Man, this week seems to be going by very slowly. I can’t believe it’s only Wednesday! It feels like at least Thursday… I’d even believe Friday.
I blame the Sasser worm. And the heat. And John Ashcroft.
Anyway, seeing as how today is yet another day when I didn’t really get to take a full lunch break, I just wanted to post a brief update now about — what else? — Powerball. Tonight’s jackpot is $170 million, and my office bought 55 tickets. I contributed $5 to the pot, meaning I stand to win roughly $15.5 million (before taxes) if we hit it big.
The odds of that happening: 1 in 2,191,396. (That’s 55 in 120,526,770, for those inclined to check my math.)
AFTERTHOUGHT: Of course, the odds above fail to take into account the possibility of somebody else also winning, resulting in a split jackpot. So the odds of me winning $15.5 million are somewhat below 1 in 2,191,396. But the odds of me winning 1/11th of a piece of the jackpot — whatever the amount of a single jackpot piece ends up being — are 1 in 2,191,396.
Now, here’s a question for the math whizzes. The odds of winning at least $3.00 (that’s the minimum prize, if you get only the “Powerball” right and nothing else) are supposedly 1 in 36.06, which, by my calculations, comes to 3,342,395 out of 120,526,770. What is the correct way to calculate our odds of winning at least $3.00, given that we have 55 tickets? It can’t be 55 times the numerator, because that would end up producing odds of 1 in 0.66 — i.e., we have a better than 100% chance of winning — and that can’t be right, since there are fully 117,184,375 non-winning tickets out there. So what is the correct way to do that math? I’m drawing a blank…. Mike? :)
UPDATE: We didn’t win. Neither did anybody, so Saturday’s jackpot is now $205 million.
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Categories: My Life
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May 5th, 2004 at 4:38:56 pm
Um, if you all bought 55 tickets of different number combinations, doesn’t that mean your chance of winning is
2,191,396:55
/
55
?
May 5th, 2004 at 4:39:59 pm
Okay nevermind, I checked the link. Your math is right but your presentation is screwy and confusing.
May 5th, 2004 at 4:44:00 pm
Just posted the “afterthought,” then read your comment. Now I’m even more confused.
Mike, save us!
May 5th, 2004 at 5:08:36 pm
I believe that math is right, but you then reverse it: In other words, the odds are 2:3 that you won’t win at least $3. Right?
May 5th, 2004 at 5:16:03 pm
My confusion stemmed from you not including some details and presenting your calculations in a linear manner. For instance, you wrote:
“Tonight’s jackpot is $170 million, and my office bought 55 tickets. I contributed $5 to the pot, meaning I stand to win roughly $15.5 million (before taxes) if we hit it big.”
Nothing was incorrect about that, but your communication was very unclear. So you should have added at some point that there were 11 contributors total.
In the next sentence, it’s very easy to read 1 in 2,191,396 as being the odds of a winning ticket; 55 tickets, hence my calculation. It would have been more clear to start with 1 in 120,526,770, then 55 in 120,526,770, and then 2,191,296. Again, your math was indeed correct, and perhaps I should have recognized that more immediately, but the non-linear presentation didn’t work for my brains that have been fried by endless meetings filled with strange acronyms with which I am not familiar. :-)
May 5th, 2004 at 5:18:24 pm
Oy, time to go lift weights. Temporary peace for my mind!
May 5th, 2004 at 5:33:29 pm
Ah HA! I always suspected Andrew of Doing all the Heavy Lifting.
“My confusion stemmed from you not including some details and presenting your calculations in a linear manner.”
That’s the Maternal chromosomes showing, Gawd bless ‘er…Woops, I’m Doomed, quick, everybody into the gene pool. :)
“I blame the Sasser worm.”
