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odds are reduced

Topic closed. 45 replies. Last post 11 years ago by anonymous77.

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Zeta Reticuli Star System
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 Posted: February 23, 2006, 9:20 pm - IP Logged

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.

Michigan
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 Posted: February 23, 2006, 9:42 pm - IP Logged

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

Poway CA (San Diego County)
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 Posted: February 23, 2006, 9:42 pm - IP Logged

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.

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 Posted: February 23, 2006, 10:12 pm - IP Logged

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).

Zeta Reticuli Star System
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 Posted: February 23, 2006, 11:36 pm - IP Logged

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 based on possible combinations of numbers, odds are not necessarily ratios.  It's not engineering, it's probabilities.

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 Posted: February 23, 2006, 11:52 pm - IP Logged

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 game but doesn't really change the odds.

Covering all Powerballs or Mega numbers does really change the odds dramatically since you eliminate one variable of the game. \$46 buys one ticket of a 5/56 game. No way you can factor out that dramatic reduction in odds.

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 Posted: February 24, 2006, 12:03 am - IP Logged

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 based on possible combinations of numbers, odds are not necessarily ratios.  It's not engineering, it's probabilities.

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

Poway CA (San Diego County)
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 Posted: February 24, 2006, 12:21 am - IP Logged

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.

Zeta Reticuli Star System
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 Posted: February 24, 2006, 12:25 am - IP Logged
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 site for lottery gives you this:

Your search - lottery - did not match any documents.

I think you have assumed odds and ratios are the same. They are not.

Poway CA (San Diego County)
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 Posted: February 24, 2006, 12:38 am - IP Logged
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 site for lottery gives you this:

Your search - lottery - did not match any documents.

I think you have assumed odds and ratios are the same. They are not.

Good point.

PA
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 Posted: February 24, 2006, 12:43 am - IP Logged

"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.

Zeta Reticuli Star System
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 Posted: February 24, 2006, 12:51 am - IP Logged

LotteryPlayer

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 of how probabilities for a “powerball” type        lottery can be calculated
http://mathforum.org/dr.math/faq/faq.prob.intro.html
An introduction to probability
http://mathforum.org/library/drmath/view/56495.html
An explanation of the differences between odds and probability
http://mathforum.org/dr.math/faq/faq.prob.world.html
A good overview of why we need to know probability- and how it applies in        the real world

Those were from

http://imet.csus.edu/imet4/tlburke/probability/main.html

Poway CA (San Diego County)
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 Posted: February 24, 2006, 1:57 am - IP Logged

"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.

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

Tennessee
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 Posted: February 24, 2006, 2:58 am - IP Logged

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

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 Posted: February 24, 2006, 3:06 am - IP Logged

LotteryPlayer

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 of how probabilities for a “powerball” type        lottery can be calculated
http://mathforum.org/dr.math/faq/faq.prob.intro.html
An introduction to probability
http://mathforum.org/library/drmath/view/56495.html
An explanation of the differences between odds and probability
http://mathforum.org/dr.math/faq/faq.prob.world.html
A good overview of why we need to know probability- and how it applies in        the real world

Those were from

http://imet.csus.edu/imet4/tlburke/probability/main.html

Thanks for the links, Coin Toss.

Much of the disagreement here seems to come from the differences between odds and probability. I've noticed that many of us have interchangeably used the words "odds" and "probability," which apparently even many of these math sites have done. My examples are in reference to probability, so let me first discuss this topic:

From reading the links that Coin Toss has provided, the examples I've previously posted appear to be correct (with some errors in semantics).

This page answers this exact question for Powerball: Powerball's Probability (Note: they mistakenly use the word "odds" in place of "probability" a few times.)

Here is another example from one of the websites:

"For example, two people have bought raffle tickets. If A wins the grand prize, then B cannot win, and vice versa. In this case, it makes perfect sense to add the probability of A winning to the probability of B winning, and the result is the probability that "either one of them" will win. If there are 1,000 raffle tickets, then A has a 1/1000 chance of winning and B has a 1/1000 chance of winning. Together, their chance of winning is exactly 2/1000." -- MathForum.org

Now let's discuss the odds if you buy 2 tickets (assuming there are 40,000,000 combinations):

The odds of each individual ticket winning is 1:39,999,999. To get the odds for holding both tickets, would you just add the odds together to get 2:39,999,999? If so, the odds are reducible to basically 1:20,000,000, so there really doesn't seem to be any significant difference between the odds and probability here (39,999,999 can't be divided by 2 without a fraction, obviously, but this isn't important because I could've simply used 40,000,001 as the number of combinations and it would've worked out with no fraction). Is this correct? I may be making some incorrect assumptions, with respect to the odds.

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