# Problem with my intuition concerning probabilities of dependent events and conditional probability

I'm having an issue understanding what's going on in the following problem.

Suppose we have a urn containing four distinct numbered balls. A match occurs if the $m^{th}$ ball is obtained on the $m^{th}$ draw.

Suppose I want to determine the probability that I will obtain a match on the second draw. Denote this event by $A_2$.

By conditional probability, we have $$\Pr(A_2) = \frac{3}{4}\cdot \frac{1}{3} = \frac{1}{4}$$

But the above probability doesn't make sense to me intuitively. I would think that the probability of $A_2$ is $\frac{1}{3}$.

Help?

Here are two ways to think about the problem . . .

The first way matches the calculations you had issues with.

Let $B$ be the event that the first ball drawn is ball $2$, and let $B'$ be the event that the first ball drawn is one of the other $3$ balls.

For the first draw, all $4$ balls are equally likely, hence $P(B)=\frac{1}{4}$, and $P(B')=\frac{3}{4}$.

If event $B$ occurs, $A_2$ can't occur, so $P(A_2|B)=0$.

If event $B'$ occurs, there are $3$ balls left, one of which is $A$, and all are equally likely, hence $P(A_2|B') = \frac{1}{3}$.

Then we get \begin{align*} P(A_2) &= P(B)P(A_2|B)+P(B')P(A_2|B')\\[4pt] &=\left({\small{\frac{1}{4}}}\right)(0)+\left({\small{\frac{3}{4}}}\right)\left({\small{\frac{1}{3}}}\right)\\[4pt] &={\small{\frac{1}{4}}}\\[4pt] \end{align*} In concept, in order for $A_2$ to occur, first $B'$ must occur, and next, given that $B'$ has occurred, ball $2$ must be the next ball of the $3$ remaining balls.

The other way is even simpler . . .

Ball $2$ must occur on some draw.

By symmetry, all draw orders of the $4$ balls are equally likely, so for $1\le k\le 4$, the probability that ball $2$ occurs on draw $k$ is $\frac{1}{4}$.

Hence $P(A_2)=\frac{1}{4}$.

The event that the second draw is a match consists of two mutually excluding events:

(A) The first draw is a match and the second draw is a match.

(B) The first draw is not a match and the result of the first draw is not the second thing and the second draw is a match.

For (A), we have $$\frac14\frac13,$$ that is we have to draw the first ball out of the four balls then we have to draw the second ball out of the three remaining balls.

For (B), we have $$\frac24\frac13$$

because we have to draw either the third or the fourth ball first and then the second ball out of the remaining three balls.

So, the result is

$$\frac1 {12}+\frac2 {12}=\frac14.$$

I would think that the probability of $A_2$ is $\frac14$.

The first ball makes the difference. Note that: \begin{align}P(A_2)=P(2_2)=&P(1_1\cap 2_2)+P(2_1\cap 2_2)+P(3_1\cap 2_2)+P(4_1\cap 2_2)=\\ &P(1_1)\cdot P(2_2|1_1)+P(2_1)\cdot P(2_2|2_1)+P(3_1)\cdot P(2_2|3_1)+P(4_1)\cdot P(2_2|4_1)=\\ &\frac14\cdot \frac13+\color{red}{\frac14\cdot 0}+\frac14\cdot \frac13+\frac14\cdot \frac13=\\ &\frac14.\end{align}

Similarly, $P(A_3)=\frac14$, because the first two balls make the difference.

Similar question: Three white and one red ball probability, where the ordered players must match the winning orders.