So each lottery is an independent event with a one in 487,635 chance in winning. is this correct? So if the same 4 numbers are played week after week , the chance of winning with those same numbers is not increased because each event is independent.

@24ballin23 He did do that, because order doesn’t matter, he divided it by 4! and he got 487635 for the combinations. And for the possibity he got 1/487635. its not wrong.

## azeenator

January 3, 2012 at 1:57 amKhan you just saved my life! 🙂

## PaidVideoReviews

January 3, 2012 at 2:40 am@Dragonnoodle1 Yes.

## Dragonnoodle1

January 3, 2012 at 3:17 amSo basicly if the order was important you would use permutations?

## Dragonnoodle1

January 3, 2012 at 3:54 amI have a test of this in an hour and I didnt really study o.O I am going to die!

## l4l01234

January 3, 2012 at 4:02 amExcellent video.

## moka22051

January 3, 2012 at 4:26 am@Ekman900509 YYYESS!!

## Ekman900509

January 3, 2012 at 4:41 amThese 4 vids learned me everything about permutation and combinations that I tried to learn with my calculus book for hours.

## PhxPride1

January 3, 2012 at 4:51 amWatched the four video set and loved it, thanks Khan.

## treefangers

January 3, 2012 at 5:00 am@24ballin23 They occur in 24 different ways; however, there are 11,703,240 permutations – not 487,635.

24/11703240 = 1/487635

Tacos are delicious.

## mitomke

January 3, 2012 at 5:48 amSo each lottery is an independent event with a one in 487,635 chance in winning. is this correct? So if the same 4 numbers are played week after week , the chance of winning with those same numbers is not increased because each event is independent.

## crystalidx

January 3, 2012 at 6:33 am@24ballin23 He did do that, because order doesn’t matter, he divided it by 4! and he got 487635 for the combinations. And for the possibity he got 1/487635. its not wrong.

## 24ballin23

January 3, 2012 at 6:37 amHis answer is wrong.

Its [24/ 487635] because the order doesn’t matter thus those winning numbers occur in 4!=24 ways.

## shdubbleorti

January 3, 2012 at 6:49 amhow would you solve one an outrageously high number such as. 300P289?

## Waranle

January 3, 2012 at 7:33 amThank you Sal, plz more videos of discrete maths 🙂

## MomoBG22

January 3, 2012 at 7:36 amthe old calc was way cooler :))