Suppose that there are n possible outcomes of a trial, with outcome $i$ resulting with probability $pi,i=1,...,n,\sum _{i=1}^{n}i=1pi=1$. If two independent trials are observed, what is the probability that the result of the second trial is larger than that of the first?

Trials with results in $\{1,2,\dots ,n\}$.

$P\left(\text{result is}i\right)=p_i$.

Two independent trials are preformed, with results$(i,j)$.

Firstly, since the trials are independent:

$P\left[\right(i,j\left)\right]=p_i·p_j$

Therefore

$P\left[\right(i,j\left)\right]=P\left[\right(j,i\left)\right]$

for every $i,j\in \{1,2,\dots ,n\}$

Now since all $(i,j)$for different pairs $i$and $j$are mutually exclusive events:

$P\left[\right\{(i,j);i<j\left\}\right]=\sum _{1\le i<j\le n}P\left[\right(i,j\left)\right]=\sum _{1\le i<j\le n}P\left[\right(j,i\left)\right]=P\left[\right\{(i,j);j<i\left\}\right]$

And since the result of the two trials can be either that the first result is larger, that the second result is larger or that they are both same (and those possibilities are mutually exclusive)

And that third probability is:

$P\left[\right\{(i,i),i\in \{1,2,\dots n\}\left\}\right]=\sum _{i=1}^{n}P\left[\right(i,i\left)\right]=\sum _{i=1}^n{p}_i^2$

Equation (1) states that

$P\left[\right\{(i,j);i<j\left\}\right]=P\left[\right\{(i,j);j<i\left\}\right]$

Therefore the red equation gives:

This transforms to:

$P\left[\right\{(i,j);i<j\left\}\right]=\frac{1-{\sum }_{i=1}^{n}P\left[\right(i,i\left)\right]={\sum }_{i=1}^n{p}_i^2}{2}$

The probability that the first result is larger is equal to the probability that the second result is larger.

Probability that they are equal is $\sum _{i=1}^n{p}_i^2$.