- #1

O Great One

- 98

- 0

(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!

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- Thread starter O Great One
- Start date

- #1

O Great One

- 98

- 0

(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!

- #2

HallsofIvy

Science Advisor

Homework Helper

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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?

- #3

chroot

Staff Emeritus

Science Advisor

Gold Member

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I bet people would still play... ;)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?

- Warren

- #4

arcnets

- 508

- 0

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?

=> 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?

Last edited:

- #5

russ_watters

Mentor

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

- #6

Loren Booda

- 3,119

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

- #7

arcnets

- 508

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

- #8

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.

- #9

Loren Booda

- 3,119

- 4

mea culpa, arcnets

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