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# Buying more tickets = reduced odds

Topic closed. 100 replies. Last post 13 years ago by Dowser.

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Member #2673
November 2, 2003
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 Posted: May 3, 2004, 8:33 pm - IP Logged

This topic came up in another thread but the thread went off in a tangent so I decided to start a new thread devoted to the following questions.

Q: Can lottery odds be reduced by buying more tickets?

A: Yes

Q: How are the new, reduced odds calculated?

A: Say that a lottery has X number of unique outcomes. Then the odds of hitting the jackpot is 1 in X. If one buys N number of tickets where no two tickets have exactly the same set of numbers, then the odds of hitting the jackpot is 1 in X/N.

Good luck,
Jake

P.S. prob987, this is in response to your request in the "Are you serious" thread

Pennsylvania
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 Posted: May 3, 2004, 8:56 pm - IP Logged

it doesn't work like that...

let's use the simplicity of the pick 3... 1,000 possible straight combinations (ignore boxes for the purpose of the discussion...)

you buy 1 ticket (1 combo out of 1,000)... you leave 999 uncovered possibilities...

you buy 2 tickets (2 combos out of 1,000)... you leave 998 uncovered possibilities... by your formula above it would be a reduction from 1:1000 to 1:500, but you leave 498 combos unaccounted for.....

they may express odds as a fraction, but that does not mean you can manipulate it with simple algebra... if it were that easy...

think of the original odds formula... Favorable Outcomes : Possible Outcomes... hence 1:1000 means one ticket gives you a chance in 1,000 of being right

2 tickets means you have 2 chances in 1000, because all 1000 possibilities still exist before draw time, they don't magically disappear... (except for here in PA where they had worked wonders with some latex paint and a syringe ;-)  to limit possible draw outcomes. )

The ONLY way to reduce odds in a single dvent game is to alter the matrix (like pulling all the even balls out of the machines).... and since we can't do that... we can't beat the odds by volume ticket purchases... it only takes one ticket to win... quality of picks over quantity of picks anyday... (too bad I still can't come up with quality picks...)

Playing more than one ticket per game is betting against yourself.

New Jersey
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January 22, 2004
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 Posted: May 3, 2004, 9:32 pm - IP Logged
Quote: Originally posted by Jake649 on May 03, 2004

This topic came up in another thread but the thread went off in a tangent so I decided to start a new thread devoted to the following questions.

Q: Can lottery odds be reduced by buying more tickets?

A: Yes

Q: How are the new, reduced odds calculated?

A: Say that a lottery has X number of unique outcomes. Then the odds of hitting the jackpot is 1 in X. If one buys N number of tickets where no two tickets have exactly the same set of numbers, then the odds of hitting the jackpot is 1 in X/N.

Good luck,
Jake

P.S. prob987, this is in response to your request in the "Are you serious" thread

Jake, I dropped you a note over at that thread telling you that I have a nice little demonstration to show that restricted wheeling in fact reduces the odds by less than the X/N case, which only holds under special circumstances.  I will specify what those conditions are.  Stop back at that thread if you have a chance.  You've motivated my thinking to come up with a more formal demonstration, and I'd love to discuss it with you, as you're a pretty good thinker.  The key is, as I hinted earlier with the necessity for scoring the first ball and the probability of that dvent and the probability of achieving it in repeated trials.

I'm a little busy right now, but remind me to get to it.  Tell me whether you'd like it posted here or there.

Thanks,

Prob

Member #2673
November 2, 2003
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 Posted: May 3, 2004, 9:39 pm - IP Logged

hypersonig,

To use your example of the pick 3 game. Say you played 500 tickets. My formula says your new odds are 1 in 2 (50% chance). Your formula says you have only odds of 1 in 500 becuase there are 500 numbers not covered.

I am willing to bet my paycheck that I am right.

Jake

Chief Bottle Washer
New Jersey
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May 31, 2000
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 Posted: May 3, 2004, 11:47 pm - IP Logged

dragon,

To add Jake's argument to the mix, he is saying that if you buy the 123 and the 124, the odds of winning are essentially cut in half, going from 1 in 1000 to 1 in 500.  Basic mathematics would say that you can reduce a fraction from 2/1000 to 1/500, but the question here is does that same reduction principle apply to the odds?

Check the State Lottery Report Card

Sign the Petition for True Lottery Drawings
Help eliminate computerized drawings!

San Diego
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May 1, 2004
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 Posted: May 3, 2004, 11:55 pm - IP Logged

Todd, I see your point, but I think you can, in this case, deduce this intuitively for the pick-3 game. Buy 1 ticket with one permutation: 1:1000. Buy 500 tickets: 1:2. Buy 1000 different tickets 1:1.

Incidentally, I said this before, I am really impressed with your coding (or the end-results of it that I can see). You use Javascript. What's the platform, what's the database?

Chief Bottle Washer
New Jersey
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May 31, 2000
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 Posted: May 4, 2004, 12:07 am - IP Logged

JavaScript is only a very small fraction, and only deals with small functions on the Web pages themselves - not with any of the code which runs the site.

