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Topic closed. 4 replies. Last post 6 years ago by RJOh.

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New Member
Wilmington, NC
United States
Member #105131
January 25, 2011
1 Posts
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 Posted: January 25, 2011, 8:21 pm - IP Logged

For example - if you buy/develop a system for a particular game with published overall odds of: 1:8.1

And you bet say - \$100 on the system.

How much would you expect to get back,

and how much over odds would you have to be - before anyone would care?

Thanx, Jason

(P.S. would have jumped on this back in college -but stats class was a long time ago..)

United States
Member #93947
July 10, 2010
2180 Posts
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 Posted: January 26, 2011, 1:16 am - IP Logged

For example - if you buy/develop a system for a particular game with published overall odds of: 1:8.1

And you bet say - \$100 on the system.

How much would you expect to get back,

and how much over odds would you have to be - before anyone would care?

Thanx, Jason

(P.S. would have jumped on this back in college -but stats class was a long time ago..)

jason-eagle,

What you would get back is called the Expected Value of the wager.

Expected Value is the Probability of Winning multiplied by the Payoff IF you Win.

We need more information to answer your question in this case.  Specifically, what the publisher means by 1:8.1.  In most Pick-3 games, for example, the published odds are 1:1000 for a straight bet, and most states pay \$500 for a winning \$1 straight ticket.  So, in these games, the Expected Value (EV) of a \$1 straight ticket purchase is:

EV = ( 1 / 1000 ) * \$500

EV = 50¢

If you bought 100 tickets with your \$100, what you would expect to get back would be:

EV = ( 100 / 1000 ) * 500

EV = \$50.00

Of course, if you had a winner on your first go around you would be ahead \$400 right off the bat!  The Expected Values calculated above are those expected over a long period of time.

--Jimmy4164

Kentucky
United States
Member #32652
February 14, 2006
7344 Posts
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 Posted: January 26, 2011, 10:37 am - IP Logged

For example - if you buy/develop a system for a particular game with published overall odds of: 1:8.1

And you bet say - \$100 on the system.

How much would you expect to get back,

and how much over odds would you have to be - before anyone would care?

Thanx, Jason

(P.S. would have jumped on this back in college -but stats class was a long time ago..)

Most lotteries give ticket odds so 1 in 8.1 means one ticket should win something for every 8.1 tickets sold. If you buy 100 tickets, you should expect to get 12 or 13 winners (100/8.1) and if the "something" you won on each ticket was free ticket, you'll lose \$87 or \$88. A profitable system for that game would be a way to win more than a \$100 for each \$100 bet.

"how much over odds would you have to be - before anyone would care?"

Just to win "even money", you'll need a return of 200%.

Chief Bottle Washer
New Jersey
United States
Member #1
May 31, 2000
23352 Posts
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 Posted: January 26, 2011, 1:20 pm - IP Logged

<Moved to Lottery Systems forum>

Please post in the appropriate forum ... thank you.

mid-Ohio
United States
Member #9
March 24, 2001
19904 Posts
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 Posted: January 26, 2011, 1:56 pm - IP Logged

For example - if you buy/develop a system for a particular game with published overall odds of: 1:8.1

And you bet say - \$100 on the system.

How much would you expect to get back,

and how much over odds would you have to be - before anyone would care?

Thanx, Jason

(P.S. would have jumped on this back in college -but stats class was a long time ago..)

Those overall odds sound similar to those of a 5/39 game that pays for a match 2.

combination size             5
basic pool size              39
(B) bonus numbers            none
smallest match               2
tickets or chances per draw  100
possible combos of 5/39 numbers = 575757
MATCHES   ODDS              WINNING COMBOS    EXPECTED WINNERS
5/5    1 : 575757           1                        0.00
4/5    1 : 3387             170                      0.03
3/5    1 : 103              5610                     0.97
2/5    1 : 10               59840                   10.39
__________________________________________________________________________
overall odds are 1 : 8.7               11.4 total expected winners
65621 winning combos = 11.39 % of possible combos

You can expect 11.4 winners by just buying 100 chances regardless of how you pick the numbers.  I assume if you bought/developed a system then you expect one of those winners to be a match 4 or better or your system haven't given you an advantage.

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

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