mid-Ohio United States Member #9 March 24, 2001 19900 Posts Offline

Posted: March 15, 2012, 9:59 pm - IP Logged

Quote: Originally posted by Cautious on March 15, 2012

I think everyone knows that if you start looking for numbers that have resulted in wins in the past you are on the wrong track.

Let's again use the MM as an example: Odds of winning with any numbers are 1 in 175,711,536. That means there are approximately 176 million plays you could make. If I check the winners in previous games winning numbers will be only a tiny portion of all those 176 million possibilities - because we haven't played 176 million games yet. Does this mean those previous winning numbers/plays are more likely than the rest of those 175,711,536 plays to win the next time? Absolutely not!

Absolutely not!

You assume something to be a fact that you do not know.

I think of MegaMillions as two lotteries, a 5/56 with 3,819,816 possible combination and a 1/46 with 46 possible combinations. The odds of hitting them both on one ticket are 1: 175,711,536. While the odds of any two combinations that have hit together ever hitting together again are unlikey, the odds of combinations repeating in one of the two games are very likely. The megaballs are always repeating and I suspect one of the five number combinations will repeat before 10% (3,800)of all the possible five combinations are drawn because it has happened in other pick5 games that I have observed.

Ohio has had two pick5 games, a 5/36 and 5/39 and the old 5/36 which had 2800 drawings had 8 combinations repeat and the new 5/39 has had almost as many drawings so far and has had 5 combinations repeat so far and the same is likely to happen in MM.

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

bgonÃ§alves Brasil Member #92564 June 9, 2010 2134 Posts Offline

Posted: March 15, 2012, 10:03 pm - IP Logged

Hello,caution,if youget 10drawsofall lotteriesin the world sincepick4.pick5pick6etc ...view allresultsof the world,and divideinto 4 parts,in 87% oflotterynumberswill be inthreesections,until theeurominhao,from todayuntil10drawsof alllotteries in the world,you can createupto simulatemorelotteries Andmake billions ofpairings,80%of the drawswill be inthreesectors orthreegroups, Are1,2,3,4,5,6,is dead,

cleveland ohio United States Member #65897 October 9, 2008 275 Posts Offline

Posted: March 16, 2012, 12:34 am - IP Logged

Quote: Originally posted by SergeM on March 15, 2012

Coin flip: p(face) = 0.5

I love math.. flip a coin you should have 50-50 chance right? the Farther you go out the closer you will get yet the more you will lose. This is a fact. Do the math.

What do you mean? Explain yourself.

Ok you have to really think about this. IF you are flipping a coin betting on a 50=50 prop then if the first 10 come out 4-6 then you are at 40% on the losing end ignore heads or tails now take the game out t0 100 the final is 42 -58 your closer having now reduced the odds to 42% thus gaining an advantage right? Wrong the other side has turned up 52-38 after the inital 10 draws. Take it out to 1000 lets say you now get 450-550 you have again increased to being within 3 more % yet you are bleeding red.

Appleton, Wi United States Member #118178 October 24, 2011 199 Posts Offline

Posted: March 16, 2012, 1:17 am - IP Logged

Quote: Originally posted by Cautious on March 15, 2012

I think everyone knows that if you start looking for numbers that have resulted in wins in the past you are on the wrong track.

Let's again use the MM as an example: Odds of winning with any numbers are 1 in 175,711,536. That means there are approximately 176 million plays you could make. If I check the winners in previous games winning numbers will be only a tiny portion of all those 176 million possibilities - because we haven't played 176 million games yet. Does this mean those previous winning numbers/plays are more likely than the rest of those 175,711,536 plays to win the next time? Absolutely not!

Cautious:

Don't want to be rude so please excuse my suspicion, but what do you plan to do with this information once revealed?

Appleton, Wi United States Member #118178 October 24, 2011 199 Posts Offline

Posted: March 17, 2012, 8:04 am - IP Logged

Quote: Originally posted by BlueDuck on March 15, 2012

Dice Game and Lottery Game Comparison

same odds 1:36

or

Why to choose or not to choose 1-2-3-. . . in Lottery

Here is a "sums chart" for a Dice Game using two die, one white and one red. The sums chart list all 36 combinations.

