particular arrangement, then my guess is that yes it is entirely possible you could end up with the same proportions. The probability will be minuscule but it is possible.*Originally posted by talzamir***Indeed so..**

so.. if someone moves a bucketful from black to red, and you can only move to black to red to white to black.. is the only way to get the proportions to be the same in a finite time to put all the beans together and then dividing by three?

Consider the following example:

Container#1 contains 15 White, Container#2 contains 15 Red & Container#3 contains 15 Black

move 1: move 10W from C#1 to C#2

C#1 contains 5W, C#2 contains 10W & 15R, C#3 contains 15B

move 2: move 10R from C#2 to C#3

C#1 contains 5W, C#2 contains 10W & 5R, C#3 contains 10R & 15B

move 3: move 10B from C#3 to C#1

C#1 contains 5W & 10B, Bag#2 contains 10W & 5R, C#3 contains 10R & 5B

move 4: move 5W & 5B from C#1 to C#2

C#1 contains 5B, C#2 contains 15W & 5R & 5B, C#3 contains 10R & 5B

move 5: move 10W from C#2 to C#3

C#1 contains 5B, C#2 contains 5W & 5R & 5B, C#3 contains 10W & 10R & 5B

move 6: move 5W & 5R from C#3 to C#1

C#1 contains 5W & 5R & 5B, C#2 contains 5W & 5R & 5B, C#3 contains 5W & 5R & 5B

Solved. Absolutely minuscule probability and I'm guessing we need to use a lottery probability calculator for this. So this now comes down to defining a "finite time". If that was defined as while the sun was still burning, no problem. If however, this must be completed before the beans rot, then it is highly unlikely.

Yet another interesting problem.

Thanks! - 01 Nov '11 09:29 / 1 editGlad you like. In a concise format,

initial orders: 15W + 15R + 15B

After move 1: 05W + 10W15R + 15B

After move 2: 05W + 10W05R + 10W15B

After move 3: 05W10B + 10W05R + 10W05R

..

After move 6: 05W05B05R + 05W05B05R + 05W05B05R

Actually move 3 suffices for what I first had in mind but apparently was careless in putting in words, as after three moves each sack has beans in ratios 0:1:2, in order of least to most. The second round makes it far prettier though, with 1:1:1 in each, which is a more stringent way to interpret what I was saying - or the only solution if the ratio is in order W:R:B rather than least:middle:most.

However.. if one requires that the sack is perfectly mixed each time.. (yes, I know that in the real world it is not doable) it would not be possible to move 10 white beans. from a sack that has beans of many colors. The above, taking 2/3 of the contents each time.. I'll use 81 beans of each kind:

Initial orders: 81W + 81R + 81B

After move 1: 27W + 54W81R + 81B

After move 2: 27W + 18W27R + 36W54R81B

After move 3: 51W36R54B + 18W27R + 36W54R81B

..and that's a dilemma. It's possible to make it so that each sack has 81 beans again, but the ratios of least : middle : most may or may not be possible to make the same without making one or more of the sacks empty. So.. is it possible to get identical mixtures that are NOT 1:1:1 , such as 1W10R70B + 1R10B70W + 1B10W70R, by mixing fully between transfers?

With two sacks it is of course trivially easy. Move any number of beans from the first sack to the second, and the same number back (yes I know in real world that can't be done either), and the ratios in less : more are always the same. But with three sacks it is not that clear. - 03 Nov '11 13:10

I, by contrast, find it fascinating that for two colours, you can be a sloppy as you like and the exact correspondence is*Originally posted by talzamir***I would hope that this of all places - a puzzle section inside a virtual chess club - is an environment for people who simply enjoy the mental challenge of puzzles. Apparently not. Sadly, some of the other ways take the fun out of this for me.***guaranteed*, while for any higher number - even for three - the solution can only be asymptotically correct.

Richard