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# what are the odds

Topic closed. 27 replies. Last post 9 years ago by jr-va.

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 Posted: December 8, 2008, 1:50 pm - IP Logged

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)

CT
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 Posted: December 8, 2008, 4:58 pm - IP Logged

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 (eve) also hit, and 311 hit twice  (all in the second half of 2008).  Anyway, do you intend to keep all these that hit in the previous few months ?

thank you everyone and good luck.

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 once from 10 picks played 29 days in a row: 25.28 %

Hit once from 10 picks played 30 days in a row: 26.03 %

.

.

Hit once from 10 picks played 69 days in a row: 50.02 %.

You need to play 10 picks for 69 days at \$690 total to have a 50.02% chance to win \$500 (\$190 less than you spent).

CT
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 Posted: December 8, 2008, 4:59 pm - IP Logged

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

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.

mid-Ohio
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 Posted: December 8, 2008, 7:14 pm - IP Logged

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.

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?

* you don't need to buy more tickets, just buy winning ones *

CT
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 Posted: December 8, 2008, 7:26 pm - IP Logged

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?

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

CA
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 Posted: December 9, 2008, 6:32 am - IP Logged

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

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

Blessed Saint Leibowitz, keep 'em dreamin' down there.....

Next week's convention for Psychics and Prognosticators has been cancelled due to unforeseen circumstances.

=^.^=

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 Posted: December 9, 2008, 8:32 am - IP Logged

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 the bet to \$1 and perhaps raise it above \$1 if there is no match by the 75th day?

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 Posted: December 9, 2008, 8:43 am - IP Logged

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?

CT
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 Posted: December 9, 2008, 12:13 pm - IP Logged

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?

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.

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 Posted: December 9, 2008, 1:27 pm - IP Logged

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

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 Posted: December 9, 2008, 1:28 pm - IP Logged

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 I would buy more tickets per drawing, my tickets are like dice rolls trying to match what the lottery machine came up with.  This time I only switched the players in my perspective, this time I'm allowing the lottery machine many and many tries to match something in the constant list of numbers.  Most lottery machine draws will be unique, it's true that possibly one or two may be identical, but that shold not be detrimental.

For example, I think of 10 numbers and then you get one try to guess one of them, and someone else gets 2 tries per day for 90 days.  Although 'each' time that the other person calls out his guess has the same chance as your single guess, the other person has a higher 'overall' chance of guessing a right number and it exceeds 50% past the 69th attempt.

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 Posted: December 9, 2008, 1:32 pm - IP Logged

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

oh I'm sorry, but hi anyway

I agree, it can get confusing and expensive

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 Posted: December 10, 2008, 8:26 am - IP Logged

oh I'm sorry, but hi anyway

I agree, it can get confusing and expensive

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

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