Game strategy with probability Consider the following game. I have two bells that ring independently at random times. Bell
A’s rings follow a Poisson process with mean rate 2 per minute. Bell B’s rings follow a Poisson process with mean rate 1 per minute. Every time bell A rings, I pay you \$1. You can shout “stop” at any time ending the game — you keep your winnings. But if Bell B rings before you shout stop, you must give the money back to me and the game ends. For example, if Bell B rings before Bell A, you get paid nothing and the game is over.
(a) Suppose your strategy of play is to stop when you get the ﬁrst dollar. What are your expected winnings?
(b) Suppose your strategy is to stop when you get \$2. What are your expected winnings?
(c) Is there a better strategy?
For (a), I can condition on which bell rings first and then calculate the expected winnings, but how to (b) and (c)?
 A: You can calculate the probability of $B$ ringing at time $t$. You can also calculate the probability of $A$ ringing twice before time $t$. Then, integrate the product of these probabilities from $t=0$ to $t=\infty$.
A: Let $N_t$ the number of times the bell rang during time $t$ (in minute). As it is a Poisson process of mean $\lambda$ we deduce that (with $\lambda=2$ for bell A or $\lambda=1$ for bell B)
$$P(N_t=k)=e^{-\lambda.t}\frac{(\lambda.t)^k}{k!}$$
Let $T_i$ the time of the i(th) ring of the bell. If the bell ring $i$ times before time $t$, at time $t$, the bell rang at least $i$ times. Hence
$$P(T_i\le t)=P(N_t\ge i)=\sum_{k=i}^{\infty}P(N_t=k)$$
In particular $P(T_1\le t)=1-P(N_t=0)$
The PDF of $T_1$ will then be :
$$f(T_1)=\lambda.e^{-\lambda.t}$$
So $T_1$ has an exponential distribution. First strategy :
$$P(T_1^A\le T_1^B)=\int_0^\infty P(T_1^A\le t)f_B(t)dt=\int_0^\infty (1-e^{-2t})e^{-t}=\frac{2}{3}$$
You mean gain is $\frac{2}{3}$\$.
Second strategy :
$$P(T_2^A\le T_1^B)=\int_0^\infty P(T_2^A\le t)f_B(t)dt=\int_0^\infty (1-e^{-2t}(1+2t))e^{-t}=\frac{4}{9}$$
You mean gain is $\frac{8}{9}$\$. This is better.
You can verify that $$P(T_{i+1}^A\le T_1^B)=P(T_i^A\le T_1^B)-\frac{2^i}{3^{i+1}}=1-\frac{1}{3}\sum_{k=0}^{i}(\frac{2}{3})^i=\left(\frac{2}{3}\right)^{i+1}$$
Then you can verify that the best strategies are : wait to gain $2$\$ or wait to gain $3$$ (same mean for both, but the second is more risky)

You have another strategy : Wait until time $t$. 
You'll win $E(N_t^A)$ with probability $P(N_t^B=0)$. hence Your mean gain is :
$$G_t=E(N_t^A).P(N_t^B=0)=2t.e^{-t}$$
$G_t$ is maximum for $t=1$ and $G_1=\frac{2}{e}\approx0.7358$\$
Too bad, this is worst than waiting for $2$\$ !
