mersenneforum.org Lottery pick-3 odds?
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 2017-05-01, 02:37 #1 schickel     "Frank <^>" Dec 2004 CDP Janesville 84A16 Posts Lottery pick-3 odds? So, I know just enough probability to get myself in trouble. For example: pick-3 games (pick 3 digits from 0-9 inclusive) have 1 in 1000 odds of winning on a straight bet, ie. I play 395 and the computer picking that number should happen every 1000 draws or so. (I like to think) I'm not naive enough to think that if a number hasn't hit in 900 draws that it is more likely to hit than any other number (assuming a fair random number generator and unbiased code*), but I don't know how to work out what the odds are that if I were to pick a number, say 734, and play it constantly that I will have to wait more than 1000 draws for it to hit. * California uses an "Automated Draw Machine" to pick winners for the Pick-3 and other games.
2017-05-01, 02:47   #2
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts

Quote:
 Originally Posted by schickel So, I know just enough probability to get myself in trouble. For example: pick-3 games (pick 3 digits from 0-9 inclusive) have 1 in 1000 odds of winning on a straight bet, ie. I play 395 and the computer picking that number should happen every 1000 draws or so. (I like to think) I'm not naive enough to think that if a number hasn't hit in 900 draws that it is more likely to hit than any other number (assuming a fair random number generator and unbiased code*), but I don't know how to work out what the odds are that if I were to pick a number, say 734, and play it constantly that I will have to wait more than 1000 draws for it to hit. * California uses an "Automated Draw Machine" to pick winners for the Pick-3 and other games.
depends on if they can be redrawn etc. at last check if they can the odds of not picking n for y drawings is (999/1000)^y I think, which would put your odds of not picking it for 1000 at about 36.77% in theory.

2017-05-01, 03:31   #3
schickel

"Frank <^>"
Dec 2004
CDP Janesville

2×1,061 Posts

Quote:
 Originally Posted by science_man_88 depends on if they can be redrawn etc. at last check if they can the odds of not picking n for y drawings is (999/1000)^y I think, which would put your odds of not picking it for 1000 at about 36.77% in theory.
If you mean redrawn within each draw, yes, it's 3 independent fields of 10 digits. In fact just within the last week, the winning number was 555 (I think.....let me check)

Last fiddled with by schickel on 2017-05-01 at 03:34 Reason: Re-thoiught the post

2017-05-01, 03:41   #4
schickel

"Frank <^>"
Dec 2004
CDP Janesville

1000010010102 Posts

Quote:
 Originally Posted by schickel In fact just within the last week, the winning number was 555 (I think.....let me check)
Sunday April 23 evening draw. Looks like it doesn't really pay to play triples.

 2017-05-01, 04:17 #5 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 90216 Posts At the risk of stating the obvious, the probability of picking any given 3 digits is (1/10)^3, so from a theoretical point of view, given the large enough sample(of draws), you will get any given 3 digit combinations very close to 1/1000th of times. However studying historical lottery results you will find out that some numbers are biased and occur more often. There are two different factors at work. 1. The draw is not truly random 2. There is observed influence in random draws, which probably has a quantum basis. In quantum physics it is a basic principle that just observing an experiment will affect the outcome of that experiment. For example in the double slit experiment, observing which slit a photon passes through will collapse its wave function and destroy the interference fringes which would be otherwise present. Last fiddled with by a1call on 2017-05-01 at 04:18
2017-05-02, 00:54   #6
schickel

"Frank <^>"
Dec 2004
CDP Janesville

212210 Posts

Quote:
 Originally Posted by a1call At the risk of stating the obvious, the probability of picking any given 3 digits is (1/10)^3, so from a theoretical point of view, given the large enough sample(of draws), you will get any given 3 digit combinations very close to 1/1000th of times. However studying historical lottery results you will find out that some numbers are biased and occur more often. There are two different factors at work. 1. The draw is not truly random 2. There is observed influence in random draws, which probably has a quantum basis. In quantum physics it is a basic principle that just observing an experiment will affect the outcome of that experiment. For example in the double slit experiment, observing which slit a photon passes through will collapse its wave function and destroy the interference fringes which would be otherwise present.
Yeah, maybe, but if it was really easy, someone would have exploited any bias by now. According to the FAQ, they have two different draw machines and two different ways to pick the winning numbers, with both being chosen before the draws are run. I imagine there would also be a test draw or two before the winning draw, so it would seem to be impossible to figure out any effect there.
Quote:
 Originally Posted by science_man_88 depends on if they can be redrawn etc. at last check if they can the odds of not picking n for y drawings is (999/1000)^y I think, which would put your odds of not picking it for 1000 at about 36.77% in theory.
Wow, that seems high to me. That's really why I'm asking, since my instincts could be wrong.

