The Poisson distribution

[O Great One]

If you have a lottery (Megamillions) and you sell 20,000,000 tickets, the probability of them all losing is given by:
(135,145,919/135,145,920)^20,000,000 = 0.862448363

A close approximation is given by:

e^-(20,000,000/135,145,920) = 0.8624413

I just learned this from a book. That's pretty cool!

 

[HallsofIvy]

I suggest that you go back and read that book again!

Did it say anything about where it got the figures
"135,145,919" and "135,145,920". For that matter, did it say anything about the legality of a lottery in which there is an 86% chance that EVERYONE loses?

 

[chroot]

Originally posted by HallsofIvy
For that matter, did it say anything about the legality of a lottery in which there is an 86% chance that EVERYONE loses?

I bet people would still play... ;)

- Warren

 

[arcnets]

This is because ln(1+x) ~ x for small x.
=> ln[(k/(k+1))^n] = n ln [k/(k+1)] = n ln [1 - 1/(k+1)]
~ -n/(k+1)
=> (k/(k+1))^n ~ e^[-n/(k+1)].
See?

However I'm puzzled because it doesn't say how many tickets there are in total. Don't we have to know this to figure out the 135,145,919?

 

[russ_watters]

did it say anything about the legality of a lottery in which there is an 86% chance that EVERYONE loses?

Ever hear of PowerBall?

 

[Loren Booda]

arcnets

This is because ln(1+x) ~ x for small x

At face value this doesn't seem like a good approximation for some similar applications, like 1/(1-ln(1+x)) where x=1.

 

[arcnets]

Loren Booda, by 'small x' I meant |x| << 1. So it's not valid for x=1, of course.

 

I assume there are 135,145,920 possible numbers/choices, with one chosen as the winner.

The rules say there is a power-number 1-52, and 5 distinct numbers 1-52 which can be in any order. So that's 52 * 52! / (5!*47!) , which is right.

 

[Loren Booda]

mea culpa, arcnets