Please help settle this basic math problem

This debate started with my basic understanding of the ill logic of buying multiple lotto tickets to "double your odds". I now refer to an article from Snopes.com, where the myth in this article explains that on the one year anniversary of 911, one of the pick-3 numbers was 9-1-1.

It goes on to say that the odds of this aren't astronomical, as the odds of picking 3 numbers (0-9) correctly are 1-1000.

After this, the article claims that since there are 2 drawings of numbers, that the odds become 1-500 roughly.

I am not talented in mathematics to challenge this directly, but I do believe there are numerous errors in the logic of this. Firstly, it's intangible that the lotto drawing (in New York of all places) would have the lotto drawing have not only 9-1-1 as a winning number, but also actually be 9-1-1 (in that order, and not say, 1-9-1 or 1,1,9). On this, I may be wrong.

However, I strongly disagree that even if the odds were 1-1000 of winning that lottery, that having 2 drawings in the same day would make the odds 1-500. It was this line in particular that alerted me of possible holes in one of my favorite websites. (Although I really like what I've seen here.)

So please, I need some backup on this; in discussing my gripes, my co-worker can't understand why buying 2 tickets in ANY lottery doesn't "double your odds". How can I at least explain THAT much? Thanks in advance!

Apologizes to anyone who clicked a moment ago, I fixed the link.

Actually, buying two tickets with different numbers for the same drawing of the same lottery DOES "double your odds". You are twice as likely to win in that draw. (There may be some semantics involved if it is a progressive jackpot, as one set of numbers may have also been picked by more people than the other, but you are still twice as likely to win a share of the jackpot.)

The 9-1-1 problem is different because you are talking about buying the same numbers for two different drawings, and in this case, you do not "double" your odds because of the possibility that 9-1-1 is drawn both times. (Otherwise, if you pick 9-1-1 for 1000 different drawings, you would be guaranteed a win - of course, this isn't true.)
The probability of 9-1-1 comes up at least once in two draws = 1 minus the probability that it does not come up in either draw = 1 - (999/1000 x 999/1000) = 0.001999, which is about 1 in 500.25, or "roughly" 1 in 500.

If you play both lotteries, then the odds become 2 in 1,000 or 1:500.

But that's assuming you play only one ticket, in each lottery, and the only winner you're concerned with is the exact match.

If you buy 2 tickets in only one of those lotteries, the odds are still 2:1,000 or 1:500.

If you but all 1,000 combnations in one lottery, guess what? Your odds of having an exact match are 1,000:1,000 or 1:1.


The trouble is, no matter how you slice it, the payout isn't worth it.

Well in the article, it states that the odds of 9-1-1 appearing are one in 500 roughly, on the basis of the drawing occurring twice that day. I find that to be under the fallacy of the odds increasing with each event.

On the subject of buying multiple tickets, are you saying that buying 10 "pick 3" numbers on a given day would increase my odds to 1/100? Again, I'm not the best at this, I just don't quite get it yet.

You bring up a number of issues.

First, the odds of the numbers being 9-1-1, in that order, on any day, including 9-11, are 1 in 1000.

Second, it seems the NY Lottery doesn't do that game any more, but Maryland does. If you ask about the odds of winning, you have to specify if you're playing it "straight" or as a "box." Straight means you must get the digits in order, so the odds would be 1 in 1000 (on any drawing). A "box" means the order doesn't matter. In the case of using two different digits, like 9-1-1 as a "box" you could win with 1-1-9, 1-9-1, and 9-1-1, so the odds of winning are 3 in 1000. In case anyone wants to know, in Maryland a $1 ticket pays $500 on an exact ticket, and $160 on a 3-way box. The exact pick is thus the better value.

Finally, we just discussed this in the Ask Marilyn thread, but if the question is what is the probability of winning at least once playing 9-1-1 straight in two consecutive drawings, the answer is 1-.999*.999 = 0.001999.

