Krdytkyu

Finni’s game:

Person A thinks of a number (1 to 10). This number is called n.

Person B says a number (1 to 10). This number is called x.

Person A tells the absolute difference of n and x. This difference is called u.

Person B says a new number (1 to 10). This number is called y.

Person A tells the absolute difference of n and y. This difference is called v.

Person B's Goal: u + v shall be as small as possible.

Which strategy should person B follow? Why?

Note:

I myself do not know the answer yet, but I am currently working on it.

First, a remark: The question may not have a well-defined answer: is B trying to make u+v as small as possible in the worst case where A gets lucky, or on average in some sense, or what? I'll assume that each player is trying to optimize their worst case.

After A announces the number u,

B knows that A's secret number n must be either x+u or x-u. (Perhaps only one of these is possible, e.g. if x=1.) B will then pick y to make max(|y-(x+u)|, |y-(x-u)|) as small as possible. What value of y does this? Always x itself, when both values are actually possible. If not, then B knows the value of n and will pick it.

So, let's suppose B picks x with 1 <= x <= 5. (The other choices are equivalent by symmetry.) Then

if A responds with something =x, in which case B now knows what n is and can hit it on the nose; worst case is where n=10 and the total difference is 10-x.

Now

as we increase x, one of these gets worse and the other gets better. The best worst case will be where they are equal: 2(x-1) = 10-x or x=4. Then the worst possible total difference is 6.

So B should pick

either 4 or 7 (or some random choice between them)

and A should pick

either 1 or 10 (better pick at random, else B can do better!)

after which

the total difference will always be 6.

I think the answer is:

Strategy for B: selects 5 or 6 as x and u+x as y

Reason:

When difference in talking is always positive so u and v are positive.

To reduce value of an add operation of two positive number you should minimize both.
Maximum differences between n and x can be 10-1=9 so middle of that is 9/2+1=5.5 that nearest values are 5 and 6.

If B selects 5 or 6 as x; he will minimize the risk.

To minimize v as B knows n=u+x then by saying u+x as y then v=0 and then u+v=u and it will become minimal.

I think B picks:

either 3/8 or 4/7.

Why?

Let's say B picks x=4, (the argument is similar for x=7) then if A had picked n=1,2,3,4,5,6,7 u=3,2,1,0,1,2,3pts respectively. B now knows the range is n=1,...,7, and on their next guess should pick either y=3 or y=5, which at worst will be at most 4pts, a total of 7pts. If not, say A chose from 8,9,10, B already knows the result, for a maximum total of 6pts (from n=10). Result: $u+vle7$.


If B picks x=3, the worst case is 7pts from A having chosen n=10 (next guess from B is 0pts).


The other two cases: If B picks x=5, worst case is 9pts (n=9 incurs 4pts, B goes for y=6 but n=1 gives another 5pts).


If B picks x=1 or x=2, worst case is at least 8pts.

For x, B should pick:

8

And or y, B should pick:

8 - u

The expected value of u + v is then:

3.7

This is probably similar to other answers, but I don't think it is expressed quite this way....

EDIT - ok so reading more closely it is the same as Gareth McCaughan's answer, but expressed a little bit differently

B should start with 4 - then if the difference is 0,1,2 or 3 B should repeat 4 and then the total will be either 0,2,4 or 6 and the unknown number was between 1 and 7

looking at the other possibilities

If the first difference is 4,5 or 6 then B knows the unknown number was 8,9 or 10 and so for B's second number B can choose 8, 9 or 10 and then the total of the two differences will be 4,5,6.

Thus the result is

the largest sum of differences is 6

and

the same effect could be had starting with 7

Note that

If B starts with 5 (or 6) then a difference of 4 initially may mean the unknown number is 1 or 9 and so B will be unsure how to minimize the total and will need to stick with 5 and get a total difference of 8

and also note

If the first guess is less than 4 (or more than 7) then the first difference could be greater than 6 and so the total of the differences could be greater than 6.

Thus in conclusion

must start with 4 (or 7)

If there were more guesses then it would be better to start with

3 - if there were 3 guesses. and probably 2 if there were 4 or more guesses.