probability of blue 3 times before red 4 times hi: interesting probability question here. you have a bag with a red ball and a blue ball in it. The rules of the game are
1) At each trial,  a ball is taken out of the bag without looking by an unbiased person.
2) At each trial, after the ball is taken out of the bag, it is returned to the bag.
3) At each trial,  the color of the chosen ball is noted.
The probability question is: What is the probability that a blue ball is picked 3 times BEFORE a red ball is picked 4 times.
 A: The players are Alicia and Beti. Alicia wins if a blue is picked $3$ times before a red is picked $4$ times. We assume the game is played until a winner is determined. 
Change the game slightly. In any case, we will play $6$ times, and Alicia wins if blue occurs $3$ or more times. Alicia wins the modified game if and only if she wins the original game. 
The probability she does not win the modified game is $\frac{\binom{6}{0}+\binom{6}{1}+\binom{6}{2}}{2^6}$. 
We could also take advantage of symmetry to get a somewhat simpler expression. For if $p$ is the probability that Alicia loses, then $p+\frac{\binom{6}{3}}{2^6}+p=1$. From this we get that the probability Alicia wins is $\frac{1}{2}+\frac{\binom{6}{3}}{2^7}$. This would be useful if we replace the $3$ and $4$ of the problem by say $20$ and $21$.  
Added: A later post claims this answer is wrong. Calculate. We find that the probability Alicia loses is $\frac{22}{64}$, which is $\frac{11}{32}$. The probability she wins is therefore $\frac{21}{32}$. That is the same number as the one obtained by  a more complicated method. 
A: It can be recognized as the event that at the first $6$ trials at least $3$ blue balls are taken out. So its probability equals $P[X\geq 3]$ where $X$ has binomial distribution with $n=6$ and $p=\frac{1}{2}$.
A: Lets solve it for the general case: what is the probability $n$ blue balls are picked before $m$ red balls.
If we denote by $p(k)$ the words consisting of $k$ letters $r$ and $n$ letters $b$ that end in $b$ then the probability we are looking for is 
$$\sum_{k=0}^{m-1}\frac{p(k)}{2^{n+k}}$$
but what exactly is $p(k)$? well the last letter $b$ is fixed so we have $n-1$ letters $b$ and $k$ letters $r$ we can order however we want. So $p(k)=\binom{n+k-1}{n-1}$.
So the probability is $$\sum_{k=0}^{m-1}\frac{\binom{n+k-1}{n-1}}{2^{n+k}}$$
