# Estimating the number of tickets bought in a lottery

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.

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Are the tickets randomly assigned or selected by personal taste? For example, if ever "1,2,3,4,5,6" were dran in Lotto, there'd be way more rank 1 winners tan the total might suggest. This would be reflected by the prize for a rank1 win would drop considerably (and be a good indicator for the relative frequency of such tickets) – Hagen von Eitzen Mar 2 '13 at 15:28
@HagenvonEitzen for now let's assume that all tickets are assigned randomly. I have collected data on the number of winners for each Lotto draw, and am trying to estimate whether certain numbers are more popular among lottery participants - but that will have to be for another question. – Vincent Tjeng Mar 2 '13 at 15:32

Note that each ticket has $6$ out of $45$ numbers, that ${45 \choose 6} = 8145060$, and that all the denominators in your table are divisors of this number. If you add up the winning probabilities, you get $P_{win}=\frac{25410}{8145060}$.
The total number of winners this particular week is $37503$ so if you make the very strong assumption that "all tickets are assigned randomly" then a reasonable estimate is that there were $37503 \times \frac{8145060}{25410} \approx 12021416.2$ tickets sold.
If that is the order of magnitude of tickets, then the proportion of them which win has a confidence interval (using two standard deviations) of about $P \pm 2\sqrt{P(1-P)/N}$ which is about $[0.003088, 0.003152]$. Divide the number of winners by the two ends of the range and you get a confidence interval for the number of tickets sold of about $[11898723,12146666]$.
Ignoring the spurious accuracy, this confidence interval is far too narrow, because in reality some numbers are more popular than others, and evidence on the number of tickets sold and prizes won shows the proportion of winning tickets varies dramatically from draw to draw: for example in the last UK national lottery (different rules) the proportion winning was about $0.0176$ while the previous draw the proportion was about $0.0212$: the 47 ball in the more recent draw is less popular than any of previous draw's balls.