Welcome Guest
You last visited December 3, 2016, 10:44 am
All times shown are
Eastern Time (GMT-5:00)

# Buying more tickets = reduced odds

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

 Page 2 of 7
San Diego
United States
Member #4520
May 1, 2004
243 Posts
Offline
 Posted: May 4, 2004, 8:43 am - IP Logged
Quote: Originally posted by hypersoniq on May 04, 2004

no, I would say it would be 500:1000

the 1000 is constant...

Are maybe all of us saying the same thing -- just with different examples? The comment above in blue comes full circle to the beginning of this thread, doesn't it?
New Member
iowa, illinois
United States
Member #4477
April 27, 2004
40 Posts
Offline
 Posted: May 4, 2004, 10:24 am - IP Logged

I don't think so...the topic of the discussion is whether buying more tickets "reduces odds".  No, it doesn't reduce odds, it increases chances.  In Jake's formula....1:x/n, it decreases the number of possible outcomes.  So in a 1:1000 game, if you bought 5 tickets, the formula says you have a 1:200 chance of winning as opposed to 5:1000.  The number of possible outcomes is not reduced to 200 by buying 5 tickets.   More tickets only amounts to more chances, without affecting the odds.  The odds don't change because only 1 of however many tickets you buy can win and each 1 ticket is up against 1:1000 odds.  Increasing chances and reducing odds are not the same thing.

Chief Bottle Washer
New Jersey
United States
Member #1
May 31, 2000
23259 Posts
Online
 Posted: May 4, 2004, 12:14 pm - IP Logged
Quote: Originally posted by hypersoniq on May 04, 2004

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...

Saying that it's 500:1000 is THE SAME THING as saying 1 in 2, because there is a 50% probability of hitting in that case.  Likewise, 2 in 1000 is THE SAME THING as 1 in 500.  I'm with Jake on that.

Check the State Lottery Report Card

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

mid-Ohio
United States
Member #9
March 24, 2001
19816 Posts
Offline
 Posted: May 4, 2004, 12:28 pm - IP Logged

So you are saying the thread should read:

Buy more tickets = Increase chances to win with same odds*

*same amount of winning/losing combinations regardless of the number of tickets sold
RJOh

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

Tx
United States
Member #4570
May 4, 2004
5180 Posts
Offline
 Posted: May 4, 2004, 1:13 pm - IP Logged

It's not how many tickets you buy, but which numbers you buy.

How you get your numbers is what will determine your odds and not so much how many tickets or numbers you buy.

New Member
iowa, illinois
United States
Member #4477
April 27, 2004
40 Posts
Offline
 Posted: May 4, 2004, 4:43 pm - IP Logged

the only example that clearly reflects "reducing odds" with the number of tickets purchased formula, is when you purchase half of the possible outcomes of a drawing.  You have a 50% chance of winning.  But in the case of the pick 3 straight combo paying \$500, would investing \$500 to "reduce your odds" in half, and having one of your 500 numbers drawn "winning"?  No, it's just a break even prize.  Odds are normally figured with the inclusion of break even prizes.  In another scenario, assume everyone that plays the Powerball is in a single lottery pool.  Because of the size of our pool, we can afford to purchase 60 million possible outcomes, about half.  Would you feel confident that one of those 60 million tickets would hit the jackpot?  Our odds would be 1:60,000,000 that one of the tickets would hit, because 60,000,000 possible outcomes were left unpurchased.  And if one did, even for a \$170,000,000 jackpot, and you selected the cash option of \$90.4 million, and after taxes,  you  broke even, what would we have gained?  Actually we could end up losing a few million after all tax obigations are paid, and each players portion of the prize money would basically equal their investment.  Even if you used one person in this scenario purchasing \$60,000,000 in powerball tickets, the break even odds are the same.

New Member
iowa, illinois
United States
Member #4477
April 27, 2004
40 Posts
Offline
 Posted: May 4, 2004, 4:58 pm - IP Logged

The only way odds are ever reduced by purchasing multiple tickets is in a drawing where more than one winning combination is drawn.  Think about it!!

mid-Ohio
United States
Member #9
March 24, 2001
19816 Posts
Offline
 Posted: May 4, 2004, 7:27 pm - IP Logged
Quote: Originally posted by wiseone2 on May 04, 2004

Even if you used one person in this scenario purchasing \$60,000,000 in powerball tickets, the break even odds are the same.

Some years back, an Australia group tried to play all the possible combinations in a Virgina lottery which took about 11M tickets.  They took control of several SevenEleven terminals for 3 days and the most they could play were 8M combinations.  They won the lottery but the point is that there wasn't enough time to  play that many tickets between drawings.  Most states don't allow any group or person to control a terminal very long if others are waiting to play.  Now that plan A is impossible, what's plan B?

