How do you calculate the probability in a pick $5$ from $1$ to $39$ lottery game ($575,757$ possible combinations), that the next drawing will contain no repeated numbers from the previous THREE drawings? One repeated number? Two repeated numbers? Three repeated numbers? Four repeated numbers? All Five?
Keep in mind that the three previous drawings will produce a number set that can be anywhere from $5$ to $15$ numbers, but of course far more likely to be closest to $15$. Also, every drawing comes from a pool of $1$ to $39$. This conditional yet independent situation has me baffled.
First calculate the chance of each possible number of numbers in the previous three draws. You will have $15$ different numbers $\frac {34\cdot 33 \cdot 32 \dots 25}{(39\cdot 38 \cdot 37 \cdot 36 \cdot 35)^2}\approx 9.968\%$ of the time. For $n$ excluded numbers, you chance is $\frac {(39-n)!\cdot 34!}{39!\cdot (34-n)!}$. Multliply them together and then add.
Added: the chance that there are $13$ distinct numbers in the previous three draws comes from three parts: $0,1,2$ duplicates in the second draw. The chance of one duplicate in the second draw is $\frac {5\cdot 5 \cdot 34 \cdot 33 \cdot 32 \cdot 31}{39\cdot 38 \cdot 37 \cdot 36 \cdot 35}\approx 0.4027$, where the first $5$ is the number of positions of the match and the second is the number to match. The chance of exactly one duplicate in the next draw is $\frac {5\cdot 9 \cdot 30 \cdot 29 \cdot 28 \cdot 27}{39\cdot 38 \cdot 37 \cdot 36 \cdot 35}\approx 0.2380$. This way of getting $13$ is then a $9.58\%$ chance. The others are figured similarly.
Based on your last comment, it sounds like you want to beat the lottery based on avoiding numbers that have come up recently. To check this, it would be easier to separate the $3857$ draws into bins based on how many numbers had come up in the previous three weeks. Then you can just use that as data and not calculate the chance of $15$ different numbers coming up.
Let's assume there are $13$ different numbers that came up in the last three draws. The chance of not duplicating one this draw is $\frac {26\cdot 25 \cdot 24 \cdot 23 \cdot 22}{39\cdot 38 \cdot 37 \cdot 36 \cdot 35}\approx 11.425\%$ The chance of duplicating exactly one is $\frac {5\cdot 5 \cdot 26\cdot 25 \cdot 24\cdot 23}{39\cdot 38 \cdot 37 \cdot 36 \cdot 35}\approx 12.983\%$ You can check them against your data.