# Elementary Probability Question

So the problem is in Tanis & Hogg, Probability and Statistical inference section 1.5 Independence of Events.

An Urn contains five balls, one marked WIN and four marked LOSE. 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 with replacement.

I can't see why the probability isn't $\displaystyle\frac{1}{5}$. The book states it is $\displaystyle\sum_{k=0}^\infty(\displaystyle\frac{1}{5}\times\displaystyle\frac{4}{5}2k)$

Thanks!

-
The probability is $\frac{1}{5}$ that you win on the first turn. But if neither you nor your opponent win on your first turns, you might still win on the second turn, etc. – Qiaochu Yuan Jun 7 '12 at 18:58

First lets find the probability that you win on your $k^{th}$ draw. Let us denote it by $p_k$. Then the probability you win is given as $\displaystyle \sum_{k=1}^{\infty} p_k$.

The probability you win on your $k^{th}$ draw can be computed as follows. If you have to win on your $k^{th}$ draw, then in each of your previous $k-1$ draws, you should have chosen one of the balls marked LOSE else you would have won before your $k^{th}$ draw. Your opponent must have also chosen one of the balls marked LOSE in each of his $k-1$ draws else he would have won before you. In each draw, the probability that a person chooses one of the balls marked LOSE is $\dfrac45$ and the probability that a person chooses the ball marked WIN is $\dfrac15$. Hence, \begin{align} p_k & = \underbrace{\left(\dfrac45 \times \dfrac45 \times \cdots \times \dfrac45 \right)}_{(k-1) \text{ times}}_{\text{You lost your first $k-1$ draws}} \times \underbrace{\left(\dfrac45 \times \dfrac45 \times \cdots \times \dfrac45 \right)}_{(k-1) \text{ times}}_{\text{Opponent lost his first $k-1$ draws}} \times \underbrace{\dfrac15}_{\text{You chose the right ball on your $k^{th}$ draw}}\\ & = \left(\dfrac45 \right)^{2k-2} \times \dfrac15 \end{align} Hence, the probability you win is $$\sum_{k=1}^{\infty} p_k = \sum_{k=1}^{\infty} \left(\dfrac45 \right)^{2k-2} \times \dfrac15 = \dfrac15 \left( \sum_{k=1}^{\infty} \left(\dfrac{16}{25} \right)^{k-1}\right) = \dfrac15 \left( \dfrac1{1-\dfrac{16}{25}} \right) = \dfrac59$$

-

Your question about the series has already been well answered. The calculation below gives an alternate way to find the probability that you win the game. In effect it bypasses summing the series.

Let $p$ be the probability that you (ultimately) win. This can happen in two ways: (i) you draw a WIN immediately or (ii) you draw a LOSE, so does your opponent, but you ultimately win.

The probability of (i) is $\frac{1}{5}$. For the probability of (ii), the first two picks must be LOSE, LOSE (probability $\left(\frac{4}{5}\right)^2$). Given that the first two picks are LOSE, LOSE, the probability you ultimately win is $p$, since it is the same as at the start. Thus the probability of (ii) is $\left(\frac{4}{5}\right)^2p$. We thus obtain the equation $$p=\frac{1}{5}+\frac{16}{25}p.$$ Solve this linear equation for $p$. We get $\frac{9}{25}p=\frac{1}{5}$, and therefore $p=\frac{5}{9}$.

-
+1 @Andre... The concept the probability you ultimately win is p, since it is the same as at the start seems to be a very useful one. (Did not think about this!) – Stat-R Jun 7 '12 at 21:55

The probability of you winning is equal to the probability that you pick the WIN ball at the first draw (with probability $(1/5)$) or you pick the LOSE ball, your opponent also picks the LOSE ball and on the next round you pick the WIN ball (with probability $(1/5)*(4/5)*(4/5)$) and so on, thus probability of you winning is $$p=(1/5)+(4/5)*(4/5)*(1/5)+(4/5)*(4/5)*(4/5)*(4/5)*(1/5)+...$$

-

As @Qiaochu pointed out... you need to account for all the possible ways you can win and then add those probabilities.

Let p = event that you won , q= event that you lose => q=1-p

You should either win on the first draw and if not then the other player should lose and then on the third draw you should win. This means you should win on the 1,3,5..... draws and the opponent should lose on the 0(if you draw the WIN ball on the first turn),2,4,6.... draw.

=> P(you win) = $p+pqp+pqpqp+pqpqpqp..... = p+q^2p+q^4p+q^6p+ ...$

This is an infinite geometric series with each subsequent element getting multiplied by $q^2$.

Therefore, P(you win) = $a/(1-r)$=$p/(1-q^2)$ =$5/9$

here, a= $p$, and $r=q^2$

-

As someone mentionned in a comments, 1/5 is the probability that you win on your first draw. What is the probability that you win on the second draw? You have to both lose first, hence the (4/5)^2 and then you have to win. Then you want to know what is the probability that you win on your 3rd, 4th etc. draw. Summing all these will give you your chance of winning the game

-