I blame -
“James Sasser -
Former U.S. Senator from Tennessee: Chairman of the Budget Committee and of the Appropriations Committee’s Subcommittee on Military Construction; former U.S. Ambassador to China”
http://www.constitutionproject.org/wp/members.html
May 5th, 2004 at 5:45:42 pm
But Dammit, I meant to buy one of them Tickets — I Saw The Sign :) — and AH FURR GOT! bah / I never bought into the innumerable Office Pools on ‘em. Screw this Sharing stuff, that’s for the Christians. Huh? Well, yeah, sure, Technically… :)
Anyways, they’re always producing bigfat Lawsuits over something or other, like, the office Bagman ripped off somebody’s tickets or some such damn thing, I dunno…of course that’s good for the Lawyers I suppose…but they don’t like to Share either…WELL except with their poor kindly old Parents…waw haw haw. / (Bah. :)
May 5th, 2004 at 6:17:52 pm
Probabilities work on multiplication, Brendan. For a first order approximation: If you’re correct that the odds of winning at least $3 on a single ticket are 1/36.06, then your odds of not winning at least $3 on a given ticket are 1 - 1/36.06 = 35.06/36.06. In order not to win at least $3, your pool must fail to win $3 on each ticket, so that’s (35.06/36.06)^55 = .2129… Therefore, your odds of winning at least $3 for the pool are 1 - (probability of NOT winning at least $3) = 1 - .2129….. = .78706… In other words, your group has a slightly less than 79% chance of winning at least $3–by buying 55 tickets…
The reason that this is a first order approximation, rather than the specific result, is due to the complicating factor that the odds of winning with 55 tickets in a single draw are different than the odds of winning with 1 ticket each in 55 draws. It’s fine for the extremely unlikely outcomes, as the deviation is quite small, but it’s different for the more likely outcomes of winning lesser prizes. The powerball appears to be a separate ball from the rest of them, designated as the powerball number on the ticket, ranging from 1-42. So, with 55 tickets, you are garaunteed to have at least some powerball number represented more than once. You would then use the Poisson distribution to determine how many of the 42 powerball numbers are likely to be represented in a random set of 55 numbers, and use that number over 55 as the probability of getting the powerball number correct. You’d also have to use the Poisson distribution to determine the probablities of hitting the other numbers, get the probabilities of striking type of winning combination, and sum all of those probabilities together to get the actual overall probability of winning at least $3. Or, instead of using the Poisson distribution, you could look through your office’s actual tickets and figure out how many overlapping winning combinations you have in your actual set. That’s more definitive (the Poisson distribution merely approximates this over sufficiently large data sets), but also more work.
If you’re interested, the Poisson distribution is {[e^(-x)](x^k)}/(k!), where e is the base of the natural logarithm (2.718281828459045…), x is the probability of a specific event, and k is the number of times it occurs. So, for instance, when you’re looking at the 42 possible powerball numbers distributed among 55 tickets, the probability of a specific number NOT occuring (in other words, occuring exactly 0 times) is {[e^(-55/42)][(55/42)^0]}/(0!) = .26994…. That means that you would expect to see 30 or 31 of the possible 42 powerball numbers (73.005…%) in a sample of 55 tickets chosen separately. If you had simply taken 1 - probability of not hitting it in 55 separate trials, you’d think that your odds of getting the powerball number were 73.429…%–a small enough difference that you might want to ignore the correction. Doing the calculations for other values of k show that you expect about 15 powerball numbers to occur exactly once, about 10 to occur exactly twice, about 4 to occur exactly three times, and about 1 to occur exactly four times on your 55 tickets.
May 5th, 2004 at 6:41:10 pm
Mike: Woohoo! I knew I could count on you. :) Thanks!
Andrew: Really, I should have said explicitly that each ticket cost $1, and then the fact that my contribution was 1/11th of the totle ($5 out of $55) would have been clear. As for the other thing… ah, whatever. :)
Dad: The main reason I entered the office pool is cuz, like, what if I didn’t enter… and they won? That would suck enough that it’s worth my $5 (well, $15 actually — we’ve played in the last three drawings; no winner yet so the jackpot keeps going up) to prevent that possibility. :)
May 5th, 2004 at 7:45:33 pm
Sure ask the science major and not the MATH major…
May 6th, 2004 at 7:20:05 am
the odds of hitting powerball are the same as the odds of getting struck by lightning twice in your lifetime. Boo
May 6th, 2004 at 8:09:52 am
Ha! Sure the odds are small (less than miniscule even), but somebody told me once (well, many people told all of us many, many times) that I can do anything I set my mind to. SO, one of these days I’m going to put my mind to winning powerball, or regular CA lotto. And I’m going to do it darnnit! I just have to start playing or something =o). (Funny little prerequisite, that…)
May 6th, 2004 at 8:19:00 am
Sure ask the science major and not the MATH major…
Sorry Dave, but you know, I gotta have Trojan pride and ask the guy with the USC degree first. I wouldn’t want to be accused of bowing down to Washington… :)
May 6th, 2004 at 11:39:15 am
Sure, the odds are similar to getting hit by lightening, but SOMEONE will win the jackpot. Normally, lotto is a tax on people who are bad at math - I don’t know about Powerball, but the house take in California is fifty percent (compare to private gambling establishments, where a ten percent hold is high). At this point, five bucks isn’t a bad risk - it’s not out of line with the reward (he said without doing the math).