Simon, Sara and I bought a lottery ticket for a 32 million jackpot to be drawn on New Years Eve. Our discussion upon looking at the ticket brought for the idea that we had no idea what our odds of winning were, so we’ve decided to try and figure it out.

It should be noted here however that we are refusing to look it up. We have to try and figure out this on our own.

So, the lottery we have a ticket for has 45 total numbers and you pick 6.

From this, we know that the total possible combinations of 6 digits is equal to:
45x44x43x42x41x40 = 5,864,443,200
(The numbers decrease because for each number selected you remove a possibility).

Right, now that would be great, but 5.8 billion is waaay to high. Even if every Australian bought 10 tickets each, someone would only win about one in every twenty draws, or twice a year.

So we thought on this for some time. I cleverly made the next hurdle. Consider if the lottery was 6 numbers and you got to choose 6. Even though their are 720 possible arrangements of the numbers 1 through 6, your odds of winning are still 1/1 because the order of the numbers do not matter.

And this is kind of where we sit. We are unsure of the math required to figure out how many unique 6 figure sets there are irrespective of their order.

I’m sure that by the time she’s read this far Catsy has already written a python script to calculate the odds. That said, we don’t want the answer, we want to know how to do the math. Any suggestions?

Also, I should mention that this ticket was purchased in the lovely seaside town of Sorrento. Simon, Sara and I had the best time EVER playing in the massive (for central Canadian me) waves.