A Game that follows Hypergeometric Distribution? I have the following problem:

Inside a box there are $2$ white and $3$ black spheres. Two friends, $A$ and $B$ play the following game: They pick, one after another, a sphere from the box without putting it back in until they pick a white sphere. The first one that picks a white sphere wins the game. Player $A$ starts first. 
  
  
*
  
*What is the probability that player $A$ wins the game.
  
*If player $A$ wins the game then player $B$ has to pay him $\alpha$ Euro, while if $B$ wins the game, $A$ has to pay him $\alpha+2$ Euro. Define $\alpha$ suh as the game is fair (meaning that winnings for each player are equal to $0$.)

Now I think that I should use the Hypergeometric distribution for this one but once again, I do not know how to define the necessary parameters in order to solve the problem. If anyone could assist me in this one I would be grateful. Thank you.
 A: I think you can avoid the formulation via the hypergeometric distribution and calculate it directly. The ways that $A$ wins the game ar the following:


*

*$A$ draws directly in the first draw a white sphere, or

*$A$ draws a black sphere but $B$ draws also a black sphere and then $A$ draws a white sphere, or

*$A$ draws a black sphere but $B$ draws also a black sphere, then $A$ draws again a black sphere and oops, there are only white spheres left, so $B$ draws a white sphere and wins for sure. So $A$ can win only in the first 2 ways listed above!


Now, the probability that $A$ wins, say $P(A)$, is equal to the sum (or translates to $+$ in probabilities) of the probabilities of the first $2$ scenarios above is:
$$P(A)=\frac{2}{5}+\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}=\frac{3}{5}$$ So $A$ wins with a probability of $\dfrac35$ and $B$ with a probability of $1-\dfrac35=\dfrac25$ (there are no ties). So, the expected winnings of $A$, say $E[A]$ are $$E[A]=α\frac35-(α+2)\frac25$$ Set $E[A]=0$ so that the expected winnings (and not winnings as you have it in your post) are equal to zero (fair game) and will find that this is the case for $α=4$.
