If you played lotto for 20 years with the same numbers each week, or played lotto for 20 years with randomly selected numbers each week, which would give you a better chance of winning the jackpot over a 20 year period? I understand that any given draw will result in an equal chance of them winning, but over a 20 year period, would you have a better chance of winning if you played the same numbers every week?
The odds are the same I believe.
Each drawing, no one number is more or less like to funnel out than any other number.
I’m sure if I’m wrong, someone will be along shortly to correct me. Gently, please.
I dunno. If you flip a coin, and call “heads” every time, it eventually comes up. Of course, you only have two choices instead of 7 million or so in a 6number lotto game. My guess is that over time, every number should come up, but there is no guarantee that it will. At least that’s my intuition, which, given the nature of probability, means that my argument is dead wrong.
I reasoned it even more simply.
Think, if you will, roulette.
Each spin the odds are 381 on any given numbered space.
My odds do not change no matter whether I always play the same number or if I move my chip around. Each spin the odds are always going to be 381.
Now I may believe that playing the same number until it hits is more likely to win me moola than changing my selection around, but oddswise it just isn’t so.
This actually came up in a stats class I took in Grad school. According to my professor, the odds stated on the tickets are set by one play, one time. Thus  they odds are set for the same numbers each play because each time the balls are drawn, it’s one play, one time. (It doesn’t matter how many “sets” of numbers you buy  the odds are the same because only one draw is made).
However!
He stated that if you choose random numbers each time you play, he said that “your” odds of winning are now squared from the original odds because now you have entered randomness on your end of the game rather than there only being randomness on the draw side of the game.
They’re equal strategies.
To give you some idea of how bad your odds are, I knew a guy who played the numbers 1,2,3,4,5 and 6 in the local 6/49 draw. It took me awhile to convince my parents, but 123456 is just as likely to come up as anything else.
From what I can discern, probability wise you are more likely to win playing the same numbers for 20 years. The reason is because when you figure your odds, you base them on the total times you play with one set of numbers. So in theory (assuming you play the lottery once a week for 20 years):
1 set of the same numbers played 1040 times
or
1040 sets of numbers played 1040 times
With playing one set of numbers the entire twenty years, your odds of winning always increase in your favor, due to the nature of the fact that those numbers are due.
Conversely playing a different set of numbers each time is the equivelant of playing a different slot machine after each pull of the arm, you may get lucky, but mathmatically the odds are not in your favor.
I can’t find the exact math on this, but I believe you can basically figure it out by a comparison of the odds of winning versus the number of times you play the number.
Made a calculation error, it should go like this:
1 set of the same numbers played 1040 times
or
1040 sets of numbers each played 1 times
That guy is gonna be pretty pissed if those numbers ever come up. I recall an interview with our local lottery dude who said a surprising number of people pick those numbers.
He may prove his point, but he’ll have to share the glory with a couple hundred of people.
Schieschkopf
It sure seems that way doesn’t it.
But the problems is:
Mathematically, there is no such thing as a number being “due” in a random system. Hopefully, someone more articulate regarding mathematics than I will come along and explain why this is. The odds are always the same no matter which numbers you play or whether you play them consistently.
Welcome, BTW.
I think we’re dealing with a fallacy here known as “The Law of Averages”. If you throw a coin and it comes up heads 20 times in a row, people will bet on tails, thinking the “law of averages” means the odds are improved that tails will come up, because it’s “overdue.” Pure rubbish.
There is only one way to improve your odds of winning the lottery: buy more tickets. There is only one way to be guaranteed of winning the lottery: buy ONE of EVERY ticket. Yes, it has been done.
I can think of two strategies for the lottery, both of which are still very unlikely to make you a winner.

If there is any lack of randomness (unlikely) then it will cause the same numbers to be picked more often so better to pick numbers which come up most. This is more rational than the ‘due numbers’ strategy as it does not require the numbers to have ‘memory’ but rather the lottery balls to have some physical abnormality. In practice any lottery in which this was a significant feature would IMHO be corrupt.

Accepting that there is not real way of improving the odds on winning, why not improve the odds on being the sole winner? In the UK’s National Lottery the twiceweekly prize is divided between all those who matched all the numbers. If you can get stats on winning lines and numbers of winners that draw, you can calculate for each number the average number of winners each time it was in the winning line. It’s a pretty good guess that a combination of those with the lowest average give the best chance of keeping all the money yourself  should the 14 million to one chance actually occur of course!
The best strategy of all is not to play at all and save your money. Remember, whenever someone wins, that’s YOUR MONEY they are pocketing.
I undertsand that in ANY GIVEN DRAW, a person with random numbers has the same chance of winning as someone who picks the same numbers week in, week out.
