probability, once fair die , 3 players. I have this problem, three players A,B and C, takes turnes rolling a fair die, the first to roll and odd prime is the winner. If A goes first, then B, then A and so forth, what is each player probability of winning expressed as a percent? 
I tried with two players, and I took into consideration that it will be the 2/6+4/6(p(a)-1) but this if the player B doesn't get a prime odd and A does it at the second try. How I can express it with 3 players and most importantly no knowing in which turn a player will win?
 A: one of them must win so $P(A)+P(B)+P(C)=1$
also 
$$P(B) = P(A)\times P(A\text{ does not win on first roll}) $$
$$P(C) = P(A)\times P(\text{ neither A nor B win on first roll}) $$
So
$$P(A)+P(B)+P(C)=P(A) ( 1+ \frac 23 +\frac 49)$$
A: Let P be the event the number tossed is an odd prime, and N the event it is not.  Then the toss patterns for which A wins are: 
P, NNNP, NNNNNNP, NNNNNNNNNP, and so on. 
These have probabilities:
$1/3$; $(2/3)^3(1/3)$; $(2/3)^6(1/3)$; $(2/3)^9(1/3)$; and so on.  
For the probability A ultimately wins, add up. We have a geometric series with first term $1/3$ and common ratio $(2/3)^3$. The sum, by the usual formula, is
$$\frac{1/3}{1-(2/3)^3}.$$
This can be simplified to $\frac{9}{19}$.  
For the probability B wins, the relevant probabilities to add are 
$(2/3)(1/3)$; $(2/3)^4(1/3)$; $(2/3)^7(1/3)$; $(2/3)^{10}(1/3)$, and so on. Add up. Again a geometric series, with each term $2/3$ times the series for A winning. 
The sum is therefore $6/19$.  A similar argument shows that the probability C wins is $4/19$.
