Given that, theoretically, each number has an equal chance of coming up, how is it that some of them do come up more often than others? Is the time-frame involved a factor?

I’ve just flipped a coin and it came up heads six times in a row and it’s not rigged. Do you want bet me on what the next one will be?

Your question is faulty.

In time, all the numbers will come up an equal amount of times. That’s just the nature and definition of probablity.

State lotteries use more than one set of numbers, randomly chosen, and frequently test the individul balls to prevent any inperfections that would result in one ball getting chosen more than another.

Hence:

“Is the time-frame involved a factor?”

So the question then becomes: What does ‘In time’ entail? I’m assuming that this would be a function of the numbers involved (lengthier if there are 54 balls than if there are 49). Have studies ever been done to determine approximately what the time-frame would be?

Sorry for the snippy response.

In theory, for a perfect random distribution: infinite time.

In practice, for a Pretty Damn Close to Random distribution…it does depend on the numbers, but IIRC, after about 100 drawings you have a 95% confidence factor in your findings.

Thanks snippy…

Sorry to be nitpicky, but I’m just that way. The number of times each number appears will not come up to be equal as the number of trials increases. But that number won’t increase as fast as the number of trials, so their ratio goes toward the average value.

This sounds confusing, so I’ll make it into an example. If you flip a fair coin 100 times, you’d expect around 50 heads, but probably not exactly 50. Probably somewhere between 45 and 55, though. You’d expect the average success to be 50% +/- 5%.

Now consider that you tossed it 10,000 times. The number of heads would likely be between 4950 and 5050. There are now 100 numbers that are in the “likely” band, so your sample’s error increases as the number of trials goes up. However, the ratio is now 50% +/- 0.5%.

So the ratio goes toward the mean as the number of trials increases, but the absolute error also increaes.

And consider also that while the numbers drawn are picked randomly, the numbers selected by the players are not. Since the prize money is spread evenly across all those who picked the winning number, there should be number choices that have a higher expected value.

The thing I don’t understand is why people feel if a given lottery vendor has had a big winner(like the big jackpot) they all think that THAT’S the place to buy your lottery tickets at.

Wouldn’t having a big winner at a given site DECREASE the odds of there being another big winner at the same given site? (The old “lightning doesn’t strike the same place twice” theory)

Chris W

PS Sorry for hijacking the thread!

A big winner at a store won’t decrease the odds of another winner there, either. The odds of winning a given lottery game are completely independent from what happened in all games that come before it; it makes no difference where you buy the ticket.

As for the OP, here’s an anology you may be familiar with–the old birthday problem. Ignoring leap days and assuming a uniform random distribution of birthdays, what’s the probability that, out of 23 people, at least two people have the same birthday (just month and day, not year)? It’s slightly over 50%, which is a lot higher than most people expect.

Another way of looking at the birthday problem, to make it analogous to the lottery problem: Say I have a lottery game, and each game I pick a number from 1 through 365 to determine the winner. Each time the game is played, each number has an equal chance of coming up a winner. However, after I’ve played 23 games, it’s likely that at least one number has come up at least twice, while most numbers haven’t even come up at all. A priori, of course, there’s no reason to expect that any *particular* number will come up more often than others. All you can reasonably conclude is that: 1) It’s likely that some numbers will come up more often than others, and 2) The chance of 1) happening for any *particular* number is just as likely as 1) happening for any other particular number.

To add to what **Cabbage** said, there is no “lightning doesn’t strike the same place twice” theory. Independent events have independent probabilities.

[hijack] Could you show me the math for that?? That is a potentially very lucrative bar bet.[/hijack]

Sure. The easiest way is to calulate the probability that no two people have the same birthday. Subtract that from one, and you’ve got the probability that at least two people have the same birthday.

So if you have 23 people, the total number of different birthday distributions is 365[sup]23[/sup] (each of the 23 people has 365 different possible birthdays).

Now count the total number of different birthday distributions so that everybody has a different birthday. Take the people one at a time. The first person has 365 different possible birthdays. The second person has to have a different birthday from the first, so he only has 364 different possible birthdays. The third person’s birthday has to be different from the first two, so he has 363. The fourth has 362…The 23rd has 343. So there are

365 * 364 * 363 * 362 * … * 343

birthday distributions among the 23 so that everybody has a different birthday.

So the probability they all have different birthdays is:

## 365 * 364 * 363 * 362 * … * 343

365[sup]23[/sup]

= about 49.27%

So the probability at least two have the same birthday is about 50.73%.

So it’s just slightly over 50%. It gets larger pretty quickly. With 30 people, it’s 70.63%; for 35 it’s 81.44%; 40–89.12%; 45–94.10%; and for 50 there’s a 97.04% chance at least two will have the same birthday. So if you’re in a crowd of 45-50 people and can find someone to take the bet that two people will share a birthday, you’ve got a damn good chance of winning.

The next winning series will be

04 06 08 11 15 19

Right?

Y’know, I once saw the lottery come up “7 3 2”… And on that day, the number “7 3 2” came up more than all of the other numbers combined! What’s up with that?

I’ve got a scientific answer for you:

Fluke!

(Hey Chronos!:))

You’re probably just noticing when the lower numbers are repeated. If a 1 is picked amongst all 6 balls, it will still be listed first in the newspaper.

In the Virginia Lotto, for example, 962589 of the 7059058 combinations has the number 1. Since it’s listed first, you notice it more often. When more than 1/8 of the combinations start with 1, you think something is up. But in truth, every other number is contained in 962589 combinations.

If you’re referring to the series I gave above, sorry to disappoint but it’s got nothing to do with noticing **NOT** saying your observation isn’t quite valid, though; it’s just a series I’ve been playing - without much success, of course - for many years. Call it wishful thinking on my part.

When I said that certain numbers come up more often than others, I wasn’t necessarily referring to lower numbers, but to the lists published in different magazines that compile such things.

I used to work in a pub which had gaming machines based on draw poker. A lot of customers wouldn’t play them if the Royal Flush had appeared recently. It was a lost cause trying to explain to them that each hand had the same probability. I even showed them the records which showed that it had often occured twice in five minutes, or not at all for a week or two.

The machines these days even display on the screen: *number of games played since last Royal Flush*. Complete nonsense of course, but machines which haven’t “gone off” for a while will have a queue at them.

As far as lotteries go, the only sound advice I’ve ever heard (though I don’t play them), is "All numbers have the same chance. The only favour you can do yourself is to pick numbers 32 or higher. They don’t have any greater chance than numbers below 31, but given the amount of “lucky numbers” based on bithdays etc which people habitually use, if you *do* win, you will be less likely forced to share the prize pool.