But when you are talking about multiple recurrences, the odds do change. For example, let’s say I toss a coin 2 times. What are the odds that I am going to get 2 heads? They are 1 in 4. Now let’s toss the coin. It’s a head. What are the odds I am going to get 2 heads? Are they still 1 in 4? No, they are now 1 in 2. I am wondering if this is the same in a lottery draw, over multiple games.
Or…
Let’s think of it this way. What are the odds that my loto numbers are going to come up if I keep picking the same one’s for the next 20 years? Well, if they come up at ANY POINT during those 20 years, I will win. But what if I change my numbers each week? I would have to pick those random numbers on the night that they come out. Not the week before, no the week after, but on THAT NIGHT. Someone who picks the same numbers every week wouldn’t have that problem.
Would this effect the odds?
Come to think of it, I could write a small program to simulate this. But my eyes are hanging out of my head, so if it hasn’t been answered by tomorrow, I’ll do it then.
No, this would only apply to picking numbers which come up twice in two games.
No. Every draw is a fresh start. Every set of numbers has the same chance on every draw.
Here’s an interesting bit on the subject: A couple of weeks ago our Lotto 6/49 here in Canada had a winning sequence of 38,43,44,45,46,47 and there were 6 winners!
Which obviously reduces each winners take considerably.
Imagine picking an “odd” combination like this for years only to find out you have to split the pot 6 ways. It’s not like it’s going to happen again in your lifetime…
That, uh, was kinda my point **Ookpik[b/].
Are these names getting stranger, or is it just me?
First of all, the idea of collecting the averages for the times x ball comes up will improve your odds a little bit, but not much…
Why? Mainly because the fact that most lottery commissions don’t use the same machine and the same balls all the time. They switch them out periodically to prevent this.
Second of all, I believe my idea is right because of the law of averages (and of gambling). It can be expressed according to this formula here (which is only accurate if you play with one set of numbers):
N = log(1  DC) / log(1  p)
(where Number of Drawings (N)
Necessary For An Event of Probability §
to Appear With The Degree of Certainty (DC))
Degree of Certainty being expressed in Percentage (i.e. 50%)
Probabilty being the odds against you (1/100000 or whatever)
It would take longer than 20 years to have a 99% probability of having your numbers picked (for a 6 ball lotto). As a matter of fact it would take a few hundred thousand years to ensure a 99% probability of your numbers being picked, and that’s playing with the same numbers. Playing with different numbers greatly skews this equation, and resets the Percentage of your numbers bing picked.
Who wants to try it out and see? ":'P
I doesn’t change your odds of winning, but it can change your E (expectation) of winning. If you pick some sort of pattern of numbers, then most likely someone else will pick out on the same pattern, so if you win, you will have to split it more ways. So if you scientifically pick a set of numbers based on lack of pattern, and lack of being commonly selected(Numbers over 31 are good because many people pick their families birthdays),and play that every week, you will be less likely to have your winning numbers matchedif you do happen to win, as opposed to a true random selection every week.
From The Assailant
Here’s some code. Three players play. One stays the same always. One picks randomly. And, just to show it really doesn’t matter, one picks randomly but only from half the possible values.
The code:
#include <stdio.h>
#include <stdlib.h>
int main () {
int i, draw;
int wins1=0;
int wins2=0;
int wins3=0;
srand48(1234); //random seed
for (i=1;i<10000000;i++) { //let's play ten million weeks
draw=(lrand48()%50); //lottery drawing (049)
if (draw==1) wins1++; //he always plays 1
if (draw==(lrand48()%50)) wins2++; //he plays random 049
if (draw==(lrand48()%25)) wins3++; //he plays random 024
}
printf("%d %d %d
", wins1, wins2, wins3);
return 0;
}
Outputs with a few different random seeds:
 199034 200017 200657
 199837 200400 199392
 200030 199667 199891
Your argument’s fault lies around your natural gut instinct to think, “I’ve been playing this same number for months now. I can’t change now! What if it comes up?!” The thing to realize, though, is that it is just as likely that the number you might change to might come up. What if it hits?! The net result: it doesn’t matter what you pick.
Another way to look at it:
Jim and Bob decide to play 20 weeks of the lottery. Jim plans to play the number 32 each week. Bob picks a string of 20 random numbers to play, one each week. Jim, then, is saying, “I want week 1 to hit number 32, week 2 to hit 32, week 3 to hit 32, week 4 to hit 32, …” Bob is saying, “I want week 1 to hit 12, week 2 to hit 43, week 3 to hit 22, …” So, they expect to win this many lotteries:
Jim: Prob(week1=32) + Prob(week2=32) + Prob(week3=32) + …
Bob: Prob(week2=12) + Prob(week2=43) + Prob(week3=22) + …
where Prob(week1=32) means “The probability that week 1 comes up 32”. They end up with the same expected number of wins, i.e., the same probability of winning – all because the draws are independent and uniform.
P
Cecil Adams on How can I pick the winning number in the lottery?