odds are reduced

Reply #45

anonymous77 — Sat, 25 Feb 2006 06:02:06 GMT

Thoth wrote:

Of course the whole debate could possibly be just a matter of the way in which people choose to define what odds are.

If you want to define your odds as the favorable outcomes within the total amount of outcomes possible, then two different 6 of 49 combinations would indeed give you the odds of 2:13,983,816 - That should be observed as 2 in 13,983,816 and could easily be written as the fraction 2/13,983,816 or 1/6,991,908 when reduced.

If you want to define odds as a RATIO... [ More ]

Reply #44

dvdiva — Sat, 25 Feb 2006 04:21:15 GMT

Odds of winning Megamillions. 1:175 million.This is really two drawings. Five white balls 1-56 or for the odds COMB(56,5) which is 1:3.82 million.

a second drawing occurs for a megaball COMB(46,1). Since it's only one number anyways it's 1:46

If you take one combination and cover all the possible mega numbers you now have one combination in a 3.82 million to one game. Same as if you had one ticket in a 5/56 game. Odds are you will lose anyways since some other combination of white balls wi... [ More ]

Reply #43

hypersoniq — Sat, 25 Feb 2006 00:19:34 GMT

the thing is you are buying a ticket for one draw, not several million draws...

favorable outcomes = # of tickets you bought, these are YOUR favorable outcomes.

possible outcomes = # of possible outcomes (discounting permutations in a non-replacement game. for the purpose of the discussion, 1-2-3-4-5-6 pays the same as 4-6-2-3-1-5 or 6-5-4-3-2-1)

they can't be manipulated by reduction, the only variable you have control over is the number of favorable outcomes. possible outcomes and fav... [ More ]

Reply #42

GuessWhat! — Fri, 24 Feb 2006 22:18:30 GMT

Recommended reading:

https://www.lotterypost.com/thread/129273.htm

Reply #41

Coin Toss — Fri, 24 Feb 2006 19:40:51 GMT

It does happen when you buy all the tickets that are possible.

You can not use this as an example becuase buying all the tickets possible assumes a solo hit. If someone else has the same combination you are now stuck millions of dollars. So you have the means to buy $175, 711,536 tickets figuring you've got the jackp[ot locked up and then if one other person has the same ticket you've just cut your winnings in half. Let's say the jackpot was $200,000,000. That'a what you think you're g... [ More ]

Reply #40

fxsterling — Fri, 24 Feb 2006 19:29:40 GMT

how about 1 winner for a drawing that sold 50 million tickets Thats 1 in 50 million or less

Reply #39

Bradly_60 — Fri, 24 Feb 2006 18:34:34 GMT

I have enough knowledge of how probability actually works to understand that playing two 6/49 combos is parallel to playing only one combo for a game with 6,991,908 total outcomes. In the same way, if you break down all the Pick 4 numbers into groups of ten you will have 1000 groups that statistically perform and follow the exact probabilities for a 1 in 1000 game like pick 3.

That is exactly what I am trying to say.

As for another point

It has to be ignored because it is totally

Reply #38

Thoth — Fri, 24 Feb 2006 18:19:51 GMT

I think odds often get confused with probability or expected statistical frequency. Normally, when I list an odd in a reduced fractional form I am thinking more along the lines of an immediate probability (expressed as a fraction) for that particular trial. I think many other people inadvertently do this also. When playing 2 different Pick 3 combinations in straight form, the ODDS are 2 in 1000 or 2 to 998, but this often gets written as the fraction 2/1000 or 1/500 when reduced. From the po... [ More ]

Reply #37

Coin Toss — Fri, 24 Feb 2006 17:32:05 GMT

Brad

If you buy one ticket the odds are that your numbers will come up 1 in every 175 million drawings

If you buy two tickets the odds are that your numbers will come up 1 in every 87.5 million drawings.

If you buy 46 tickets the odds are that your numbers will come up 1 in every 3.9 million drawings.

