Archeological evidence indicates that the ancient (pre-Roman) Etruscans played dice using a dodecahedral die having 12 pentagonal faces numbered 1 through 12 (figure above). One could simulate such a die by drawing a random card from a deck of cards numbered 1 through k. for your own personal value of k, begin with the largest digit in the sum of the digits in your student ID number. This is your value of k unless this digit is less than 5, in which case subtract it from 10 to get your own value of k. a.) John and Mary draw alternately from a deck of shuffled k cards. The first one to draw an ace-the card numbered one- wins. Assume that John draws first. Use the formula for the sum of a geometric series to calculate (both a rational number and a four place decimal) the probability J that J

Step 1
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 $(1-<!--\; -\; -->p{)}^{3}p$.
In order to win on the seventh draw, A, B, C must all fail twice, and then A must win. This has probability $(1-<!--\; -\; -->p{)}^{6}p$.
Step 2
Similarly, the probability A achieves her win on the tenth draw is $(1-<!--\; -\; -->p{)}^{9}p$. 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 $p+(1-<!--\; -\; -->p{)}^{3}p+(1-<!--\; -\; -->p{)}^{6}p+(1-<!--\; -\; -->p{)}^{9}p+(1-<!--\; -\; -->p{)}^{12}p+\cdots <!--\; \cdots \; -->.$
The above is an infinite geometric series. Recall that the infinite geometric series $a+ar+a{r}^2+a{r}^3+\cdots <!--\; \cdots \; -->$ has sum $\frac{a}{1-<!--\; -\; -->r}$ (if $|r|<<!--\; <\; -->1$). 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 ${\displaystyle \frac{1}{3-<!--\; -\; -->3p+p^2}}$.
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 ${\displaystyle \frac{p}{1-<!--\; -\; -->(1-<!--\; -\; -->p{)}^3}}$. So the probability B wins is $\frac{(1-<!--\; -\; -->p)p}{1-<!--\; -\; -->(1-<!--\; -\; -->p{)}^3}.$.
A similar argument shows that the probability C wins is $\frac{(1-<!--\; -\; -->p{)}^{2}p}{1-<!--\; -\; -->(1-<!--\; -\; -->p{)}^3}.$.

Step 1
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 $(1-<!--\; -\; -->p{)}^{2}a$.
Step 2
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.