TheConcept
Member
Hi Frank. If I use the example of a 49 ball lottery with the bonus ball, the probability of a number not being drawn is 42/49. The probability of a number not being drawn after n draws is (42/49)^n. (^n = to the power of n)Frank said:I'm pretty sure there is a mathematical law that determines what the average cycle span is for a particular lottery matrix. It's definitely an exponential decay curve if you plot balls still to come out in the cycle against draws into the cycle. As Icewynd has already demonstrated, the probability of NOT matching the balls already drawn in the cycle ( hence adding to the count of unique balls drawn) reduces the further into the cycle you go. This is a classic property of exponential decay. Somewhere the exponent of the natural constant e contains an expression involving the size of the lottery ( how many balls) , and how many are drawn. Working out what this expression is, may be beyond me, but I'd love to know what it is.
The number of balls left to come out can be worked out by
49x(42/49)^n), so for n=1 (after one draw), there are 42 balls statistically (and in reality!) left to come out. After two draws (n=2), it is 36, etc.
This is an exponential function, so the value will never be zero. In the past, I've tried working out the number of draws for the number of balls being 0.5 (the lowest number that rounds to 1) as an attempted approximation. This gives a value of 29.7 draws.
For those with Excel, use the calculation =LOG(0.5/49,42/49)
The above calculation can be used to work out the number of draws before there are b balls left to come out
=LOG(b/49,42/49)
If anyone wants to adapt the above formula, the second fraction is the probability of a ball not being drawn for the lottery, so e.g. for the National Lottery 59 balls, 6 balls drawn, it would be =LOG(b/59,53/59).
Ion Saliu mentions something with e and the degree of certainty on his site here:
http://saliu.com/Saliu2.htm
I haven't worked out yet how to work out the median number of draws for all balls to have appeared at least once for a particular lottery. Homework tonight?