SUMS CHART for dice game

2 3 4 5 6 7 8 9 10 11 12

1-1 1-2 1-3 1-4 1-5 1-6 2-6 3-6 4-6 5-6 6-6

2-1 2-2 2-3 2-4 2-5 3-5 4-5 5-5 6-5

3-1 3-2 3-3 3-4 4-4 5-4 6-4

4-1 4-2 4-3 5-3 6-3

5-1 5-2 6-2

6-1

Any of the 36 combinations has the same chance of being rolled. 1-1, 6-6, or 3-4 can all hit and they all have the same chance.

...but the object of the dice game is to roll a seven or eleven, to win. Many chances (eight) for the right winning combination to pop up.

Why can't lotteries be like that?

It is my hope that I can prove to myself, at another time, that "7"s hit more times in dice for the same reason, that "sums combinations" of 150 hit more times in a 6/49 Lottery.

Here is a "sums chart" for a drop ball Lottery Game using 9 balls. 2/9 odds 1:36. The sums chart list all 36 combinations.

Choose 2 different numbers from 1 through 9

SUMS CHART for 2/9 lottery

3 4 5 6 7 8 9 10 11 12 13

1-2 1-3 1-4 1-5 1-6 1-7 1-8 1-9 2-9 3-9 4-9

2-3 2-4 2-5 2-6 2-7 2-8 3-8 4-8 5-8

3-4 3-5 3-6 3-7 4-7 5-7 6-7

4-5 4-6 5-6

14 15 16 17

5-9 6-9 7-9 8-9

6-8 7-8

If this lottery could be won by "rolling a 10", any one of the 4 combinations would be a winner.

Why can't Lottery Games be more like dice games?

Just a few thoughts,

BlueDuck

Correction.

"It is my hope that I can prove to myself, at another time, that "7s" hit more times in dice for the same reason that "sums combinations" of 150 hit more often in a 6/49 Lottery" should read:

It is my hope that I can prove to myself that "sums combinations" of 150 hit more often in a 6/49 Lottery Game for a different reason than why "7s" hit more often in dice.

Using the "dice game" analogy to support the theory of "Most Probable Sums" doesn't work.

Elgin, IL United States Member #68867 January 1, 2009 1224 Posts Offline

Posted: March 17, 2012, 3:18 pm - IP Logged

Quote: Originally posted by Cautious on March 12, 2012

Much misinformation has been posted about the odds of winning the lotto with specific strategies. A common mistake, seen many times, promotes the false idea that the odds of winning the lotto by selecting a sequence of numbers (2, 3, 4, 5, 6; 7) is lower than selecting non-sequenced numbers (8, 15, 24, 31, 48; 5). This idea is false. Any numbers you pick, sequenced or not, have exactly the same probability of winning. This statistical fact seems to be counter intuitive to many people. I’ve seen posts on lotto websites claiming that you will not see a sequence of numbers win the lotto in your life time. That is true, but it’s also true for any non-sequenced set of numbers – the odds are exactly the same.

Let’s look at this issue another way. Suppose the lottery was based on a set of pictures of things. Each number would be replaced by a picture. To play, you’d pick pictures instead of numbers. With pictures of things there would be no illusion of any sequences being somehow special and unlikely to win, so I guess most people would understand that all picture picks would have exactly the same odds of winning. Now consider that the numbers we use to play lotto are just symbols; they have no numeric value as far as the lottery process is concerned. The lottery would work exactly the same with any symbols or pictures. The lottery selection of a winner wouldn’t operate any differently if we used pictures of trees, people, animals, dots, dashes, or other symbols; and there’s no difference when we use those marks that we see as mathematic symbols. Those marks that we see as numbers could just as well be chicken scratchings as far as the lottery process is concerned – for lottery purposes there’s no sequence to those numbers, they’re all just random pictures.

Take a look at any of my posts in Pick 5 and/or Jackpot Games - They all begin with "Combos & Patterns". We all know what the odds for a given game are, but certain combinations and patterns come in more often then others. Take the Wisconsin Badger 5 game - To date there were 3,316 games played - I will give you a list of only the 5 of 5 that hit more than once. These games make up 97 of the 3,316 games to date.

