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Below I have created an excel spreadsheet showing real data of the 6/49 national lottery in Canada. The "ODDS 1 IN:" column are the official odds of winning what is shown in the "Prize Category" column.

The "PLAYERS" column is one I created by multiplying the "Winners" by the "ODDS 1 IN:" column. Then, when I average all the numbers in the "PLAYERS" column, I get an estimate of the number of players in this particular Lotto 6/49 game.

Based on my calculations, there were roughly 6,034,755 players in this particular game on this particular day if each person bought only 1 ticket. Knowing some people buy more than one ticket, this means the number of players was actually smaller.

I wanted to know if my reverse engineering estimate is an accurate calculation of the number of players per game? If not, how do I calculate a better estimate?

Thank you!

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Cannon444
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    I would disregard the 4.661 million players number as an anomaly, and take the average of the 5 other values instead. It makes sense to disregard this number as its coming from only 2 data points. Overall, your method is good, except of course for the issue that you found roughly the number of tickets sold, not number of players. There is no way at all using this data to find out how many players there are. – Moosh Dec 02 '20 at 18:30
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    I wouldn't average in the first one. It's obviously an outlier. – saulspatz Dec 02 '20 at 18:31

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Take this answer with a grain of salt, as I'm not to familiar with statistics, but I think this is a good way of getting a much more accurate estimate:

The biggest issue here is that you are taking an average when you really should be doing something like a weighted average, since getting 6.055 million people as an estimate based on a sample of 729,000 people is much much more significant than getting 6.4 million from a sample of 116 people, or getting 4.6 million from a sample of 2.

You could disregard the 3 smallest samples, and average over the 3 largest ones instead, but you still have the issue that 700,000 is much bigger than 77,000.

So how do you do something like a weighted average? Well it will require something a bit more advanced than what you've done. For a given number of tickets sold, you can generate the probability of seeing the number of winning tickets that we've seen. Using this, you can find the number of ticket sold which maximises this probability.

Moosh
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