That is odds...that is math...how can that be ignored?

Brad

It has to be ignored because it is totally inaacurate. Using that theory, you would reach a number of tickets th... [ More ]

Reply #36

LosingJeff — Fri, 24 Feb 2006 14:56:24 GMT

This is ridiculous! With odds such as the Powerball buying 1000 tickets are no greater than 1 in hitting the big 1. Come to Indiana if you want astronomical odds of trying to win. With total combos played in eight years (somewhere between 1-2 million combo numbers) I am yet to hit a strt combo. Those odds just dont compute, unless, of course, the numbers are being filtered. Any comments

Reply #35

Bradly_60 — Fri, 24 Feb 2006 14:45:58 GMT

I think we should just agree to disagree.....haha

I just can't understand the concept that if you buy two tickets you don't cut your odds down to 1 in half the amount of one ticket. It doesn't make sense to me.

Odds are just that odds.....they don't say what is going to happen they say what should happen.

If you buy one ticket the odds are that your numbers will come up 1 in every 175 million drawings

If you buy two tickets the odds are that your numbers will come up 1 in every 87.5... [ More ]

Reply #34

LOTTOMIKE — Fri, 24 Feb 2006 14:37:18 GMT

he was reportedly last seen at a burger king in ames,iowa.....

Reply #33

truecritic — Fri, 24 Feb 2006 14:33:37 GMT

>What are the odds of the odds actually being correct?

About the Beatles? Or about Elvis?

Elvis has a slight edge...he was reported alive at various times and places

Reply #32

sirbrad — Fri, 24 Feb 2006 09:00:23 GMT

What are the odds of the odds actually being correct

Reply #31

LOTTOMIKE — Fri, 24 Feb 2006 08:14:12 GMT

oops....i forgot to add elvis.

Reply #30

truecritic — Fri, 24 Feb 2006 08:09:01 GMT

LOTTOMIKE

I wouldn't say that the odds are quite that bad. With the Beatles, you'd have to bring them back to life first and that probability is 0 (zero).

Reply #29

LotteryPlayer — Fri, 24 Feb 2006 08:06:36 GMT

Quote: Originally posted by Coin Toss on February 24, 2006 LotteryPlayer

Here's sone links:

LINKS TO HELP YOU LEARN ABOUT PROBABILITY:

http://www.calottery.com/

The official California Lottery website.

http://mathforum.org/dr.math/faq/faq.comb.perm.html

A helpful web page on calculating permutations and combinations

http://mathforum.org/library/drmath/view/56117.html

A helpful discussion of odds vs. probability

http://mathforum... [ More ]

Reply #28

LOTTOMIKE — Fri, 24 Feb 2006 07:58:05 GMT

,with those odds the chance of me winning is about as good as the beatles getting back together....

Reply #27

CalifDude — Fri, 24 Feb 2006 06:57:56 GMT

It worked for Sir Issac Newton! (Tends that is.

Reply #26

Coin Toss — Fri, 24 Feb 2006 05:51:14 GMT

LotteryPlayer

Here's sone links:

LINKS TO HELP YOU LEARN ABOUT PROBABILITY:

http://www.calottery.com/

The official California Lottery website.

http://mathforum.org/dr.math/faq/faq.comb.perm.html

A helpful web page on calculating permutations and combinations

http://mathforum.org/library/drmath/view/56117.html

A helpful discussion of odds vs. probability

http://mathforum.org/library/drmath/view/56122.html

An explanation... [ More ]

Reply #25

sirbrad — Fri, 24 Feb 2006 05:43:50 GMT

Play what is HOT, not what is not!

What is HOT tends to stay HOT. What is COLD tends to stay COLD

That is what I thought also, then sure enough 3 cold numbers came up and 3 average ones. I bet pretty heavy too. Tends is a very questionable term.

Reply #24

CalifDude — Fri, 24 Feb 2006 05:38:42 GMT

Good point.

