A Probability question: last ball in the bag Players are pulled up to pick a ball out of a hat containing $14$ red and $1$ blue. 
If the odds of drawing the blue ball are $1/15$ what are the odds of every person not drawing the blue ball and leaving it for the last person to draw? 
Initially I thought you would multiply the probability of not drawing the blue ball for each person and multiplying them together so i did 
$$\dfrac{14}{15} \times \dfrac{13}{14} \times \dfrac{12}{13} \times \cdots$$
But when I roughly calculate that, I get the same as $1/15$. 
Is this correct? 
 A: Yes, the answer is $\frac{1}{15}$, and there is a much easier way of reaching it by re-thinking the question. Essentially, the $14$ people drawing balls are uniquely determining which ball the final person will choose. Though the final person really makes no choice at all, the final person recieves a randomly selected ball. And from the final person's perspective, all balls are equally likely (which is important).
There are $15$ balls that can be left in the hat, one of them is blue, therefore the probability is $\frac{1}{15}$.
A: Rephrase it like this: 
There is a blue ball together with $14$ red balls. 
The balls will be numbered with $1,2,\dots,15$ randomly. 
So for a fixed ball the numbers have equal probability to become the number of that ball.
Now what is the probability that the blue ball will be numbered with $15$?
Yes, of course: $$\frac1{15}$$
A: The probability that you have calculated ($\frac {14}{15}*\frac {13}{14}*\frac {12}{13}*...*\frac {3}{4}*\frac {2}{3}*\frac {1}{2}*$) can be written rather neatly as $\frac {14!}{15!}=\frac 1{15}$ which is your answer.
A: The $14$ in $\frac{14}{15}$ is cancelled against the $14$ in $\frac{13}{14}$, and the $13$ in $\frac{13}{14}$ is cancelled against ... and so on. In the end, you're left with only the $15$ in $\frac{14}{15}$ and the $1$ in $\frac12$, so the final answer is indeed $\frac1{15}$.
