Probability that the second number picked is greater than the first Given the set {1,2,3,4,5} we randomly pick one element. Without replacing(returning) the number we picked, we pick another random element. Find the probability that the second number picked is greater than the first.
 A: Yes, we can find the probability and it is $\frac 12$.  The pedestrian way is to list all the pairs, of which there are $20$, and count that $10$ of them have the second number greater.  You can also argue by symmetry that the first number is just a likely to be the greater as the second to show that the answer is $\frac 12$.  
You can certainly compute the probability that the second is greater conditioned on what the first is.  The probability that the second is greater given that the first is $2$ is $\frac 34$ because $3$ of the $4$ remaining numbers are greater than $2$.  This is another route to come up with the overall $\frac 12$ answer.  You can average this conditional probabilty over all the possible first numbers.
A: We have that


*

*$E_1=1 \implies P_1(E_2>E_1)=1$

*$E_1=2 \implies P_2(E_2>E_1)=\frac34$

*$E_1=3 \implies P_3(E_2>E_1)=\frac12$

*$E_1=4 \implies P_4(E_2>E_1)=\frac14$

*$E_1=5 \implies P_5(E_2>E_1)=0$


therefore
$$P(E_2>E_1)=\frac15 \cdot \left(1+\frac34+\frac12+\frac14+0\right)=\frac12$$
A: Let call $X$ the first number and $Y$ the second number you pick.
If $X = 1$, there are 4 possibilities for $Y$ that all work.
If $X = 2$, there are still 4 possibilities. But if $Y=1$, then it doesn't work, so only 3 work.
If $X = 3$, only 2 work.
If $X = 4$, only 1 work.
If $X = 5$ only 0 work.
The number of possibilities that work is $\sum_{k=0}^4 k = 10$.
The number of all possibilities is $4*5 = 20$.
The probability is $10/20 = 50\%$.
A: The easiest way to see that the answer is .5 is to realize that we are picking two non-equal numbers A and B and there is no reason to suppose that we pick A first and B second rather than B first and A second.  Thus A > B and A < B are both equally likely.
This observation also has the advantage of extending to a set of numbers of any size.  Provided none of the numbers repeat, it doesn't matter if your set contains 5 numbers or 50 numbers or 500 numbers -- the probability that the second number is greater than the first is always 50%.
A: 
Illustrated above are all the possible ways you could select two numbers from {1,2,3,4,5} without replacement.  (The diagonal is excluded because it requires picking the same number twice).
Of the 20 possible selections, 10 have the second choice as higher than the first, and 10 are the other way around.  Hence, the probability is 10/20 = 50%
It is also trivial to see that this percentage will hold true for any ordered set of any size (assuming there are at least two elements), due to the symmetry of the diagram.
