We look at the three-person game, since it is more interesting. We also assume that the deck is shuffled between draws, in order to simulate tossing a fair $k$-sided die. You will find it easy to adapt the idea to the simpler $2$-person game, and if you wish, to a $d$-person game.
Let our players be A, B, and C. We suppose A tosses first, then (if necessary) B, then, if necessary, C, then, if necessary A, and so on.
To make the notation simpler, let $p=\frac{1}{k}$.
Player A can win on the first draw, the fourth, the seventh, the tenth, and so on.
The probability she wins on the first draw is $p$.
In order for A to win on the fourth draw, A, B, and C must all fail to win on their first three draws, and then a must win. The probability of this is $(1-p)^3p$.
In order to win on the seventh draw, A, B, C must all fail twice, and then A must win. This has probability $(1-p)^6p$.
Similarly, the probability A achieves her win on the tenth draw is $(1-p)^9p$. The probability A achieves her win on the thirteenth draw is $(1-p)^{12}p$. And so on.
So the probability that A wins is $p+(1-p)^3p+(1-p)^6p +(1-p)^9 p+ (1-p)^{12}p+\cdots.$ The above is an infinite geometric series. Recall that the infinite geometric series $a+ar+ar^2+ar^3+\cdots$ has sum $\frac{a}{1-r}$ (if $|r|\lt 1$). In our case, $a=p$ and $r=(1-p)^3$, so the probability A wins is $\frac{p}{1-(1-p)^3}.$ This can be "simplified" to the less attractive expression $\dfrac{1}{3-3p+p^2}$.
We could go through a very similar calculation for the probability that B wins. However, there is a shortcut. Suppose that A fails to win on her first throw (probability $1-p$). Then effectively B is now first, so has probability of winning $\dfrac{p}{1-(1-p)^3}$. So the probability B wins is $\frac{(1-p)p}{1-(1-p)^3}.$ A similar argument shows that the probability C wins is $\frac{(1-p)^2 p}{1-(1-p)^3}.$
Another way: We can avoid summing an infinite series. Let $a$ be the probability that A ultimately wins. As discussed earlier, the way for B to win is for A to fail on her first draw. Then effectively B is the first player. So the probability B is the ultimate winner is $(1-p)a$. Similarly, the probability C is the ultimate winner is $(1-p)^2a$.
But it is (almost)clear that someone must ultimately win, the probability the game goes on forever is $0$. It follows that $a+(1-p)a+(1-p)^2 a=1,$ and therefore $a=\frac{1}{1+(1-p)+(1-p)^2}.$ A little calculation shows that this is the same answer as the one obtained earlier.