Probability to win a chance game The game is quite simple, let's put it in marbles terms :
There's a bag of 25 marbles, 1 is white.
Each user picks one, and doesn't put it back.
I've figured the probability of each pick to be the winning pick, but I'm struggling to figure the probability of a game to be won after N clicks. My math lessons are very far away...
So :
Pick 1 : 4% chance to pick the white marble
Pick 2 : 4.17%
...
Pick 24 : 50% chance to pick the white marble.
What's the probability of the game being won after 10 picks?
I know I can't just add all the probas I've calculated, but I'm running out of ideas and don't have the vocabulary to ask google properly.
 A: I'm guessing pick white means you won. 
So $A = \{\text{Win on $i$th draw}\}$,then
for example, end on second draw is that you got a non white on the first draw and a white on the second. This is
$$\frac{24}{25}\cdot\frac{1}{24} =\frac{1}{25}$$
For end on third draw, you draw two non white twice, then the winner. This is
$$\frac{24}{25}\cdot\frac{23}{24}\cdot\frac{1}{23} = \frac{1}{25}.$$
Following, similar logic gives
$$P(A) = \frac{24!\cdot(25-i)!\cdot1}{(24-(i-1))!\cdot 25!} = \frac{1}{25}.$$
Thus, the game is equally likely to end on any draw.

If you are asking for the probability that the game takes more than 10 turns, well that's the same as saying that is fails to end in the first 10 turns. Thus
$$P(\text{Fail first 10}) = \frac{24\cdot23\cdot22\dotsb 16\cdot15}{25\cdot24\cdot23\dotsb 17\cdot 16} = \frac{15}{25} = \frac{3}{5}.$$
A: With $M$ marbles. 
End at time 1 is $\frac{1}{M}$. 
End at time 2 is: we did not end at time 1 and we ended at time 2: 
$(1-\frac{1}{M})\cdot\frac{1}{M-1}=\frac{M-1}{M}\cdot\frac{1}{M-1}=\frac{1}{M}$
End at time 3: 
$(1-\frac{1}{M})(1-\frac{1}{M-1})\frac{1}{M-2}=\frac{1}{M}$ 
and so on until we hit time $M$.  
A: The game is equally likely to end at Pick $1$, Pick $2$, Pick $3$, and so on.  So the probability it ends at or before Pick $10$ is $\dfrac{10}{25}$.
To see that the game is equally likely to end at any pick, imagine that the balls are arranged in a line at random, with all positions for the white equally likely. Then the balls are chosen in order, from left to right. It is clear that the white ball is equally likely to be in positions $1$, $2$, $3$, and so on.
It looks as if you got the $4.17\%$ by calculating $\dfrac{1}{24}$. This is the probability that the second pick is white, given that the first ball was not white.
To use conditional probabilities to find the probability the second pick is white, note that if the first is white, the probability is $0$, while if the first ball is not white, the probability is $\dfrac{1}{24}$. Thus the probability the second pick is white is
$$\frac{1}{25}\cdot 0+\frac{24}{25}\cdot \frac{1}{24}.$$
This simplifies to $\dfrac{1}{25}$, exactly the same as the probability Pick $1$ is white.
We could use a similar conditional probability argument to show that the probability Pick $3$ is white is $\dfrac{1}{25}$. But this is the hard way of doing things. The easy way is described in the first two paragraphs.
A: Taking win to be drawing the white ball,
if you are asking for P(win on 10th pick), it is simply $\dfrac1{25}$,
because colors have no preference for positions.
If you are asking for P(win after 10th pick)= 1 - P(win in first 10 picks),
and since winning on any pick has the same probability, 
$Pr = 1 - \dfrac{10}{25} = \dfrac35$
