Stumped on basic statistics question. The question:
Alfonso and Colin each bought one raffle ticket at the state fair. If $50$ tickets were randomly sold, what is the probability that Alfonso got ticket $14$ and Colin got ticket $23$?
According to this, the answer should be $ \frac{1}{2450}$ which presumably comes from $\frac{1}{50}\times \frac{1}{49}$. But it seems that the order does not count. I did not assume that Alfonso got ticket $14$ first then Colin got ticket $23$ second.
Update: What is wrong with this reasoning.
When I said that I did not assume order, I meant that it's possible


*

*Alfonso got ticket $14$ first, then  Colin got ticket $23$,

*Colin got ticket $23$ first, then Alfonso got ticket $14$.


Both of these possibilities are possible before the tickets are given out, so we can make an 'or' statement.
Label the event Alfonso got ticket $14$ by $A_{14} $ and Colin got ticket $23$ by $A_{23}$. Then by the addition rule
$ \Pr(\text{ ($A_{14}$ first and $C_{23}$ second)  or  ($C_{23}$ first and $A_{14}$ second}) ) 
\\ = \Pr(A_{14}) \times \Pr(C_{23} \mid A_{14}) + \Pr(C_{23}) \times \Pr(A_{14}\mid C_{23}) = \frac 1 {50} \times \frac 1 {49} \times 2.$
I realize that once the tickets are sold, then only one of $ \{ A_{14}C_{23}~ , ~ C_{23}A_{14} \}$ must occur, but before the tickets are sold both possibilities are plausible. Why would the probability change before and after the tickets are sold.
 A: It's not about the order in which they get the tickets, it's about who gets which ticket. When you write $1\over{}_{50} C_2$, you are saying that there is only one acceptable pair of tickets out of ${}_{50} C_2$ possible pairs. But after you choose the correct pair, you still have to say who gets ticket $14$ and who $23$, and there is only one way to do that correctly out of two possibilities.
A: "I did not assume that Alfonso got ticket 14 then Colin got ticket 23" - order in this sense has nothing to do with time.  It is a question of whether or not

Alfonso got ticket 14 and Colin got ticket 23

is different from

Alfonso got ticket 23 and Colin got ticket 14.

And clearly it is different, and only the first is asked for in your problem.  There are $50\times49$ ways in which tickets could be allocated to A and C, with order (in the sense I have just explained) being important; and only one successful case; so the probability is
$$\frac1{50\times49}\ .$$
A: To say that order does not count means there is no difference between the following:


*

*Alfonso got ticket $14$ and Colin got ticket $23$,

*Alfonso got ticket $23$ and Colin got ticket $14$.


You found the probability that between them they got a certain set of two tickets. After they get those two tickets, there still the choice of which gets which, and there are two choices, each with (conditional) probability $1/2.$
You can also look at it like this:
$$
\Pr(\text{Alfonso got }\#14) \times \Pr(\text{Colin got }\#23\mid \text{Alfonso got }\#14) = \frac 1 {50} \times \frac 1 {49}.
$$