Economy class Belgium Member #123700 February 27, 2012 4035 Posts Offline

Posted: March 17, 2012, 3:42 pm - IP Logged

Quote: Originally posted by nickbrownsfan on March 16, 2012

Ok you have to really think about this. IF you are flipping a coin betting on a 50=50 prop then if the first 10 come out 4-6 then you are at 40% on the losing end ignore heads or tails now take the game out t0 100 the final is 42 -58 your closer having now reduced the odds to 42% thus gaining an advantage right? Wrong the other side has turned up 52-38 after the inital 10 draws. Take it out to 1000 lets say you now get 450-550 you have again increased to being within 3 more % yet you are bleeding red.

a) 4-6 for 10, that can be 0-10 or 3-7 or anthing. b) You can be winning 10 times of ten times. c) If you bet tail for 100 times, you might eventually win 55 times and lose 45 times. d) On 10 000 times, you probably were right around 5 000 times, if you stuck to the same bet; but you can also have lost every single bet by betting on the same chance.

If 50 of 100 firemen die in the fire in some city, that doesn't mean that none dies in the fire in another city. You have got the truth wrong.

Ventura California United States Member #124382 March 12, 2012 12 Posts Offline

Posted: March 17, 2012, 5:07 pm - IP Logged

Quote: Originally posted by SergeM on March 17, 2012

a) 4-6 for 10, that can be 0-10 or 3-7 or anthing. b) You can be winning 10 times of ten times. c) If you bet tail for 100 times, you might eventually win 55 times and lose 45 times. d) On 10 000 times, you probably were right around 5 000 times, if you stuck to the same bet; but you can also have lost every single bet by betting on the same chance.

If 50 of 100 firemen die in the fire in some city, that doesn't mean that none dies in the fire in another city. You have got the truth wrong.

cleveland ohio United States Member #65897 October 9, 2008 275 Posts Offline

Posted: March 17, 2012, 10:10 pm - IP Logged

Quote: Originally posted by SergeM on March 17, 2012

a) 4-6 for 10, that can be 0-10 or 3-7 or anthing. b) You can be winning 10 times of ten times. c) If you bet tail for 100 times, you might eventually win 55 times and lose 45 times. d) On 10 000 times, you probably were right around 5 000 times, if you stuck to the same bet; but you can also have lost every single bet by betting on the same chance.

If 50 of 100 firemen die in the fire in some city, that doesn't mean that none dies in the fire in another city. You have got the truth wrong.

Perhaps I didnt explain myself to well so I will direct you here.

cleveland ohio United States Member #65897 October 9, 2008 275 Posts Offline

Posted: March 17, 2012, 10:42 pm - IP Logged

So wanted to add how was my post wrong?

I love math.. flip a coin you should have 50-50 chance right? the Farther you go out the closer you will get yet the more you will lose. This is a fact. Do the math.

So if you want to disagree with those 2 links then please do so and provide evidence to do so as well and prove that I am wrong.

You don't understand what is written on your pages.

1) I know the math behind and don't need your links. 2) I was a dealer for a year. 3) I play myself. 4) I program myself.

I say that you have got it wrong. We are talking about YOU! You are just one SAMPLE.

Take a coin, flip it on the table until you get a series of ten times face. After the tenth face continue flipping until the series breaks. Write down every outcome and post it. Use the math of your links and show me the math probability of that.

cleveland ohio United States Member #65897 October 9, 2008 275 Posts Offline

Posted: March 18, 2012, 11:40 am - IP Logged

Quote: Originally posted by SergeM on March 18, 2012

You don't understand what is written on your pages.

1) I know the math behind and don't need your links. 2) I was a dealer for a year. 3) I play myself. 4) I program myself.

I say that you have got it wrong. We are talking about YOU! You are just one SAMPLE.

Take a coin, flip it on the table until you get a series of ten times face. After the tenth face continue flipping until the series breaks. Write down every outcome and post it. Use the math of your links and show me the math probability of that.

The gambler's fallacy can be illustrated by considering the repeated toss of a fair coin. With a fair coin, the outcomes in different tosses are statistically independent and the probability of getting heads on a single toss is exactly ^{1}⁄_{2} (one in two). It follows that the probability of getting two heads in two tosses is ^{1}⁄_{4} (one in four) and the probability of getting three heads in three tosses is ^{1}⁄_{8} (one in eight). In general, if we let A_{i} be the event that tossi of a fair coin comes up heads, then we have,

.

