Let’s say you’re playing Pick 3… you choose three numbers, 0-9, and the balls are replaced into the bowl after each draw. If you buy a ticket with the numbers 2-2-2, are you making the odds worse for yourself? In other words, are the chances of the same number coming up three times better, worse, or the same as three different numbers coming up?
I understand the whole thing about, well, there are ten numbers to be chosen on each draw, so the odds of getting three 2s are the same as getting 3-6-1 or whatever. But at the same time, it seems like there’s got to be something in there about the same number coming up being a lesser probability.
Nope, it really is the same probability to get 2-2-2 as it is to get 3-6-1. Any particular set of 3 numbers has the same probability as any other particular set of 3 numbers. Now, if the mechanics of the ball-picking make it more likely to get a ball from the bottom of the pile, and the old ball gets replaced back on top, that’s different and it changes the probabilities. But if you’re truly randomly picking the next number, they all have the same chance.
The chances are the same, assuming the number that comes up is really random.
There may even be a slight chance the chances are BETTER, if there’s something making it not truly random – for example, if it’s one of those air-blowy machines where the ping pong balls bounce around, if one of them weighs less than the other, it’s more likely to show up.
But as long as there is a truly random chance of each ball coming up each time, the chances are the same no matter which three numbers you pick.
I think the odds are the same, no matter what three numbers you pick. I know it seems like the chances of drawing something like 2-2-2 are less than other numbers, but I think its just because when a significant number like that happens to come up, we take note of it and remember it.
What I think is going on here is that when someone sees 2-2-2 they think “Wow the odds of that are low” and in fact they kind of are. 2-2-2 is as likely as any other 3 digit number. But the odds of drawing three of the same number are in fact 1 in a 100 or 1%. Thaty is where I think the “feeling” comes from, your feeling is not that 2-2-2 is less likely than any other three digit number it is that this pattern is less likely than not seeing the pattern.
If you are dealt all thirteen spades, you think it’s practically a miracle. But the odds of getting that hand are exactly the same as (for example)
Spades: K 9 6
Hearts: A J 4 3
Diamonds: Q 6 4 2
Clubs: 8 2
The above hand appears to be nothing special, except that each card is clearly defined, so your chances of ever getting it are astronomically low. :eek:
The chances of getting a triple are lower than not getting a triple (there are ten combinations representing triple numbers and 990 representing other, non-triple patterns) - that’s what throws us off the scent a bit, I think.
But of course that doesn’t give us any useful information on picking a likely combination - they’re all equally likely (assuming a fair game, etc)
That’s the odds of drawing “2-2-2” or “3-3-3”, but the odds of drawing three of the same number (any number), are ten times that, or 1%, as Gangster Octopus said.
Another point is that the numbers 1, 3 and 6 could come up in any order, giving you six different combos - so a permutation of 1, 3 and 6 will come up six times as often as 2-2-2, but 3-6-1 specifically will not.
One way to get a good perspective of the chances of winning a lottery (say five numbers, with balls labeled from 1 to 39), is to think about how amazed you would be if the numbers “1-2-3-4-5” came up. Yet that combination is no less likely than the specific combination that you picked, no matter how “random” your pick looks.
I think the odds appear low because one thinks “What are the chances that the same number will come up versus it not happening”.
Which is correct becuase 1% of the time the three digits will be the same, 99% of the time they won’t.
But the “not happening” combinations amount to 990 other numbers.
But if you’re forced to pick 10 numbers; 000, 111, 222, 333, 444, 555, 666, 777, 888, & 999
or 10 other “specific” numbers; 345, 854, 611, 074, 804, 846, 022, 583, 737, 968
then your odds of you winning with a number from either set remain the same, 1%.
I don’t think the OP is suffering from a false sense of randomness.
What he’s asking is… let’s say you have a 9 balls in a lottery. 1 1 1, 2 2 2, 3 3 3. Is 222 any less likely than 123?
Well, if you think about it in terms of a consecutive series - the first ball is a 2 (1/3rd chance), so you now have 2/8 chances to draw a 2 with the remaining ball. You do. Now you have 1/7 chances of drawing the last 2. It appears that, from that perspective, the odds of that happening are 1/3 * 1/4 * 1/7 = roughly 1.1%.
On the other hand, the 123 drawing appears to be 3/9 * 3/8 * 3/7 = roughly 5.1%.
In other words, if there are a fixed number of each type of ball, every time you draw one of those numbers, it becomes less likely that the number will be drawn again.
And… this is fundamental stuff to me, but I’m up past my bed time and am unable to think clearly, so I’m not able to figure out the error here. I’ll revisit this tomorrow if no one has offered an explanation.
Some of the explanations in this thread seem faulty, though. If you roll a die 3 times, you’re as likely to get a single number all three times as any other combination of numbers, because one die roll doesn’t affect the next. However, when you’re taking lottery balls away from a lottery as you pick them, you’re affecting the future probabilities. I think some people are trying to explain the former, rather than the latter.
He said “the balls are replaced in the bowl”, so I think you’re the one trying to explain the wrong model, dude. And you’re right, this is fundamental stuff, in this case RTFOP.
Aha, you’re right. Is that standard lottery procedure? I don’t watch the lottery drawings much, but I seem to recall them taking off a number of balls with no replacement. Then again, maybe I’m thinking of one of those 1-50 type lotteries where there’s one ball for each number.
Anyway, yes, any combination in that scenario is equally likely.
The drawings I remember seeing on the local news as a kid had no replacement… but they used a different machine for each number, so no number was dependent upon another.
This is a really good point. According to the North Dakota Lottery FAQ page I just found, the order is irrelevant. Thus, for our drawing of 3 numbers out of 10 possible, if you pick 3 identical numbers on your ticket you have a 0.1% chance of winning. If you pick 3 different numbers you have a 0.6% chance of winning. Wow! By choosing different numbers you increase your odds by a factor of 6! The more numbers drawn, the higher the benefit.
Sorry I am multi-posting, but I am finding more information out (as I do not play the lottery myself, considering it to be a special tax on people who are bad at math). The Mega Millions lottery, which appears to be the world’s biggest, has you pick 5 numbers in the range 1-56 and a last number in the range 1-46. The first five can be in any order, but the last number must be last. So, you can duplicate your last number with any of the others with no penalty, but if any of the first five are the same your odds go down.
Best case scenario:
Five different numbers, then anything for the last.
Odds:
(5/56)(4/56)(3/56)(2/56)(1/56)*(1/46)=120/46/(56^5)=0.00000047%
Worst case scenario:
Five identical numbers, then anything for the last:
Odds:
(1/56)(1/56)(1/56)(1/56)(1/56)*(1/46)=1/46/(56^5)=0.0000000039%
So, picking all different numbers for the first 5 increases your chances 120 fold! I’d still rather bet on getting hit by lightning, but that is a heck of an increase.
As the OP has been answered, would anyone object if I were to hijack this thread on Monday when I am at work? It is sometimes possible to play the lottery with a positive expectancy. Interested readers with access can read “Parimutuel Betting Markets: Racetracks and Lotteries” from the Journal of Economic Perspectives.
