Will Ben win?

Calculating probability of Ben's win without using geometric series:

Let $p$ be the probability of the first player winning on the first throw, and $p_{w}$ be the probability of the first player winning the match.

Since the players throw equally well, it is clear that one of two things must happen:

1) With probability $p$ , Alex wins on the first throw.

2) With probability $(1-p)$ , Alex misses the triangle on the first throw. Conditioned on this miss, Ben will win with probability $p_w$ .

Thus $p_w = p + (1-p)(1-p_w).$ We may solve this for $p_w$ :

$p_w = \frac{1}{2 - p}$

(All numbers below are approximations.)

Plug in $p = 0.413497$ (the ratio of the area of an inscribed equilateral triangle to the area of the circle containing said triangle) and we get

$p_w = 0.630317$ .

This is the probability that Alex wins (remember: Alex goes first). The probability that Ben wins is $1 - p_w = 0.369683,$ which rounds to $0.37$

Sum of infinite series qp+qqqp+qqqqqp+... =q/(1+q)=0.3697

p=(area of triangle)/(area of circle)

=~0.4135

q=1-p

Without loss of generality, let the length of a side of the triangle be 2 units. An application of the Pythagorean theorem shows that the height of the triangle is sqrt(3), so that the (Area of the triangle) = (1/2) (Base) (Height) = sqrt(3). If R is the radius of the circle, another application of the Pythagorean theorem shows that (sqrt(3)- R)^2 +1^2 = R^2, so that R = 2/sqrt(3). Hence, the (Area of the Circle) = (Pi) R^2 = 4 Pi/3. Therefore, the probability of a player hitting inside the triangle on any throw is (Area of Triangle) / (Area of Circle) = 3 sqrt(3) / (4 (Pi)), which is approximately equal to .413497. Call this probability y. Then, the probability of missing on any given throw is n = 1 - y, which is approximately .5865.

Ben can win in any of the following ways. Alex can miss his first throw, then Ben can hit his first (with probability n y). Or, Alex can miss, Ben can miss, Alex can miss, then Ben can hit (with probability y (n^3)). Or, Alex can miss, Ben can miss, Alex can miss, Ben can miss, Alex can miss, then Ben can hit (probability y*(n^5)). And so on. Summing over all of these probabilities gives:

Probability Ben wins = y n + y (n^3) + y (n^5) + ... = n y/(1-n^2) = .37. The penultimate equality follows from summing the geometric series.

Let $P_a$ be Pr(Alex win), $P_b$ be Pr(Ben win)

$P_a + P_b = 1$

If Alex missed the first trial, then Ben will be first player on the game.

$P_b$ = Pr(Alex missed at first trial) * Pr(The first player win) = $P_m$ $P_a$

$P_b / P_m + P_b = 1 \Rightarrow P_b = \frac{P_m}{ 1 + P_m}$

$P_b = \frac{4\pi-3\sqrt(3)}{8\pi-3\sqrt(3)}$ = 0.3697