what are the odds

Reply #27

jr-va — Wed, 10 Dec 2008 13:26:06 GMT

btw, why am I undulating ? what does it mean :)

anyway, I am just trying to make a good system to play pick 3 with, any thoughts or suggestions or assistance is welcome. Nothing is complete yet and ready to try, but so far I am thinking of using a constant set of about 10 numbers, and try for an exact match

Reply #26

jr-va — Tue, 09 Dec 2008 18:32:38 GMT

oh I'm sorry, but hi anyway

I agree, it can get confusing and expensive

Reply #25

jr-va — Tue, 09 Dec 2008 18:28:52 GMT

the 69th draw is not necessarily the end, it's closer to being the beginning of the $1 wager, when the chances are about 50%. By the 69th draw I should probably start betting the normal $1 per bet.

I see it easier thinking about it as the lottery trying many attempts to hit one of my 10 constant picks, like someone rolling dice for about 80 to 100 times to try to get a certain combination. The overall odds improve the more times the dice get rolled. It used to be for the same reason that... [ More ]

Reply #24

jessicalelle — Tue, 09 Dec 2008 18:27:32 GMT

well...i guess im not that good in maths, i have a hard time understanding this topic lol

Reply #23

Stew12 — Tue, 09 Dec 2008 17:13:49 GMT

Well the 50% odds are that the lottery will hit one of your 10 picks *somewhere* within a 69 draw period. If you say the odds are 50% (.5) that your numbers will hit within that period, you have 1/69th the chance for each day. 50% / 69 = .725%, or something like that. I'm not exactly sure on that calculation, but it works out to be somewhere around the standard 10:1000 percentage (1%), which is what you would expect for any day having 10 picks out of 1000 possible combinations.

Reply #22

jr-va — Tue, 09 Dec 2008 13:43:20 GMT

sorry, I meant draw, not day, since I would be allowing the lottery to use any and every official roll of its dice, so to speak.

Sounds like a conceivably reasonable system could be worked out from this, couldn't it? Remember, the first days no money is at stake, while the odds improve. For example, 50 cent bets only after the 30th or 40th draw, and $1 at the 69th draw, and $1.50/bet after the 80th draw

Reply #21

jr-va — Tue, 09 Dec 2008 13:32:48 GMT

alright, so keeping the set of ten numbers constant would be step 1.

Step 2 would be not to start betting money for a few days, the goal revolving around the fact that the probability of winning is around 50% on the 69th day.

I realize that playing from day 1 to reach the 69th day is too much money, but that is the purpose of not putting any money into it in the beginning. Perhaps gradually start at 50 cents per bet and then as we get closer to the 50th or 60th day start increasing th... [ More ]

Reply #20

johnph77 — Tue, 09 Dec 2008 11:32:55 GMT

You are correct - but these are odds, not possibilities. An even-money proposition does not guarantee you a win.

Reply #19

Stew12 — Tue, 09 Dec 2008 00:26:53 GMT

It definitely is, I was saying if you have 14 chances at 10:1000 it is not a simple (10*14):1000 chance of winning. With that calculation 100 days of 10 plays (10*100:1000 = 1:1) would *guarantee* you a win, which is not correct

Reply #18

RJOh — Tue, 09 Dec 2008 00:14:32 GMT

Isn't it more like having 10:1000 chances of winning 100 times than having a 1000:1000 chances of winning which it would be only if you played all 1000 possible combinations at one time

Reply #17

Stew12 — Mon, 08 Dec 2008 21:59:59 GMT

Your calculation is a bit off, in this scenerio if you played 10 games for 100 days you would have a 1000:1000 chance to win.

Reply #16

Stew12 — Mon, 08 Dec 2008 21:58:01 GMT

The lottery's chance to miss 10 picks:

28 days in a row: (((99/100)^28)*100) = 75.47 %

29 days in a row: (((99/100)^29)*100) = 74.72 %

30 days in a row: (((99/100)^30)*100) = 73.97 %

.

.

69 days in a row: ((((99/100)^69)*100) = 49.98 %

(99/100)^X = .5 -- X = 69 days for the lottery to be reduced to a 50% chance to miss all 10 picks each day.

Your chance is 100 minus the lottery's chance, leaving you with:

Hit once from 10 picks played 28 days in a row: 24.53 %

Hit o... [ More ]

Reply #15

jr-va — Mon, 08 Dec 2008 18:50:32 GMT

thanks johnph77,

that's what I was thinking, about 1 in 31/2 except when I try this system it will be days and nights, so that the 28 draws will be in 14 days (2 weeks).

The latest revision was to actually wait some of the days without betting any money, in effect improving my odds as it reaches the subsequent 14 day period. (Of course, an early win could be missed once in a blue moon, but I would just get a new set of #s and start the sytem over

Reply #14

johnph77 — Mon, 08 Dec 2008 18:41:49 GMT

If you're playing 10 games for 14 consecutive days on a straight-only ticket, your odds of winning are 140::1,000, or 1::7.14.

If you're playing the same way over 28 consecutive days, your odds of winning are 1::3.57.

If you're playing a box or straight/box ticket your odds will vary depending on whether your ticket has a double or not.

In any case your expected rate of return - i.e. payout percentage - is still the same, usually around 50%, depending on what the payouts are.

gl

j... [ More ]

Reply #13

jr-va — Mon, 08 Dec 2008 18:40:21 GMT

Ok, I'm playing 538, 053, 534, 384, 084, straight (exact)

and 84X, 03X, 0X3, X84 pairs.