Quote: Trevelyan

Well in the article, it states that the odds of 9-1-1 appearing are one in 500 roughly, on the basis of the drawing occurring twice that day. I find that to be under the fallacy of the odds increasing with each event.

The probability of 9-1-1 appearing on that day is, as I said earlier, about 1 in 500.25. The probability of 9-1-1 appearing in any particular draw is 1 in 1000.

Quote: Trevelyan

On the subject of buying multiple tickets, are you saying that buying 10 "pick 3" numbers on a given day would increase my odds to 1/100? Again, I'm not the best at this, I just don't quite get it yet.

You may still be confusing "particular day" with "particular draw".
For a particular draw, yes, buying 10 pick 3 numbers increases the probability to 1/100 - if the numbers are all different.
Let's say you buy the numbers 000, 111, 222, 333, 444, 555, 666, 777, 888, and 999 for a particular draw. There are 1000 different numbers, of which 10 are winners, so your odds are 10 in 1000, or 1/100.
If, on the other hand, you buy the numbers 000, 111, 222, 333, 444, 555, 666, 777, 888, and 000 again, you only have 9 numbers out of the 1000 that can win, so your odds are 9/1000, or about 1/111.

If there are multiple draws on the same day, you have to allow for the fact that the same numbers can be drawn more than once on that day.

Some very informative responses there.


I'll check out the thread. Thanks for your help guys!

The probability of any 3 digit combination is in fact 1-1000 (0.001 or 0.1%). This should be self evident because there are 1000 numbers to choose from. If it is a fair lottery every number should be equally probable. If you want to think about it as 3 digits you can calculate it as 0.1 x 0.1 x 0.1 to arrive at the same result.

The "double the odds" statement is "roughly" correct when you are talking about a low probability event like a lotto ticket. Think about a coin flip that has a 50% probability. If you make two flips you don't double the chance of heads because you have to account for the situation when both flips were heads. Here are the possible outcomes:

Heads Tails
Heads Heads
Tails Heads
Tails Tails

3 of the 4 combinations contain heads. If you are betting on heads you have a 50% chance to win on one flip but with two flips you have a 75% chance which does not "double your odds" from the first flip.

Now extend this logic to 3 flips. There are 8 possible outcomes (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT). Of these, 7 of 8 contain a head. Another way to think about this is that the probability of all 3 flips being a tail (NOT heads) is 0.5 x 0.5 x 0.5 or 0.125 or 1/8. Thus the probablility of at least one head is 1 - 1/8 = 7/8.

If the probability of winning the pick 3 with one number is 0.001, the probability of NOT winning is 0.999. Using the same method we used for the coin flip, the probability of getting at least one win by picking 2 numbers is 1 - (0.999 * 0.999) = 0.001999 which is not "double the odds" of winning with one pick, but it is pretty close. And of course there is always the 0.000001 probability of winning BOTH.

If you are talking about a tiny probability event like a lotto drawing, buying 2 tickets does double your chances of winning to a very high approximation.

Actually just ignore my entire previous post because 2 numbers in the same lottery are not independent (DUH). Too bad I spent time making a clear explanation for a different problem. Buying 2 numbers in a lottery exactly doubles your chances.

I didn't ignore your post jml, but I still get conflicting answers on the subject of this former lotto pick;

If the lotto is twice in the same day, I don't believe it to be under the same EVENT, given that a 2nd random set of numbers are drawn. To me, the fact that 2 drawings occurred (once in the morning and evening) should only increase the odds only slightly, and not halve the odds entirely.

The way I understand this, I'm comparing the odds to buying two 1/49 tickets and "doubling" my chances there as well, which I know isn't the case; you're taking 2 stabs in the dark. Picking 9-11 as your number, and having it drawn 2 SEPARATE times (on the same day) should still be around 1-1000.

I may have not phrased my questions properly, but I'm definitely paying attention to your answers.

I'm working night shift, so I gotta get some sleep. Again, thanks for the input :)