Ruin time for a two-input "risk only" slot machine Imagine a "risk only" slot machine that takes 'coins' corresponding to some real number fraction of a dollar $p$, returns the coin with probability $p$, and eats the coin with probability $(1-p)$.  For example, a dime would be eaten with a probability of 90%, a nickel with probability 95%, and so forth.
So let's keep feeding the machine two kinds of coins, $A$ and $B$, with fractional dollar values of $p_A$ and $p_B$, respectively.  I have $n_A$ coins of type $A$ and $n_B$ coins of type $B$.  Each time I use the slot machine, I randomly select a coin, ignoring its type, and place it in the machine.  I stop feeding coins into the machine when I run out of either type.  
CLARIFICATION - By "randomly select a coin" I mean that we select a coin from the population of all coins uniformly and randomly.  For instance, if we have $100$ dimes and $567$ nickels, we'd draw a dime with probability $\frac{100}{667}$.
At this stopping point, what is the probability of ending with only coins of type A or only coins of Type B?  Provided we end with coins of one type / denomination, what probability distribution and expectation do we have for the number of remaining coins of this type / denomination?
I'd also be curious on the number of coins of either type we needed to feed to the machine to reach this end-state?  E.g. how many times did we feed the machine a dime, and how many times did we feed the machine a nickel before stopping?
**
If it helps, I can provide some simulation data.  For example, starting with $100$ dimes and $100$ nickels:
$n_A = 100$
$n_B = 100$
$p_A = 0.10$
$p_B = 0.05$
We achieve the following results for $10^4$ trials:
The mean number of times we place a dime in the machine $= 109.721$ 
(Median $ = 110$)
The mean number of times we place a nickel in the machine $= 104.42$
(Median $ = 104$)
The number of times we end with only dimes: $5669$
The number of times we end with only nickels: $4331$
The average number of dimes at the end state (conditioned on running out of nickels first): $2.18328$ 
(Median $= 2$)
The average number of nickels at the end state (conditioned on running out of dimes first): $1.80513$ 
(Median $= 1$)
**
Let's do another simulation starting with $82$ copies of hypothetical 75 cent coins and $432$ copies of 5 cent nickels, and again perform $10^4$ trials:
$n_A = 82$
$n_B = 432$
$p_A = 0.75$
$p_B = 0.05$
We achieve the following results for $10^4$ trials:
The mean number of times we place a 75 cent coin in the machine $= 268.213$ 
(Median $ = 267$)
The mean number of times we place a 5 cent nickel in the machine $= 454.734$
(Median $ = 455$)
The number of times we end with only 75 cent pieces: $9999$
The number of times we end with only 5 cent nickels: $1$
The average number of 75 cent coins at the end state (conditioned on running out of 5 cent nickels first): $14.9384$ 
(Median $= 15$)
The average number of nickels at the end state (conditioned on running out of dimes first): $1$ 
(Median $= 1$)
 A: The setting is ripe for Poissonization... First consider the path followed by the system, that is, denoting by $X$ the number of coins of type A left and by $Y$ the number of coins of type B left, the set of states visited by the process $(X,Y)$.
Assume that $(X,Y)=(x,y)$. One draws a coin of type A with probability $x/(x+y)$ and the machine eats it with probability $$a=1-p_A,$$ hence $(X,Y)$ moves to $(x-1,y)$ in one step with probability $ax/(x+y)$. Likewise, $(X,Y)$ moves to $(x,y-1)$ in one step with probability $by/(x+y)$, where $$b=1-p_B,$$ otherwise $(X,Y)$ stays at $(x,y)$ in one step. Thus, the next state visited by $(X,Y)$ is $(x-1,y)$ or $(x,y-1)$ with probabilities proportional to $(ax)$ and $(by)$ respectively.
Here is a process in continuous time with the same paths. Assume that each coin $i$ of type A has a lifetime $T_i$, exponential with parameter $a$, and that each coin $j$ of type B has a lifetime $S_j$, exponential with parameter $b$. All the lifetimes are independent. At any given time, assuming that $(x,y)$ coins are still alive, the next coin to die is of type A with probability proportional to $(ax)$ and of type B with probability proportional to $(by)$.
Thus, both processes starting from $(x,y)$ hit the line $Y=0$ (no coin of type B left) before hitting the line $X=0$ (no coin of type A left) when $T\gt S$, with
$$
T=\max\{T_i\mid1\leqslant i\leqslant x\},\qquad S=\max\{S_j\mid1\leqslant j\leqslant y\}.
$$
To compute $P(T\gt S)$, note that $P(S_j\lt t)=1-\mathrm e^{-bt}$ for every $t\gt0$ and every $j$, hence $$P(S\lt t)=(1-\mathrm e^{-bt})^y.$$ Likewise, $$P(T\lt t)=(1-\mathrm e^{-at})^x.$$ This allows to compute the density of $T$, which yields
$$
P(T\gt S)=\int_0^\infty xa\mathrm e^{-at}(1-\mathrm e^{-at})^{x-1}(1-\mathrm e^{-bt})^y\mathrm dt.
$$
Equivalent formulas are
$$
P_{x,y}(\text{last coin of type A})=\int_0^1 x(1-u)^{x-1}(1-u^{b/a})^y\mathrm du,
$$
and
$$
P_{x,y}(\text{last coin of type A})=\int_0^1(1-(1-u^{1/x})^{b/a})^y\mathrm du.
$$
The first example in the question is when $x=y=100$, $a=0.9$, $b=0.95$, then $P(\text{last coin of type A})\approx0.565953$.
The second example in the question is when $x=82$, $y=432$, $a=0.25$, $b=0.95$, then $P(\text{last coin of type A})\approx0.999634$.
One can continue with this approach to get the distribution of the number of coins at the end and of the number of times one places a coin in the machine. A useful remark to do so is that the continuous time processes $(X_t)$ and $(Y_t)$ describing the numbers of coins of each type still alive at time $t$ are actually independent, $X_t$ jumping from state $x$ to state $x-1$ at rate $(ax)$ and $Y_t$ jumping from state $y$ to state $y-1$ at rate $(by)$.
