# How am I supposed to approach this probability question?

An urn contains 5 balls, one marked WIN and four marked LOOSE. You and another player take turns selecting a ball at random from the urn, one at a time. The first person to select the WIN ball is the winner. If you draw first, find the probability that you will win if the sampling is done (a)With replacement; (b) Without replacement.

My initial thought was that a would simply be 1/5. We're replacing the balls so every time we draw, it's like we have a clean slate, or as if we are only drawing one time. My professors solution sheet has the answer for part a) as 5/9. I have no clue how he got there. Part b) I honestly don't know where to start.

• Hint: consider "with replacement", and the general case of $n$ balls, one marked "win", and you go first. Then for $n=2$, you have a $0.5$ probability of immediately selecting the winning ball, so the other player gets no chance at all. There's also a $1-0.5=0.5$ probability you don't immediately win, followed by a $0.5$ probability the other player wins, so a $(1-0.5)*0.5=0.25$ probability for him on his first draw. So the first player to draw has an obvious advantage. Continue iterating that to 2nd draw, etc. And then consider the $n=5$ case. – John Forkosh Sep 21 '16 at 1:08

A)Probability of winning by drawing the first WIN ball = p = 1/5.
Probability of not winning by drawing a LOSE ball = q = 1-p.

Now, Probability of you winning, given that you draw first =
probability of you drawing the WIN ball in 1st attempt + probability of you drawing the WIN ball in second attempt*probability of the other person drawing a LOSE ball + so on...
p + (p)(1-q)^2 + p(q-1)^4 + so on...

which is the sum of a GP

B) probability of you winning = probability of selecting winning ball for the first time + probability of you selecting winning ball for the second time*probability of opponent drawing LOSE ball for the first attempt... so on till the number of balls left is 1.

These hints should be enough. I don't want to solve your homework for you.

Let $w$ be the probability that the first player wins the game, given that each single draw produces the win-ball with probability $p$.

The first draw results in an immediate win with probability $p$ ($={1\over5}$ in the case at hand), and in a momentary loss with probability $1-p$. But if in this second case the the other player draws a lose-ball as well the chances of the first player become completely intact again, whereas an instant win by the second player makes the first player's loss definitive. This story translates into the following equation for $w$: $$w=p+(1-p)(1-p) w\ ,$$ leading to $$w={1\over2-p}\qquad\bigl(={5\over9}\bigr)\ .$$ If the balls are not replaced you can argue as follows: Assume that the balls are put in a random order, and that each player has to take the leftmost remaining ball. Then the first player has three chances to get the win-ball, hence in this case $w'={3\over5}$.