Now suppose that we have just tossed four heads in a row, so that if the next coin toss were also to come up heads, it would complete a run of five successive heads. Since the probability of a run of five successive heads is only ^{1}⁄_{32} (one in thirty-two), a believer in the gambler's fallacy might believe that this next flip is less likely to be heads than to be tails. However, this is not correct, and is a manifestation of the gambler's fallacy; the event of 5 heads in a row and the event of "first 4 heads, then a tails" are equally likely, each having probability ^{1}⁄_{32}. Given the first four rolls turn up heads, the probability that the next toss is a head is in fact,

.

While a run of five heads is only ^{1}⁄_{32} = 0.03125, it is only thatbefore the coin is first tossed.After the first four tosses the results are no longer unknown, so their probabilities are 1. Reasoning that it is more likely that the next toss will be a tail than a head due to the past tosses, that a run of luck in the past somehow influences the odds in the future, is the fallacy.

For example, a fair coin toss is a Bernoulli trial. When a fair coin is flipped once, the theoretical probability that the outcome will be heads is equal to 1/2. Therefore, according to the law of large numbers, the proportion of heads in a "large" number of coin flips "should be" roughly 1/2. In particular, the proportion of heads after n flips will almost surelyconverge to 1/2 as n approaches infinity.

Though the proportion of heads (and tails) approaches 1/2, almost surely the absolute (nominal) difference in the number of heads and tails will become large as the number of flips becomes large. That is, the probability that the absolute difference is a small number, approaches zero as the number of flips becomes large. Also, almost surely the ratio of the absolute difference to the number of flips will approach zero. Intuitively, expected absolute difference grows, but at a slower rate than the number of flips, as the number of flips grows.

Economy class Belgium Member #123700 February 27, 2012 4035 Posts Offline

Posted: March 18, 2012, 12:07 pm - IP Logged

Quote: Originally posted by nickbrownsfan on March 18, 2012

The gambler's fallacy can be illustrated by considering the repeated toss of a fair coin. With a fair coin, the outcomes in different tosses are statistically independent and the probability of getting heads on a single toss is exactly ^{1}⁄_{2} (one in two). It follows that the probability of getting two heads in two tosses is ^{1}⁄_{4} (one in four) and the probability of getting three heads in three tosses is ^{1}⁄_{8} (one in eight). In general, if we let A_{i} be the event that tossi of a fair coin comes up heads, then we have,

.

Now suppose that we have just tossed four heads in a row, so that if the next coin toss were also to come up heads, it would complete a run of five successive heads. Since the probability of a run of five successive heads is only ^{1}⁄_{32} (one in thirty-two), a believer in the gambler's fallacy might believe that this next flip is less likely to be heads than to be tails. However, this is not correct, and is a manifestation of the gambler's fallacy; the event of 5 heads in a row and the event of "first 4 heads, then a tails" are equally likely, each having probability ^{1}⁄_{32}. Given the first four rolls turn up heads, the probability that the next toss is a head is in fact,

.

While a run of five heads is only ^{1}⁄_{32} = 0.03125, it is only thatbefore the coin is first tossed.After the first four tosses the results are no longer unknown, so their probabilities are 1. Reasoning that it is more likely that the next toss will be a tail than a head due to the past tosses, that a run of luck in the past somehow influences the odds in the future, is the fallacy.

For example, a fair coin toss is a Bernoulli trial. When a fair coin is flipped once, the theoretical probability that the outcome will be heads is equal to 1/2. Therefore, according to the law of large numbers, the proportion of heads in a "large" number of coin flips "should be" roughly 1/2. In particular, the proportion of heads after n flips will almost surelyconverge to 1/2 as n approaches infinity.

Though the proportion of heads (and tails) approaches 1/2, almost surely the absolute (nominal) difference in the number of heads and tails will become large as the number of flips becomes large. That is, the probability that the absolute difference is a small number, approaches zero as the number of flips becomes large. Also, almost surely the ratio of the absolute difference to the number of flips will approach zero. Intuitively, expected absolute difference grows, but at a slower rate than the number of flips, as the number of flips grows.

Stop the copy-paste! Did you do the 100 coin flips?