Calculating Conditional Probability of Survival

#1
I’m looking to analyse a Survival data event relating to lottery draw data. The event being is the time taken for one of the six winning numbers to be drawn again.
A table of the data is shown as attached
I’d be looking to say, find the conditional probability of this event happening in the next draw, given that ‘failure’ hasn’t occurred for 2 draws.
Could you please advise the best technique to evaluate the likelihood ?
 

Attachments

#4
It would be any one of the six numbers. I had thought of cox regression but it didn't seem like the correct scenario although i'm happy to be proved wrong
 
#5
I’d be looking to say, find the conditional probability of this event happening in the next draw, given that ‘failure’ hasn’t occurred for 2 draws.
It's not really clear to me what you are looking for.
As I read it, you have six numbers drawn originally. In the next two draws, none of these original numbers appeared again. Given that none have appeared for two draws, what is the probability that one (at least) of the original numbers will appear in the third draw?
Have I interpreted this correctly? kat
 
#6
Or do you mean you have six numbers drawn originally. What is the probability that at least one of these numbers appears some time in the next three draws?
or something else?
 
#8
Hi Kat, For example, if none of today's six winning numbers are drawn in the next draw and then none of the following group of six winning numbers are drawn in the next draw, then what is the probability of the third group of winning numbers having one hit in the following draw. So essentially, each group of six winning numbers is viewed as a cohort.
 
#9
Each day is independent, so the probability of a hit on the third day does not depend in any way on whether or not there have been hits on the two previous days.
So what is the probability of getting at least one hit on the third (or any particular) day?
I'm assuming here that there are 40 numbers as there are where I live. The probability of no match for the first number drawn is 34/40 because there are 40 numbers of which 34 do not match. For no match for the second 33/39 because one number has been used up. For the third, fourth, fifth and sixth the probability of no match is 32/38, 31/37, 30/36 and 29/35. Multiply these together to get the probability of no match of about 35%. So the probability of at least one match on the third day is about 1 - 35% = 65%. You will need to adjust this calculation to allow for how many numbers in your lottery. kat