Just to expand on what's been said, in case it's not clear.
"you have to consider the number of pairs of people, not just the number of people.
A group of 23 people can make 253 possible pairs. A can be paired with B, C, D and so on, for 22 pairs. B can be paired with C, D, E and so on, for 21 pairs. The process repeats until V can be paired with W, making only one pair. That makes the total 22+21+20 ... +2 +1. Any particular person in a group of 23 has only 22 chances to find another person in the group with the same birthday as them. When you take all of the possible pairs, there are 253 chances that one person will have the same birthday as one of the other 22. With an average of about 365.25 days per year, the chances of sharing a birthday are about 1 in 365.25, so the chances of not sharing a birthday are 364.25 in 365.25. The chance that none of the 253 pairs will share a birthday is 364.25/365.25 raised to the 253rd power, or 49.98%, leaving a 50.02% chance that there will be a pair with the same birthday.
It's exactly the same with the ever-increasing set of winning combinations. If there have been 3700 combinations drawn the first combination can match any of the next 3699. The 2nd can match any of the next 3698, and so on. You can add that up the long way: 3699+3698+3697 ... +3 +2 +1 to see that there are 6,843,150 pairs. You can also calculate it more easily as (3700*3699)/2 = 6,843,150. Making that generic, we get (n * (n-1))/2 = number of pairs for n drawings.
"I'm curious as to the origin of the formula [n*(n-1)/1,151,514] that calculates the expected total number of duplicates"
Notice that the formula starts out the same as the formula for calculating the number of pairs: n* (n-1). Instead of dividing by 2, it's divided by the number of possible combinations. That means the number of repeats expected as a result of simple probability is (number of pairs / number of combinations) * 2. In general, the higher the number of drawings the more likely it is that the actual number of repeats will closely match the expected number.
"I imagine the relationship is not a straight line but a curve that accelerate as the number of drawings increases."
n* (n-1) is the same as n^2 - n. The number of pairs is a function of the square of the number of drawings, so it's close to a geometric progression. After 100 drawings there will be 4,950 pairs. After 200 drawings there will be 19,900 pairs, or almost 4 times as many. Double it again, to 400 drawings and there will be 79,800 pairs, which is a bit more than 16 times as many as after 100 drawings. That means that the chances that the will be a repeat somewhere in the drawing history is an accelerating curve. The chances that a particular drawing or a particular set of drawings (such as the next 87) will produce a repeat is a different matter, and is a simple function of the percentage of combinations that have already been drawn. When the percentage of combinations that have already been drawn is very low the linear relationship that Murgatroyd describes will be true. As the percentage of combinations that have been drawn increases and repeats start occurring, the relationship will depart further and further from a straight line. At the extreme, if all P3 combinations have been drawn the chances of a repeat will remain the same even if the number of drawings increases by a factor of a million.