<s>[INST] Whose probability of winning is higher? There are $10$ blue marbles and $5$ red marbles in a black bag. Two players  $A$  and  $B$  take turns taking non-refundable (without replacement) $1$ marble each time (A takes first). The game ends when someone gets a red ball and that person loses.  Whose probability of winning is higher?
Please any help. I honestly don't know how to approach this one. Thank you.
 [/INST] You can calculate the winning probability of the person to move recursively as a function of the number of blue balls in the bag, since as long as the game is still going, the number of red balls will always be $\ 5\ $.  If $\ w_n\ $ is the probability that the person to draw will win when there are $\ n\ $ blue balls in the bag, then
\begin{align}
w_0&=0\ \ \text{ and}\\
w_n&=\frac{n\big(1-w_{n-1}\big)}{n+5}
\end{align}
for $\ n\ge1\ $.  This gives the probabilities listed in the following table
\begin{array}{c|c|}
n&w_n\\
\hline
1&\frac{1}{6}\\
\hline
2&\frac{2}{7}\left(1-\frac{1}{6}\right)=\frac{5}{21}\\
\hline
3&\frac{3}{8}\left(1-\frac{5}{21}\right)=\frac{2}{7}\\
\hline
4&\frac{4}{9}\left(1-\frac{2}{7}\right)=\frac{20}{63}\\
\hline
5&\frac{5}{10}\left(1-\frac{20}{63}\right)=\frac{43}{126}\\
\hline
6&\frac{6}{11}\left(1-\frac{43}{126}\right)=\frac{83}{231}\\
\hline
7&\frac{7}{12}\left(1-\frac{83}{231}\right)=\frac{37}{99}\\
\hline
8&\frac{8}{13}\left(1-\frac{37}{99}\right)=\frac{496}{1287}\\
\hline
9&\frac{9}{14}\left(1-\frac{496}{1287}\right)=\frac{113}{286}\\
\hline
10&\frac{10}{15}\left(1-\frac{113}{286}\right)=\frac{173}{429}\\
\hline
\end{array}
Thus, for the game given, with $\ 10\ $ blue marbles in the bag, the probability that the player going first wins is $\ \frac{173}{429}\approx0.403\ $. Before the game starts, the probability that the player going second will win is $\ \frac{256}{429}\approx0.597\ $, and so he or she is more likely to win than the player who goes first.
For what it's worth, you can obtain the expression
$$
w_n=\frac{(-1)^nn!}{(n+5)!}\sum_{j=1}^n\frac{(-1)^j(j+4)!}{(j-1)!}
$$
for the solution of the recursion for $\ w_n\ $.  However, using this expression to calculate $\ w_n\ $ is no more efficient than using the recursion directly.
</s>