I’m wretched when it comes to calculating mathematical odds. Is the probability of the lottery drawing 9-1-1 on Sept. 11 just 1 in 1,000? Higher? Lower? Do we need to take into account the number of dates that can actually be rendered as 3-digit numbers (without initial zeroes, etc.)?
Yes, the odds are only 1 in 1000.
Since there is a drawing every day (2 in fact, but for the sake of argument, we’ll only say 1), there is a 100% probability of a drawing on 9/11. As such, when 9/11 comes around, there is a 1 in 1000 chance of any 3-digit number (including 9-1-1) being drawn.
Zev Steinhardt
I’m trying to wrap my head around this too.
I know that it being Sept. 11 makes it no more or less likely for 911 to come up. Can you leave it right there and get your answer? Is the match just a weird coincidence and only noticeable because you’re looking for it?
Say you flip a coin 100 times. Each try, it’s 50-50 heads or tails. Suppose you get 99 heads in a row. The 100th time, it’s still 50-50. But wouldn’t the odds of getting 100 consecutive heads be really low? Is that extra variable (the streak, not individual draws) applicable to the Sept. 11 thing since you have to match the day?
The chance that that exact thing will happen is one in a thousand (or two in a thousand if there are two drawings). But think about it another way - what is the chance that something, anything, amazingly coincidental will happen on that day? It’s not nearly as long. The lottery was just one of many potential things that you could have heard about. All the others didn’t pan out, though.
Well, you can only match the date 10 months out of the year (1/1-9/30, plus 10/1-9, 11/1-9, 12/1-9.
So, on any of those days, the odds are 1/1000 that the date will be picked. Since there’s 300ish days out of the year that it’s possible to pick the date, one would expect the date to be picked once in just over 3 years.
Feel free to correct my math.
No. That would only apply if the second variable, the date, were randomly assigned. But it isn’t.
I’ll give you two examples, so that you can understand.
The draft lottery was conducted in this fashion:
Two drums were filled with 366 balls. The first set of balls had the numbers 1-366. The second had all the dates in the year.
A government official would pick a ball from each drum. Suppose they picked #44 and Apr. 27. That meant that the draft number for Apr 27 was 44.
The odds of a match coming up that way (such as #1 and Jan 1, or #366 and Dec. 31) are 1 in 133956 (366*366).
However, suppose the draft were done differently. Suppose instead of two drums, they only had one drum, with numbered balls from 1-366. In this scenario, it is determined that the first number drawn will apply to Jan. 1, the second to Jan. 2 and so until the last ball drawn applies to Dec. 31. In this scenario, one variable (the date) is already known. As such, it reduces the odds to 1 in 366.
The same thing applies to this lottery. One variable (the date) is already fixed. So, matching the number to the date becomes only 1 in 1000.
A last example: We all know the odds of throwing boxcars on a standard pair of dice is 1 in 36. However, suppose you don’t throw both dice at once, but only one at a time. Now, you’ve thrown a 6 on the first die, thus “setting” the first variable. Now what are the odds of achieving boxcars? One in six.
Zev Steinhardt
With two drawings, the probability of getting any number is 1/1000 + 1/1000 - (1/1000 * 1/1000), by the inclusion-exclusion principle and the assumption of independence. Not quite 2/1000, but close.
No, the odds would be 1/366. There is 100% chance that SOMETHING will be picked for the first ball, and then a 1/366 chance that the second ball will match the first. The odds would be 1 in 133956 if you were looking for a specific match.
Yes, you’re correct. Sorry about that.
However, my main point still stands regarding the NYS lottery.
Zev Steinhardt
I was trying to work out the odds in a given year of that particular lottery game’s drawing having a winning number that matched that day’s date.
The odds of a particular date expressed as a 3-digit number coming up in the drawing is 1 in 1000. There are two drawings per day, every day of the year, but matching the date is only possible 300 days of the year. So, 600 1 in 1000 chances.
Using those numbers I got the odds of this happening on any day in a given year to be about 45%… so you’d expect this to happen every couple of years. Am I correct? I’m not sure.
It doesn’t matter that it happened on 9/11 because every day in the year has special significance for someone.
-fh
It was windy in NYC that day too . . .
DaLovin’ Dj
What I found interesting was that today on the NBC evening news, they reported that “according to a mathematician the odds were one in a thousand.” So they needed to go to an expert to tell them the “official” answer for what is really a very basic math question.
Another factor is that New York isn’t the only state to have a daily three-digit lottery. Let’s say that there are about 50 such drawings in the country: Some states have two, some don’t have any, but on average, it’s probably about one per state. That brings the odds down to the neighborhood of 1 in 20 (a bit less than that, actually, but I’m approximating). Yes, 1 in 20 is noticeable, but not exactly a sign of the Apocalypse.
Yes, and in the Yankees game that day, the Yanks won 5-4 (5+4 =**9 **) in 11 innings. What are the odds?
Given the number of essentially random numbers that could have come up on September 11:
- Lottery numbers
- Sports scores
- Stock indexes, gains, and losses
- Weather conditions
- Random results of all types
I would argue that it was virtually CERTAIN that “911” would appear somewhere.
It has been correctly pointed out that I got my sums wrong. Accordingly, I have revised my statement in the other thread. The correct odds are not 500 to one, but 500.25012 to one. Details are over there.
What happens when you factor in the odds of any given day when that happens being a historically signifigant day? Since much fewer days qualify, it should take much longer for it to occur. If 1 in 2000 historically signifcant (let’s say US government recognized) days will match the lottery and there are 10 holidays (making that up - how many are there?) then it should happen once in every 200 years.
If you factor in the amount of drawings that take place one year after an event later recognized by the government the numbers get even bigger. How often does something worth naming a day for it happen? Once every decade? If so then it should happen once every 20,000 years. How long has humanity been around?
Spooky.
DaLovin’ Dj
What are the odds that somebody is rigging the New York Lottery?
And, due to having Sept. 11 on the brain, unthinkingly rigged the payoff for the numbers 9-1-1?
Call me cynical, but…
I am an engineer with a strong background in statistics. I’m not here to argue math, everything you folks have said is correct.
But sometimes stats don’t tell the whole story.
I find this to be annoyingly coincidental. First of all, 1 in 1000 is still pretty long odds. For it to happen on such a sensative date, in the state the where the whole thing took place, is a little too much for me to swallow.
Probability only works when humans aren’t meddling with the process. I’m skeptical. This reeks of tampering.
Just my opinion.
You’re in VA, right? Well, 9-1-1 was pulled in two drawings in less than two weeks in your own state (8/26/02 and 9/09/02 evening drawings), so has the VA lottery been tampered with as well?
Just curious