How many chips can A get at least? (Game, chip distribution)

[f.borke]

There are 2020 small boxes with 1,2,...,2020 chips and 2 large empty boxes. Each turn B will choose one of the two large boxes and A will choose a nonempty small box and will put the chips in it into the box chosen by B. After 2020 turns all small boxes are empty and A chooses one of the large boxes and receives the chips in it. How many chips can A get at least?

 

[joshuaa]

let the large boxes be 1 and 2

my question is
if B chose box 1 in the first turn, does he have to choose box 2 in the 2nd turn?

or

in each turn, B randomly choose one of the two large boxes?

if B randomly select the large boxes, there is a probability that he doesn't select box 2 for 2020 turns

finally, A may choose box 2 which is empty

so A can get at least zero chip

if zero is not accepted as at least

let us assumed that B chose box 2 one time while A puts 1 chip inside it

then

A gets at least 1 chip

 

[Steven G]

let the large boxes be 1 and 2

my question is
if B chose box 1 in the first turn, does he have to choose box 2 in the 2nd turn?

or

in each turn, B randomly choose one of the two large boxes?

if B randomly select the large boxes, there is a probability that he doesn't select box 2 for 2020 turns

finally, A may choose box 2 which is empty

so A can get at least zero chip

if zero is not accepted as at least

let us assumed that B chose box 2 one time while A puts 1 chip inside it

then

A gets at least 1 chip

As this is a math help forum we prefer that students solve their own problems with hints/suggestions from the helper, we never prefer (OK, almost never) to just give out complete solutions. Please read our guidelines and other posts so you can understand what we do here. Thanks

 

[Deleted member 4993]

As this is a math help forum we prefer that students solve their own problems with hints/suggestions from the helper, we never prefer (OK, almost never) to just give out complete solutions. Please read our guidelines and other posts so you can understand what we do here. Thanks

I fully concur with Jomo. We wait:

for OP to show some effort showing basic understanding of "Find" and "givens" of the problem

~24 hours (so that the urgency of "test" cheating and "HW" cheating has subsided.)

 

[Deleted member 4993]

I fully concur with Jomo. We wait:

for OP to show some effort showing basic understanding of "Find" and "givens" of the problem

~24 hours (so that the urgency of "test" cheating and "HW" cheating has subsided.)

Did I say that?!

There must be full-moon out to-night!!

 

[pka]

There are 2020 small boxes with 1,2,...,2020 chips and 2 large empty boxes. Each turn B will choose one of the two large boxes and A will choose a nonempty small box and will put the chips in it into the box chosen by B. After 2020 turns all small boxes are empty and A chooses one of the large boxes and receives the chips in it. How many chips can A get at least?

This is a poorly put question. If B selects the large boxes alternately, \(1,2,1,2\cdots\) and A chooses the small boxes in order \(1,2,3,4, \cdots\)
The large box 1 contains the sum of the first 1010 odd numbers and the large box 2 contains the sum of the first 1010 even numbers.
It would make much more sense if B tossed a coin to select a large box and A had some similar scheme.

 

[Steven G]

I do not fully concur with Subhotosh. This is my opinion, I prefer not to give out answers because doing so does not help out the student at all. I always feel that students come first!

 

[f.borke]

This question obviously assumes that A will try to get as many chips as possible, so if B chooses always the same box, A will get all chips. As we cannot expect that this happens however, the number we're looking for will be at least n(n+1)/4. However,we can see easily that A can often get even more than n(n+1)/4 no matter how B plays (e.g. for n=3 the number we're looking for will be 4). We see that for n=1,2,3 the solutions will be 1,2,4. How will this series continue?