There’s obviously something wrong with my math, but:
There are 40,664,815,074 differing number combinations in Powerball (1/69x1/69x1/60x1/69x1/69x1/26).
Let’s say that 400,000,000 tickets are sold, probably an overestimate…
This would seem to imply that there will be a winner only about one in a hundred Powerball events.
But, obviously, there are winners much more often.
What’s wrong with my figuring?
For starters, your combinations are wrong. You can’t have repeat numbers, so each “ball” slot has less and less possibilities. More like 1/69 x 1/68 X 1/67 …
The order of drawn balls doesn’t matter, so the odds of getting all five numbers are:
69!/5!=11,238,513
Then there are 26 numbered powerballs:
26X11,238,513=292,201,338
Just for people following along, this is 69!/(69-5!)(5!).
And for those who want a fuller explanation, find a site or book that explains combinations. Here or here are a couple of examples.
You are both wrong. There are five numbers plus the powerball number which is chosen separately. The powerball is between 1 and 26. The other five are between 1 and 69 and don’t repeat.
The final number that Bill Door gave is correct (unless I misunderstand how Powerball works), but Snarky Kong was right to correct how it was calculated.
Definitely an overestimate. If this site is to be believed, the number of tickets sold for most drawings is in the range of 10 to 20 million, and (after a quick scan) has only gone above 100 million three or four times since January 2017. This means that you would expect a winner only one time out of 10 or 20.
Also, a fun fact: suppose the number of tickets sold was exactly equal to 292,201,338 (the number of possible outcomes), and suppose that everyone chose their numbers randomly. The probability of nobody winning is approximately 1/e ≈ 36.8% — still pretty substantial.
More generally (in the limit when the number of possible tickets is huge) I think the probability of exactly k winners is 1/(e·k!) — i.e. 36.8%, 36.8%, 18.4%, 6.13% for 0, 1, 2, 3 winners.
Bill Door has it right. You multiply the number combinations from 1-69 with 26.
It’s 292,201,338 different combinations total.
How does that term (69-5!) parse out? 5! is 120, so (69-5!) should be (-51).
And (69-5!)(5!) would be (-6120)
So 69!/(69-5!)(5!) would be -2.79681185037277992054324891974064 X 10[sup]94[/sup]
The parenthesis is out of place. It should be 69! / ((69-5)! x 5!) or equivalently:
(69 x 68 x 67 x 66 x 65) / (5 x 4 x 3 x 2 x 1)
Yes. The first number out is 1 of 69. to expand:
The second number is one of the 68 remaining.
The third number is one of the 67 remaining.
etc.
6968676665 is the same as 69!/64! since it’s
(69x68x67x66x65x64x63x62…x1)/(64x63x62…x1) = 69x68x67x66x65
But order does not matter:
1,2,3,4 is the same as 1,2,4,3 and 1,4,2,4 and 4,1,2,3 and 2,3,4,1 and so on for the purposes of determining if a ticket wins.
In fact, given a fixed 5 numbers, you can arrange them 5! different ways and a ticket with those 5 numbers wins no matter what order they were drawn.
So: winning 5-numbers: 69!/(64!x5!) distinct draw sets.
For powerball, add in 26 choices - total winner combinations = (26x69!)/(64!x5!)
I hope my brackets are all correct.
If I’m following you here (and probability statistics is not my strength by a long shot), this would be true if each person who bought a ticket picked random numbers independent of those picked by others.
But I have been told (perhaps erroneously) that when you let the lottery computer pick the numbers, you get a ticket that is randomly chosen from the set of tickets that have not yet been picked. If that is true and the number of tickets purchased was exactly equal to the number of possible outcomes, then the probability of nobody winning would be 0.
Or am I confusing something here?
The pick of the lottery computer for your number to enter is random, with no restrictions about the number like whether it’s already chosen that number for somebody else:
Yeah, there have been cases even within the last year of multiple winners for the same drawing using automatically generated picks, so clearly there are no additional restrictions based on previously generated tickets.
Anecdotally I have heard that those who choose their own numbers tend to cluster their picks around “birthday” numbers; i.e. 1 through 31. If true; that would mean that you could reduce the odds of having to share a prize by picking from the set of numbers 32 through 69.
When I lived in Florida, I read that about 20,000 people every single week play 1-2-3-4-5-6, which meant that if that combo came up, each person would get about $300. Obviously a bad strategy. Which means at least 20,000 people can’t recognize a good strategy from a bad one. Hence, the expression that the lottery is a tax on people who are bad at math. But it’s actually worse than that. Almost everyone has a poor understanding about how the lottery works, even people who made A’s in Algebra and Calculus. I submit that virtually everyone who plays the lottery can’t recognize a good strategy from a bad one. If they could, they wouldn’t be playing the lottery; they’d be playing something else, like Roulette.
FWIW, The New WKRP in Cincinnati had an episode where Herb Tarlek picked an obvious combination, got a winning ticket, and then had to share the prize with thousands of people.
You have to recognize what Lottery really is, random numbers. As mentioned, most people tend to choose numbers based on Birthdays, kids’ ages, important dates, etc. Patterns are being made and very seldom met but if hit spread thin among people with similar idea.
Select a random set of numbers that’s odd looking or a quick pick and your chance increases.
Playing the lottery can be rational, if the entertainment value is worth the cost of the ticket. But most of the entertainment value comes from daydreaming about what you’d do with the oodles of cash you’d get if you won. And if even winning would only net you $300, then that doesn’t support much daydreaming, so playing 1-2-3-4-5-6 is surely irrational.