Reply #23

Coin Toss — Fri, 24 Feb 2006 05:25:44 GMT

LotteryPlayer 20 years of casino work. (Sorry I didn;t set up a link to it, lol). Like I've tried to tell you, odds are odds, not necessarily ratios.

People hung up on ratios and even with degres in probabilities can't beat a crap table, you think they're going to beat a lottery?

One more time, lottery odds are based on the number of possible combinations

Seriously think about one ticket eliminating 87,000,000 combinations.

Impossible.

By the way, searching the math.com sit... [ More ]

Reply #22

CalifDude — Fri, 24 Feb 2006 05:21:48 GMT

I find it so amazing that so many people think that buying TWO tickets has reduced your odds in half!

As so many other people have correctly pointed out ... if the odds are say 1:40,000,000 that means that there are a total of 40 million combinations and by buying one ticket you have ONE of those 40 million possible combinations covered. There are still 39,999,999 combinations left so your odds now are 1:39,999,999.

Reply #21

LotteryPlayer — Fri, 24 Feb 2006 05:03:42 GMT

If you buy 2 MegaMillions tickets, your chances of winning, by using a ratio, are 2:175,711,536 which is equal to the ratio of 1:87,855,768.

Your argument seems to suggest that ratios are valid in some instances, but invalid in others. You may be right, but I'm unaware of such a scenario. Please provide a link that backs your claims.

Here is a link that backs my claims:

For example, the ratio 2:4 is equal to the ratio 1:2.

Math.com: Ratios

Reply #20

dvdiva — Fri, 24 Feb 2006 04:52:56 GMT

I buy tickets covering all Mega numbers so it's one ticket in a 5/56 game or 1:3.8 million to one. There are two variables at work, the white numbers and the Mega numbers. In a 6/49 game not much reduced the odds.

You can wager that numbers under 20 won't come out and purchace your numbers accordingly, say $30. Odds are that assumption is wrong and you loose on every number. If you are correct then you have 30 tickets in a 1:118,000 game. This ultimately just changes your perception of the ga... [ More ]

Reply #19

Coin Toss — Fri, 24 Feb 2006 04:36:53 GMT

If you buy 1 ticket for MegaMillions, the chances of you of winning, by using a ratio, are 1:175,711,536. If you buy 2 tickets, your chances of winning are 1:87,855,768.

NO! NO! NO! NO! NO!

That is what we are trying to tell you.

1 ticket, 1 in 175,711,536.

2 tickets, 1 in 175,711,535.

Think about it, you are saying that second ticket covered 87,855,767 combinations with one line of numbers!

(175,711,535

- 87,855,768

------------------------

Lotteries are ba... [ More ]

Reply #18

LotteryPlayer — Fri, 24 Feb 2006 03:12:12 GMT

If you buy 1 ticket for MegaMillions, the chances of you of winning, by using a ratio, are 1:175,711,536. If you buy 2 tickets, your chances of winning are 1:87,855,768.

If you want to determine your chances of winning in the form of percentages, then the chances would be 0.0000005% with 1 ticket and 0.000001% with 2 tickets (answers are derived from dividing the number of tickets by the number of combinations).

Reply #17

CalifDude — Fri, 24 Feb 2006 02:42:42 GMT

Reply #16

truecritic — Fri, 24 Feb 2006 02:42:16 GMT

If you don't play MM, you have 0 (zero) chance of winning.

If you purchase 175, 711, 536 combinations, you have an absolute certainty of winning. Represented by 1 (one).

Any tickets you purchase inbetween 0 and 1 represent your chances of winning. So if you bought 50% of the tickets would that not change the odds to 1:1?

It wouldn't affect the payoff itself because the prize could be over or under $175,711,536.00

Reply #15

Coin Toss — Fri, 24 Feb 2006 02:20:30 GMT

All right, I'm looking at the back of a Mega Millions playslip.

Odds against hitting 5 + 1 = 1 in 175, 711, 536

So people here are saying that by playing one ticket, one set of numbers those odds are reduced by half. Absolutely no way. Play one ticket and that makes 175,711, 535 combinations to go.

