21 Dares

Throughout the last year, a certain game has popped up in my mind multiple times, haunting me- generalising 21 dares. I am pleased to report that the problem is solved.

Introduction

Recall the famous game known as “21 dares”, wherein 2 people may add an integer between 1 and 3 (starting from 0) and whoever says 21 loses. Thus it’s really a game of first to 20. Furthermore, it’s easy to win this game with the following logic:

“I must get to 20, and to guarantee getting there I must get to 16.”
“To get to 16, I must get to 12.”
And so forth until you realise that you must get the other person to start first, so that no matter what you can always just say multiples of 4 and win.

Similarly, any game of $x$ dares admits a guaranteed win for either the first or second player (if $x-1=n(y+1)$ , the second player wins, otherwise the first player does, where $n$ is an arbitrary integer and $y$ is the amount one can move). The natural question to ask is: what if there are more than 2 players?

3 or more players

Saying multiples of 4 in 21 dares is all fun and games until there are 3 players. The question, as we shall uncover, now becomes not when does player $x$ win?", butis it possible for any player to guarantee a win?” This is because:

Now that there are 3 players, to win you can either say 19 or 20 because then it is guaranteed that between the other 2, they will say 21.
The problem is how do you get there, even when the other 2 people are trying to beat you.
If you say a number between 14 and 18, then the other two players have the capacity to get to 20 and if you say any number between 13 and 8, they have the capacity to force you to say a number between 14 and 18 and so on. Basically, if they want to, the other two people can always make you lose.

Thus the question is: when can someone guarantee a win in $x$ dares where there are more than 2 players?

We shall answer this in a series of theorems:

Theorem 1:

In a game of $x$ dares, with $n>2$ players (where the amount of numbers one can add ensures that everybody plays), it is impossible to guarantee a win when the number one can add is bigger than 2.

Proof:

Since there are $n$ players, the number you want to say, $r$ , is in the range:

$x-(n-1)\leq\sigma\leq x-1.$

Thus, we require the player before us to say a number $r_1$ within:

$x-y-(n-1)\leq r_1\leq x-2.$

It is only possible to force this when you say $x-y-2(n-1)$ (otherwise the others can just add 1 every time and stay out of the range). This only works when the maximum that they can add, $y(n-1)$ , still stays less than $x-1$ . Thus it is impossible to win when the following inequality are true:

$x-y-2(n-1)+y(n-1)\geq x-1,$

which leads to the following inequality in terms of $y$ and $n$ :

$x-y-2(n-1)+y(n-1)\geq x-1$

$x+y(n-2)-2(n-1)\geq x-1$

$x+ny-2y-2n+2\geq x-1$

$ny-2y \geq 2n-3$

$y(n-2)\geq 2n-3$

$\boxed{y\geq\frac{2n-3}{n-2}}$

When $n=2$ , we see that it is possible to win whatever $y$ is. When $n=3$ , we can win when $y<3$ – $y=1,2$ . Elementary calculus shows us everything else we need to know:

$\frac{d}{dn}\frac{2n-3}{n-2}=-\frac{1}{(n-2)^2}.$

The derivative shows that the function is strictly decreasing for $n>2$ and L’Hôpital’s rule gives us:

$\lim_{n\to\infty}\frac{2n-3}{n-2}=2.$

Combining those two facts gives us:

$2<\frac{2n-3}{n-2}<3,\, n\in (3,\infty)$

because the function is strictly decreasing, but has a horizontal asymptote at $n=2$ . Therefore, $y$ can be 2 or 1 for all $n>2$ as required.

Disclaimer

This does not mean we can guarantee a win for any game of $x$ dares with $n$ players if anyone can only add at most 2 every turn. The next problem is how do we guarantee that we can say $x-y-2(n-1)$ .

Theorem 2:

When $n>2$ , one can only guarantee that you can say $x-y-2(n-1)$ if:

It is available on your first (or second) go.
$y\leq 1$

Proof:

Using similar reasoning to before, we require the person before us to say a number $r$ in the range:

$x-2y-2(n-1)\leq r\leq x-y-2(n-1)-1.$

Again, with similar reasoning to before, we therefore must say $x-2y-3(n-1)$ (otherwise they add 1 and stay out of the range). The only times this works is when they stay in this range even after adding $y(n-1)$ . This holds when:

$x-2y-3(n-1)+y(n-1)\leq x-y-2(n-1)-1$

$y(n-3)-3(n-1)\leq-y-2(n-1)-1$

$3(n-1)-y(n-3)\geq y+2(n-1)+1$

$n-1-y(n-3)\geq y+1$

$n-2\geq y(n-2)$

$\boxed{1\geq y\iff y\leq 1}$

as required.

Thus, we may combine our work into one (satisfying) theorem:

Theorem 3:

In a game of $x$ dares, with $n$ players, where one can add at most $y$ in one turn, (all of them are positive integers, $n\geq 2$ and everyone is guaranteed at least one go) one is able to guarantee a win if any of the following conditions are satisfied:

There are 2 players
There are any amount of players, but everyone can only add 1 each turn
You can say $x-y-2(n-1)$ on your first or second and $y\leq 2$ .

21 Dares

Published by gregoriousmaths

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