Age distribution when meeting I have a question regarding Poisson process. 
I will tell the story in the context of a player-monster game.
Consider a player who is born at $t=0$. He will win the game if he can survive until $t=1$. Time is continuous. At each instant, he could be stopped by a monster with Poisson rate $\lambda$. Then he enters a fight. The monster could win with probability $p$. If so, the player loses one life. 
The player has 2 lives in total. So he can lose only once. 
Assume I am one of the monsters. Conditional on I meet the player, can I infer the probability distribution of the player's progress in terms of time $t$? If so, what's that?
If the player is beaten by me this time, what's the probability he will win?
Please let me know if my question is well defined or I miss something important to make it do-able.
Thanks a lot!
^_^
 A: The player meets monsters at the rate $\lambda$, and with probability $p$ they are winning monsters, so the player meets winning monsters at the rate $\lambda p$. We can extend the game up to $t=1$, with monsters appearing after the player's death meeting a dead player. Then, since the Poisson monsters appear uniformly in time, the probability distribution over $t$ that you're looking for is proportional to the probability that the player is still alive at time $t$. This is the probability that he has not yet encountered two winning monsters at rate $\lambda p$, which is $\mathrm e^{-\lambda pt}(1+\lambda pt)$. Normalizing over $[0,1]$ yields the probability density
$$
f(t) = \frac{\mathrm e^{-\lambda pt}(1+\lambda pt)}{\int_0^1\mathrm e^{-\lambda ps}(1+\lambda ps)\,\mathrm ds}\;.
$$
If the player is beaten by a monster, the probability that he will win is the probability that he hadn't been beaten yet, which is
$$
\frac{\mathrm e^{-\lambda pt}}{\mathrm e^{-\lambda pt}(1+\lambda pt)}=\frac1{1+\lambda pt}\;,
$$
times the probability that he won't be beaten for the rest of the game, which is $\mathrm e^{-\lambda p(1-t)}$. Integrating this product with the density yields
$$
\int_0^1f(t)\frac1{1+\lambda pt}\mathrm e^{-\lambda p(1-t)}\mathrm dt=\frac{\mathrm e^{-\lambda p}}{\int_0^1\mathrm e^{-\lambda ps}(1+\lambda ps)\,\mathrm ds}=\frac{\lambda p}{2\mathrm e^{\lambda p}-2-\lambda p}\;.
$$