As far as having the bonus number pinned all that would prove is $46 dollars was spent to have every Mega Millions number. All that would guarantee is a $2 win.

Reply #14

Thoth — Fri, 24 Feb 2006 02:09:59 GMT

As I stated in my second posting - perhaps the real debate is how different people perceive the odds. I know that technically the odds are 1 in 1000 or 1:999 for a single straight number to win or 2 in 1000 or 2:998 for two different straight numbers. This does not mean that the fractional reduction is not valid - especially when thinking long term statistics and long term probability.

Many states list their boxed odds for Pick 3 as 1 in 167 for a no-match number instead of the 6 in 1000 or... [ More ]

Reply #13

LotteryPlayer — Fri, 24 Feb 2006 01:50:21 GMT

I don't understand your argument, hypersoniq. I'm not saying your argument is wrong -- I just don't understand it.

As Bradley_60 said, you determine your chances of winning, in the form of a ratio, by dividing the odds by the number of tickets you've bought.

For example, if you buy 2 tickets for a game with odds of 1:10, the odds of you winning are 1:5.

Here's a good way to visualize this:

Take 10 separate squares of paper and write a number on each them, 1 through 10. Now make five... [ More ]

Reply #12

hypersoniq — Fri, 24 Feb 2006 01:09:20 GMT

in PURE numbers (such as 5th grade math) fractions are reduceable

have you taken 5th grade science? the odds are not pure numbers, they now represent a quantity of something... I'll keep it in the pick 3 realm for this...

odds as posted 1:1000 ____ there are 1,000 possible combos... from 000 to 999, they will pick only 1

you buy 000(1:1000), they can pick 000-999, but you leave 001,999 uncovered...

you buy 000 and 999(2:1000), they can STILL pick 000-999, but you are left uncovered f... [ More ]

Reply #11

Badger — Fri, 24 Feb 2006 00:41:53 GMT

Your numbers will come up once in every 7 million draws. Of course you'll be dead by then.

Reply #10

Badger — Fri, 24 Feb 2006 00:39:20 GMT

Quote: Originally posted by Thoth on February 23, 2006 However:

Buying one Pick 3 ticket gives you odds of 1 in 1000 or 1/1000, buying two Pick 3 tickets equals odds of 2 in 1000 or 1 in 500 as a reduced fraction...and yes, 500 combos arent simply going to disappear. But, 100 Pick 3 tickets equals odds of 100 in 1000 or 100/1000 ....or 1/10 as a reduced fraction. Either way you look at it, 100 in 1000 or 1 in 10 is still exactly 10% and is also the average amount of all games that you would... [ More ]

Reply #9

Bradly_60 — Thu, 23 Feb 2006 23:38:24 GMT

Now 47 tickets would get you down to odds of 1 in 3.9 million in Mega Millions.

This is how lotteries are. That is why the lotteries odds are so high so people aren't winning it left and right. First we started with 1 in 70 million odds and we outgrew that. Then it went up to 1 in 120 million and now its at 1 in 175 million.

Go and find someone that has any math background and ask them what they think. That is the beauty of math...there is usually only one answer. 2/8 is the same as... [ More ]

Reply #8

Coin Toss — Thu, 23 Feb 2006 23:29:51 GMT

If you buy one ticket you should hit the jackpot every (for a mega million example) 175,000,000 draws. Now if you buy two tickets you are expected to hit the jackpot once every 87.5 million draws. This is the correct math. It doesn't matter how many losing combinations there are it matters what your odds are of hitting the jackpot.

Brad

Brad, you want to buy a bridge? some prime Florida land?

Your odds of hitting the jackpot are reduced by 1 for every different set of numbers... [ More ]

Reply #7

Bradly_60 — Thu, 23 Feb 2006 19:58:25 GMT

Don't be so pessimitic.

I think we are confusing two different topics here. If you buy 2 tickets your odds of winning the jackpot do cut down in half.

