One two three(added details)?

Question

line 1: 1...9
line 2: 1...9
line3: 1...9
person A chooses 1 number from each line
person B chooses 1 number from each line(but whithout knowing what person A chose)
example:person A chose:7,6,9
person A chose:6,4,2
In order to win person B need to get exactly 1 number respectively that A chose.
example of a wining game:
Person A :4,5,8
Person B:2,5,6
example of a game without a win:
Person A :4,5,8
Person b :2,5,8
p(wining)=243/1000
if person B can choose more than 1 set of numbers(each set =3 numbers)

is it possible to make sure B will win the game with 5 diffrent sets of numbers?
if so,how?
and:give an example for sutch 5 sets.
p.s
243*5>1000 so in my opinion it's possible

to Bramblyspam:
0 is a possibility as well
(10 digits,0-9)

please help me figuring out if it's possible and how to build such a set if posibble

to:math guy

if 243*5>1000 than the probability goes over 1
=100% success

Answer 1

First of all, your odds are wrong. Note that the digits must be between 1 and 9 - no zeros allowed. Hence, there are 9*9*9 = 729 possible combinations.

I get the odds of winning to be 152/729, which still is more than 20%, which is compatible with the idea that it's possible to make a set of five combos that would guarantee B a win.

I haven't yet figured out what such a set would look like, though.

LATE EDIT: It's impossible. Here's why.
Take any set of five combos, for your supposed solution.
There will be at least four possible first digits that aren't in any of the combos.
There will be at least four possible second digits that aren't in any of the combos.
There will be at least four possible third digits that aren't in any of the combos.

Thus, no matter what your proposed solution set, it can't account for a situation where A picks a first, second, and third digit from among the four possible 1st/2nd/3rd digits that aren't in your proposed solution set.

Answer 2

It actually is not possible. Even though (as you point out) 243*5>1000, this simply goes to show that it is PROBABLE that within 5 tries, player B will win.

In order to prove that there MUST be a way to GUARANTEE a win for B, we must be able to KNOW one of person A's choices after the 4th guess. Imagine that Person B gets to choose 5 numbers from each row and only needs AT LEAST (instead of exactly) one of the numbers that Person A chose. You should be able to compute that Person B would still only win 82.9% of the time (1-(5/9)^3). Since this relaxes the rules for Person B (making it more likely that he would wind) and it is still not guaranteed, then under the rules you set out, it would be evel LESS likely than this that Person B will win.

I hope this helps!