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.
 A: In any given round, $A$ goes first with an even number $b \geq 0$ of blue marbles and five red marbles in the bag.
Given the game reaches this round, it proceeds in one of three ways:

*

*$A$ draws a red marble with probability $\frac5{b+5}$. $B$ wins.

*$A$ draws a blue marble with probability $\frac{b}{b+5}$, and $B$ draws a red marble with probability $\frac5{b+4}$.  $A$ wins.

*$A$ draws a blue marble, $B$ draws a blue marble, and then another round occurs with fewer blue marbles in the bag.

That is, the probability of $B$ winning in this round is $\frac5{b+5}$ and the probability of $A$ winning in this round is $$\frac{b}{b+5}\cdot\frac5{b+4} = \frac5{b+5}\cdot \frac{b}{b+4} < \frac5{b+5}.$$
The game must end at some round and the probability that $A$ will win in any particular round is always strictly less than the probability $B$ wins in that round, so we may conclude that the probability that $A$ wins this game is strictly less than the probability $B$ will win.
A: Two approaches suggest themselves to me.

*

*An iterative solution: $A$ looses a game with 10 blue and 5 red marbles if (1) $A$ draws red on the first draw, or (2) $A$ and $B$ both draw blue and then $A$ looses a game with 8 blue and 5 red marbles.  And so on.


*A combinatorial solution: $A$ losses if the first red marble is in an odd position down the drawing queue.  $B$ wins if the first red marble is in an even position.
You just need to compare (1) probabilities of winning in each successive round, or (2) series of binomials without actually calculating their values.
A: It is obvious that B wins. The first ball has a 1/3 chance of being red, so 1/3 of the time A loses. The second ball is "about" 1/3 chance of being red, so then about 2/9 of the time B loses at the first take, which is less than A's chance. Then similar for the next round. Anyway, it's "obvious" that B wins this game most of the time.
You can work out the probabilities like
$$
P(A)={5\over15}+{10\over15}{9\over14}\left({5\over13}+{8\over13}{7\over12}\left({5\over11}+{6\over11}{5\over10}\left({5\over9}+{4\over9}{3\over8}\left({5\over7}+{2\over7}{1\over6}\right)\right)\right)\right)\\
$$
and similar for $P(B)$
and this comes out to about 0.597 for $P(A)$ and 0.403 for $P(B)$.
I didn't work out a tidy format for this but just computed it from that formula.
Code to calculate
A: 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.