The way you are looking at it, which I still feel is a unmathed asnwer, is that when you buy two sets of numbers there are still pretty much the same odds as having one set.

You are not comparing your numbers to the amount of tickets that are left when you buy a ticket....you are comparing it to what are your odds of hitting a jackpo... [ More ]

Reply #6

Coin Toss — Thu, 23 Feb 2006 19:42:44 GMT

Buying one Pick 3 ticket gives you odds of 1 in 1000 or 1/1000, buying two Pick 3 tickets equals odds of 2 in 1000 or 1 in 500 as a reduced fraction...

Not even close. Just as hypersoniq pointed out, you are reducing the number of combinations left by one, not by half.

The odds against any Pick 3 are 1:999. When you buy one you have 1 potential winner and 998 losers. When you buy another combination (play a different ticket) you have two potential winners and 997 losers, not:

1000... [ More ]

Reply #5

Thoth — Thu, 23 Feb 2006 14:50:11 GMT

Of course the whole debate could possibly be just a matter of the way in which people choose to define what odds are.

If you want to define your odds as the favorable outcomes within the total amount of outcomes possible, then two different 6 of 49 combinations would indeed give you the odds of 2:13,983,816 - That should be observed as 2 in 13,983,816 and could easily be written as the fraction 2/13,983,816 or 1/6,991,908 when reduced.

If you want to define odds as a RATIO of favorable out... [ More ]

Reply #4

Bradly_60 — Thu, 23 Feb 2006 14:49:19 GMT

YES YOU DO REDUCE THE FRACTIONS!!!!!!! Seriously....did anyone take grade school math?

2/8 is the same as 1/4 no matter how many zeros you put on the end of it. If you have two of the combination of numbers from a pool of 14,000,000 you numbers will come up 1 in every 7,000,000 draws.

Why is it so hard for people to understand that. I think people need to go back to the 5th grade and take math over again.

Brad

Reply #3

Thoth — Thu, 23 Feb 2006 13:56:40 GMT

However:

Buying one Pick 3 ticket gives you odds of 1 in 1000 or 1/1000, buying two Pick 3 tickets equals odds of 2 in 1000 or 1 in 500 as a reduced fraction...and yes, 500 combos arent simply going to disappear. But, 100 Pick 3 tickets equals odds of 100 in 1000 or 100/1000 ....or 1/10 as a reduced fraction. Either way you look at it, 100 in 1000 or 1 in 10 is still exactly 10% and is also the average amount of all games that you would win over a long period of time if you continually play... [ More ]

Reply #2

Badger — Wed, 22 Feb 2006 11:05:06 GMT

Yes, exactly. I don't know why, but seems a lot of people think they better their odds considerably by buying a few different combinations over if they would have only purchased one combination. It's a real drop in the bucket and does almost nothing to improve your odds.

Reply #1

hypersoniq — Wed, 22 Feb 2006 10:56:21 GMT

can't reduce the fraction

one ticket = 1:14,000,000, there are 14,000,000 possible combos and you have one covered leaving 13,999,999 uncovered

two tickets = 2:14,000,000, there are 14,000,000 possible combos and you have TWO covered leaving 13,999,998 uncovered... 2 ticklets does NOT make 7,000,000 possible combinations just disappear.

three tickets = 3:14,000,000, blah-blah-blah , leaving 13,999,997 uncovered

odds are reduced

stevieray — Wed, 22 Feb 2006 10:05:01 GMT

Of course odds are reduced by half ( or you have twice the chance of winning) if you buy two tickets. Try not to think of 49 balls in a tub and 6 are selected, instead think of numbers 1 to 13,983,816 in a large hat.

The only reasons balls are used is to a) allow you to select these numbers on one sheet of paper, b) allow other prizes for 3, 4 or 5 numbers to be given, c) give some excitement when the balls are drawn and allow punters to have some connectiion with the numbers (birthdays, ages... [ More ]