So, if there's a ~37% chance of not drawing one number in 1000 draws, what would be the odds of missing a win in the following scenario: take 10 different numbers and play them for 100 draws.

Intuition would seem to say that 10(numbers)x100(draws) = 1000 "trials" which would seemingly make it a certainty that you would hit once, but I don't know enough about how to calculate the odds in that case.

2017-05-02, 01:06   #7
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

20E216 Posts

Quote:
 Originally Posted by schickel Yeah, maybe, but if it was really easy, someone would have exploited any bias by now. According to the FAQ, they have two different draw machines and two different ways to pick the winning numbers, with both being chosen before the draws are run. I imagine there would also be a test draw or two before the winning draw, so it would seem to be impossible to figure out any effect there.Wow, that seems high to me. That's really why I'm asking, since my instincts could be wrong. So, if there's a ~37% chance of not drawing one number in 1000 draws, what would be the odds of missing a win in the following scenario: take 10 different numbers and play them for 100 draws. Intuition would seem to say that 10(numbers)x100(draws) = 1000 "trials" which would seemingly make it a certainty that you would hit once, but I don't know enough about how to calculate the odds in that case.
I think that would make it into (990/1000)^100 = 0.36603234127322950493061602657251738619 ( used PARI/GP both times I think)

probability would depend on so many things see:

https://www.youtube.com/watch?v=lP58mP8Wchc standupmath video ( matt parker)

that may help you. it's also why the monty hall problem stumbles so many as the video they made on that shows.

Last fiddled with by science_man_88 on 2017-05-02 at 01:07

 2017-05-02, 13:27 #8 science_man_88     "Forget I exist" Jul 2009 Dartmouth NS 2·3·23·61 Posts they give example odds for one play at the bottom of http://www.calottery.com/play/draw-g...-3/how-to-play but there's a lot to think about some numbers it doesn't matter which you play etc. numbers with all the same digit only have one way of sorting them so any= exact and the two main things become the same ( down to the prize money or using the straight/box option to win both prizes). those that have 2 digits in common ( something like 27% of the numbers total if you don't include the overlap with the all three digits the same in the set) have only 3 ways to arrange themselves, whereas numbers with no digit in common have 6 ways to arrange themselves. let x stand in for the same digits those with two or more digits the same include ( per digit x with some overlap e.g for x=6 xx6 x6x and 6xx are all the same number ( 666 the case where all three are the same digit technically)) x0x x1x x2x x3x x4x x5x x6x x7x x8x x9x xx0 xx1 xx2 xx3 xx4 xx5 xx6 xx7 xx8 xx9 0xx 1xx 2xx 3xx 4xx 5xx 6xx 7xx 8xx 9xx if you take away the overlap you have 28 possibilities remaining ( including 666 111 etc only once not three times). edit:doh I now see that this is already explained in that link. Last fiddled with by science_man_88 on 2017-05-02 at 13:40
2017-05-02, 20:23   #9
chalsall
If I May

"Chris Halsall"
Sep 2002

10,861 Posts

Quote:
 Originally Posted by science_man_88 if you take away the overlap you have 28 possibilities remaining ( including 666 111 etc only once not three times). edit:doh I now see that this is already explained in that link.
Just to share, you do understand that the house always wins (statically)?

There is a reason lotteries are not popular where gambling is legal.

2017-05-02, 20:57   #10
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts

Quote:
 Originally Posted by chalsall Just to share, you do understand that the house always wins (statically)? There is a reason lotteries are not popular where gambling is legal.
my point was one of best case scenarios because some numbers get better odds in the setup than others do.

xxx will have the best win with straight/box combo
xxy xyx and yxx forms still have best win for them in straight/box combo but are at lower number of possibilities than the forms xyz, xzy, yzx, yxz, zxy, and zyx.

2017-05-02, 21:10   #11
chalsall
If I May

"Chris Halsall"
Sep 2002

10,861 Posts

Quote:
 Originally Posted by science_man_88 my point was one of best case scenarios because some numbers get better odds in the setup than others do.
It's a bit like trying to teach calculus to a mouse.

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