A national lottery has the format where $7$ numbers are chosen from $45$ without replacement. The first $6$ numbers chosen constitute the "winning numbers", while the last number chosen is the "additional number". Participants purchase tickets containing six numbers each.
Prizes and the published odds of winning for the lottery are as follows, where $(n,m)$ denotes how many of the numbers on your ticket match the winning and additional numbers respectively. Take note that a ticket is only considered to win from the highest group it can win - for example, a ticket winning in Group 1 is not considered to have won any of the lower groups.
- Group 1: (6,0) - $1/8145060$
- Group 2: (5,1) - $1/1357510$
- Group 3: (5,0) - $19/678755$
- Group 4: (4,1) - $19/271502$
- Group 5: (4,0) - $703/543004$
- Group 6: (3,1) - $703/407253$
Now, the lottery board publishes statistics on how many winners there are in each group. For instance, during the last draw, the number of winners were as follows: $(1,7,280,875,14347,21993)$.
Using these numbers, what is a good way for me to estimate the number of tickets bought, with a confidence interval?
If I just look at the information from one set of probabilities and number of winners, I will be able to get an estimate (for example, I would estimate that there were 8145060 tickets bought if I had only considered Group 1 prizes). How can I combine all the information I have using Bayesian statistics to generate a better estimate?
Currently, I'm looking at treating the number of winning tickets in a given group as a Binomial Distribution, and then apply the Agresti-Coull interval, but I'm not sure how to combine all these intervals.