# Calculating the probability of winning the lotto

1. Feb 10, 2014

### trap101

So I have a question I was going over in a text book about calculating the probability of winning the 6/49 lotto. Which means you have to pick 6 correct numbers out of 49, order does not matter. So I got that:

P(L) = 6/49 * 5/48* 4/47 * ....* 1/44 = 1 / 13,983816

but then there is a comment afterwards that I can't figure out the solution:

"So if you were to play once a year, on the average you would win once every 13,983,816 years. If you played once per day, given that there are 364.25 days in a year,on average you would win once every 268,920 years.

How do they obtain the 268,920 years? What am I missing in being able to convert it over?

Thanks

2. Feb 10, 2014

### tiny-tim

hi trap101!
13,983,816 / 268,920 = 51.9999

sooo … i guess they meant once a week!

3. Feb 10, 2014

### trap101

Thanks Tim, but I'm still not clearly getting it. So dividing those two sets of years provides me with 51.99, which I suppose means 52 weeks. I think I'm getting confused with which unit where.

4. Feb 11, 2014

### Ray Vickson

The expected number of times you must play until you win is 13,983,816. If you played once a year, then on average you would need to wait about 13 million years until you win. If you played once per week (52 times per year) you would need to wait about 13,983,816/52 ≈ 269,000 years, etc.

5. Feb 11, 2014

### tiny-tim

hi trap101!

(just got up :zzz:)

the units are plays per win (or play.win-1)

so you multiply plays per win * time per play (= time per win) …

the lotto always has a play per win of 13,983,816

if time per play is 1 year, then time per win = 13,983,816*1 years

if time per play is 1/52 year (ie 1 week), then time per win = 13,983,816*1/52 years

6. Feb 11, 2014

### trap101

Ahhh. Thank you gentlemen. It really comes down to documenting every aspect so I could get these unit troubles under control.