Two-player game in $19$ rounds

Question

Aashna and Radhika see the integers $1$ to $211$ written on a blackboard. They alternate turns and in every step each of them wipes out any $11$ numbers until only $2$ numbers are left on the blackboard. If the difference of these $2$ numbers (by subtracting the smaller from the larger) is $\geq 111$, the first player wins, otherwise the second. If you were Aashna, would you chose to play 1st or 2nd and why?

By intuition only – I don’t see how I can prove it:

Since there are 19 turns ($19\times 11=209$ numbers + 2 on the blackboard), I would choose to play 1st. In my 1st move I would wipe out numbers 101 to 111 since these are the only ones that do not have a pair to meet the rule. Then for whichever numbers Radhika would remove, I would respond by removing the numbers that would have a difference of 111, for example for 92 I wipe out 203 etc. If Radhika chose to remove pairs with difference equal to 111, there would still be a single number, for which I would remove its pair and then I would also remove pairs with difference 111 or 112 and so on.

Does this method guarantee a win?

Answer 1

There are $100$ pairs with difference $111$: $\{1,112\}$, $\{2,113\}$, $\ldots$, $\{100,211\}$. As you noticed, there are $11$ numbers that are unpaired are $101$, $102$, $\ldots$, $111$. Since the second player can eliminate at most $9\cdot 11$ numbers, the second player does not touch at least one pair of difference $111$. Thus, if the first player does not play stupidly by corrupting these pairs, then the first player can always win.

You have recommended a good strategy. In the first move, the first player proceeds by removing the $11$ unpaired numbers $101$, $102$, $\ldots$, $111$. Every turn after the second player played, if the number $x\in\{x,y\}$ with $|x-y|=111$ was remove but $y$ was not, then the first player responds by removing $y$. Since $11$ is odd, the number of such $x$'s that the second player removed must be odd. Thus, the first player will remove an odd number of such $y$'s. If the first player has not completed the turn (i.e., fewer than $11$ numbers have been played), then there are an even numbers left to remove. Then, the first player remove both numbers form some pairs with difference $111$ until the turn is finished. Therefore, by the end of the $(2k+1)$-st turn, only $11k$ pairs of difference $111$ have been killed (and other pairs are completely untouched).

Answer 2

If you chose to play first you would have 10 turns. -> 10 x 11 = 110

If you chose to play second you would have 9 turns. -> 9 x 11 = 99

If you are second and remove every high number starting with 211 while the first removes the low numbers you would stop at 211 - 99 = 112.

The last left numbers would be 112 and one number (between 1 and 112).

The first would then only win if that number is 1, because 112 - 1 = 111 (which is >= 111).

If it was the opposite and you would play first you could remove 110 numbers and you would win if the difference is >= 111. There are only 211-111 = 100 numbers with a difference >= 111. Remove all of them and you win.

So your chance of winning by playing first is 100%.