Days thru Wednesday, Virginia.

(I must note, though, with the new idea I had as stated earlier, I really need 10 numbers that I can keep the same for up to a month if not more. So these don't necessarily count for that experiment.

Reply #12

jr-va — Mon, 08 Dec 2008 17:49:55 GMT

thanks, very well, I'm playing a few (day only) till Wednesday.

Thursday I'll re-examine. For doubles I'm just playing various pairs.

As I stated earlier, I don't trust the double and single frequencies any more, Virginia just

had a long run of singles. So with pairs I can partially cover doubles.

Alright, well I'm taking my tickets now. Thank you Dead aim.

Reply #11

Dead_Aim — Mon, 08 Dec 2008 16:29:10 GMT

Jr,

That may be the very filter you need to eliminate the drawings you feel are less fit. It will narrow the playing field somewhat allowing you to focus more attention and money of other combos.

This is a fresh set of numbers I put them together just before I posted them.

Reply #10

jr-va — Mon, 08 Dec 2008 15:30:48 GMT

thank you all

Stew12, thank you too, that is a handy way of calculating this. Now, what if I started with 50 cents or even no money, and if I didn't have a match in 28 draws then play for a dollar, the probability would be awful close to 50/50 ? (At the same cost, if I waited through a whole 28 days of failures before staking money ?)

Dead aim thank you so much. Did you filter these numbers? and how long have you used them, I ask because I noticed 584 (day) has hit, and 944,399... [ More ]

Reply #9

Dead_Aim — Mon, 08 Dec 2008 14:46:45 GMT

Hi jr,

Maybe these will help you out. Choose form the left charts if you feel singles are due and from the right charts if you feel doubles are due. I think you will find these helpful. I use them for a week at a time.

VA MID VA MID

534 538 053 584 054 533 388 055 554 844

058 384 034 038 084 588 344 033 800 004

VA EVE VA EVE

198 194 319 148 318 199 944 311 118 488

314 948 398 394 348 144 988 399 433 338

Reply #8

Stew12 — Mon, 08 Dec 2008 02:17:20 GMT

You can calculate your odds more easily by calculating how likely the lottery is to miss your numbers.

Multiplication rule of independent events

If you have 10 picks, the lotto has a 990/1000 chance of missing your numbers (99/100) each day for 28 days.

(99/100)^28 = .7547 (or 75.47%)

This leaves you at a 24.53% chance to win $500 over a 28 day period with 10 tickets for each day.

Long story short: $280 purchase with 24.53% chance to win $500. Wouldn't take that myself personally.... [ More ]

Reply #7

jr-va — Mon, 08 Dec 2008 00:09:22 GMT

I understand it involves a whole new drawing, but I still think the odds can be better by playing for 28 draws. It is expensive, though.

I think the way I was thinking is better described if we turn the system around and ask

What are the odds of having the lottery pick a number in my set of ten if I give them 28 tries?

So, even if it is a whole new drawing, as long as I keep my 10 numbers the same throughout their 28 tries, the odds of having an exact match should still be about 1

Reply #6

RJOh — Sun, 07 Dec 2008 22:52:01 GMT

jr-va,

You are confusing the odds of winning a pick3 playing 1000 combinations at one time with the odds of winning playing 1000 combinations over a period of time.

The difference is if you play all 1000 possible combinations at one time, you can only be wrong 999 times and one of your picks will be right however if you play 10 combinations 100 times, you can be wrong all 1000 times.

Reply #5

Shawn67 — Sun, 07 Dec 2008 22:29:41 GMT

As each new day is a whole new draw, and since any number can be called each drawing, I understand that your chances would not change, they would still be 1:100 for each drawing.

Reply #4

RJOh — Sun, 07 Dec 2008 22:24:08 GMT

To win $500 in a pick3 game you have to hit a straight with a $1 ticket. Playing 10 straights for 28 days cost $280 and you're more likely to spend more than $500 by the time you win that much unless you're lucky because your odds of winning (1:100) are the same every day, they don't improve as you lose unless you spend more.

Reply #3

jr-va — Sun, 07 Dec 2008 22:08:34 GMT

thanks, RJ and Shawn

so yes, it is 1 in 100,

but then if I keep those same ten numbers for 28 draws, should I have a 1 in 3.6 chance of winning the $500 exact

Reply #2

RJOh — Sun, 07 Dec 2008 22:02:55 GMT

There are only 10 different numbers(0-9) in a pick3 game which cover all 1000 possible combinations. If you play 10 combinations then your chances of a straight hit are 10:1000 or 1:100. If you're talking box hits, double or triple then the odds changes. The odds are usually printed out on the back of the play slips and are available at the lottery websites.

Reply #1

Shawn67 — Sun, 07 Dec 2008 21:58:45 GMT

In Pick-3, if you got 10 unique combos, then yes, your odds would be 1:100 of having the winner. However, it has been my experience that QP's quite frequently give repeats, which increases your odds considerably.

what are the odds

jr-va — Sun, 07 Dec 2008 21:53:23 GMT

If I play random numbers on pick3, quick picks for example, for a $500 prize:

what are the odds if I pick 10 numbers to play day and night for 14 days? 1 in 3.57?

If I play 10#s the odds are 1:100 , correct?

and then by letting the machine roll out 28 numbers against my set of 10 does that mean

divide 100/28=3.57 or 1 in 3.57 ? (if my 10#'s are the same throughout the 2 week period