I use 100% Microsoft technologies, including Windows Server 2003, MS SQL 2000, ASP (on IIS 6.0), and a small amount of ASP.NET.  Also several VB.NET and VB 6.0 programs which run some of the automated stuff.

I'll dventually publish full documentation of the site technologies.  (Too busy working on new features at the moment!)

Check the State Lottery Report Card

Sign the Petition for True Lottery Drawings
Help eliminate computerized drawings!

San Diego
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May 1, 2004
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 Posted: May 4, 2004, 12:13 am - IP Logged

That's what it looked like. My company specializes in MS technologies also, SQL 2000 or 2003, we did a lot of work in ASP.NET, but are now completely converted to .NET and C#.

New Member
iowa, illinois
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April 27, 2004
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 Posted: May 4, 2004, 2:31 am - IP Logged

Todd, in your example of buying two tickets in a game with odds of 1:1000, you are correct in that if you bought two tickets, it will equal 2:1000, but it does not reduce the odds in half, obviously.  Odds are expressed in ratios.  The definition of a ratio is a fraction expressed in its lowest terms.  That is the rule.  So no matter how many tickets you buy, the 1000 portion of the ratio stays the same, does not change.  So in a game where the odds ratio is X:1000,  X is the number of tickets purchased and 1000 being the denominator expressed in its lowest terms.  So I'm in agreement with hypersonig and Jake needs to win the lottery to make up for his lost paycheck (smile).

New Member
iowa, illinois
United States
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 Posted: May 4, 2004, 3:09 am - IP Logged

For those of you that want to reduce the odds in the lottery by purchasing more tickets.....look at it this way.....you can subtract from the number of possible outcomes by the number of tickets you purchase, but you certainly cannot divide.  Example.....1:1000, buy 10 tickets with different numbers=1:990.  What this means is that you have a 1:990 chance of losing, not winning.  Chances of winning would be 10:1000.  Hope this helps.

Dump Water Florida
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 Posted: May 4, 2004, 3:12 am - IP Logged

Of course the ratio remains the same, you only have two combinations out of a thousand, but each now can be said to face off against 500 lines each.  If you keep going doubling the amount of tickets you go from 1 in 1000 to 1000 in 1, full circle.  The Hyper Geometric Distribution doesn't go away just because some don't care for how it is described in this case. ;-)

Take 6/49 lotto, there is no question playing 24 numbers reduces the odds of having all six winning numbers among your 24 to roughly 1 in 100 and if you were playing full wheels probability would suggest within 150 draws you would be virtually assured of having at least one jackpot maybe more.  It all comes round.

The only true odds reduction that isn't counter balanced comes from buying more unique combinations.  BobP

Columbia,Pa
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January 27, 2004
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 Posted: May 4, 2004, 3:45 am - IP Logged

To wheel 24 numbers in a 6/49 game reduces the odds to one in 134,596

Chas

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iowa, illinois
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 Posted: May 4, 2004, 3:54 am - IP Logged

All that says BobP is that your chances of having the 6 winning numbers amongst your 24 is 1 in 100.  But your chances of having the six numbers in one grid are not 1 in 100.  I don't know what the odds ratio is for 6/49 lotto, but say it's 1: 20,000,000.  Your 24 numbers only add up to 4:20,000,000 chance of winning.  Say your numbers are (1-2-3-4-5-6-), (7-8-9-10-11-12), (13-14-15-16-17-18), (19-20-21-22-23-24).  Are you saying that if you played those numbers you would win the jackpot at least once within 150 draws?  This is from the Minnesota Lottery's website FAQ section:

Q: Are there any statistics regarding which picking techniques are winners? Do quick-picks win more often than hand-picked numbers?

A: The percentage of quick-picks varies a great deal from game to game and also with the size of the jackpot. In Powerball, the quick-pick percentage normally runs about 70 percent, and about 70 percent of the winners are quick-pick players.

The bottom line is that how you pick your numbers has absolutely nothing to do with being a winner. The numbers 1-2-3-4-5*6 have exactly the same chance of winning as any other set of six numbers. The only thing is, if you pick your own numbers, and you base them on something logical, artificial or limiting (like dates), you are more likely to have to split the pari-mutuel jackpot prize because others may be using the same numbers. The most commonly played number combination is 7-14-21-28-35*42. If you won with that number, you would share the jackpot with many, many other players.

Columbia,Pa
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 Posted: May 4, 2004, 4:10 am - IP Logged

Unless your talking pick 3 or pick 4 games where payouts are set, any calculations at buying more tickets to reduce the odds is useless. As wiseone2 stated your dealing with a parimutuel game. Number of winning tickets sold divided by the jackpot.

So why bother with ticket reduction odds?

Chas

Pennsylvania
United States
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April 6, 2003
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 Posted: May 4, 2004, 4:19 am - IP Logged
Quote: Originally posted by Jake649 on May 03, 2004

hypersonig,

To use your example of the pick 3 game. Say you played 500 tickets. My formula says your new odds are 1 in 2 (50% chance). Your formula says you have only odds of 1 in 500 becuase there are 500 numbers not covered.

I am willing to bet my paycheck that I am right.

Jake

no, I would say it would be 500:1000

the 1000 is constant...

Playing more than one ticket per game is betting against yourself.

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