RJOh

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

New Member
iowa, illinois
United States
Member #4477
April 27, 2004
40 Posts
Offline
 Posted: May 4, 2004, 7:38 pm - IP Logged

So RJOh, how much was the jackpot they won?  One of my points in the scenario is that anyone that purchased that many tickets can still lose.  And if they won....would they make a profit after taxes?

Columbia,Pa
United States
Member #3522
January 27, 2004
372 Posts
Offline
 Posted: May 4, 2004, 8:13 pm - IP Logged
Quote: Originally posted by keystonechas on May 04, 2004

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

Can anyone answer my orginal question? Why bother with ticket reduction odds if your dealing with a parimutuel game?

Chas

New Jersey
United States
Member #3457
January 22, 2004
248 Posts
Offline
 Posted: May 4, 2004, 11:13 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

I keep meaning to get back to you on this and will soon, but not tonight.

I will not have a chance to do so.  I will say this, your statement holds true for a type of lottery that is generally not played, where all of the possible outcomes are each listed on a single ball, one of which is selected after randomization.  Such a lottery might have 100 balls each numbered 1 to 100, thereby ever outcome is unique.

Lotteries as played in most places to do not work like this.  They are actually on the outcome of five or six independent dvents.  For this reason any two tickets may actually give you only a slight advantage over what you had with a single ticket.

More to come...I've been stressed and haven't had much time for fun.

mid-Ohio
United States
Member #9
March 24, 2001
19816 Posts
Offline
 Posted: May 4, 2004, 11:18 pm - IP Logged

Quote: Originally posted by wiseone2 on May 04, 2004

So RJOh, how much was the jackpot they won?  One of my points in the scenario is that anyone that purchased that many tickets can still lose.  And if they won....would they make a profit after taxes?

Several poster have mentioned the story.  As I remembered the value was high enough that they made a profit.  I don't know if Virgina had a cash option at that time, the jackpot was over \$25M.  They didn't just play any lottery as a spokesman for the group stated on a TV show "How They Do That".  They picked out a lottery, made out the tickets, hired some people that would play the tickets at several locations and waited for the jackpot to reach a certain level.  Virgina changed their lottery laws after that and today you hardly ever see a lottery cash value exceed the odds of winning it.  As I said in an earlier post, it is almost impossible to play even half the possible combinations of a lottery any more.

RJOh

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

Dump Water Florida
United States
Member #380
June 5, 2002
3102 Posts
Offline
 Posted: May 5, 2004, 12:45 am - IP Logged

What's fixed is the total number of lottery combinations.

If there are 13.9 million combinations we like to say the
odds are 1 in 13.9 million because one dollar buys one
combination giving the purchaser one chance to win.

Fly to a state where you get two combinations for a dollar
the total number of combinations haven't changed, but the
odds of 1 in 13.9 million are 1 in 6.9 million because now
you have two combinations in hand for one/dollar.

In the same sense, buying two tickets for two dollars also
cuts the odds in half for what each ticket faces, but does
not change the odds for one/dollar.

Of course it's still just two combinations out of 13.9
million, but it is the same division you would use if you
were in charge of any kind of team facing a larger team,
you would divide your numbers into theirs and come up with
a figure for how many each of your team would be facing.

To be precise, the total number of lottery combinations
never change.  The odds of winning do change for you
depending on how you play, but generally can't be computed
until after the draw.

For example, the decision to put ten numbers into play
rather then six reduces the odds from 1 in 13.9 million
to one in 66,589.60 of having all six among the ten as
opposed to all six among six.

To fully cover 10 numbers takes 210 combinations and 210
times 66,589.60 gives us 13.9 million so in the background
the odds didn't change overall, but they certainly did for
your chance of having the numbers among yours.  Actually
even those odds didn't change, they were the odds you took
or chose to accept by putting 10 numbers into play.

So the odds don't change, but there are odds for everything
possible within the lottery and you can choose to play at
one of those odds positions and by doing so be playing at
better or worse odds of winning then the person ahead and
behind you in line.

BobP

Australia
Member #3084
December 22, 2003
328 Posts
Offline
 Posted: May 5, 2004, 3:03 am - IP Logged

Consider a 6/49 Lotto Game:

Possible Combinations of 6 Integers: 13,983,816 (only)

Tickets purchased  Probability of having Winning Ticket
0                                            0 ie impossible
1                                            1/13,983,816
2                                            2/13,983,616
13,983,816                             1 ie certain (also qualify as a fool because of
House Margin.)
13,983,816                            1 duplicates don't count
27,967,632                            1 (Probability of having 2 winning tickets is 1;
also qualify as the biggest fool of the century
as will have to share First Prize with
at least oneself.)

Colin

Australia
Member #3084
December 22, 2003
328 Posts
Offline
 Posted: May 5, 2004, 3:07 am - IP Logged

Clarification:

13,983, 816 different tickets etc

Colin

 Page 2 of 7