Atlantic Mine, Michigan United States Member #416 June 23, 2002 1614 Posts Offline

Posted: February 24, 2006, 9:45 am - IP Logged

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 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?

Greenfield United States Member #3587 February 2, 2004 557 Posts Offline

Posted: February 24, 2006, 9:56 am - IP Logged

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?

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10344 Posts Offline

Posted: February 24, 2006, 12:32 pm - IP Logged

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 that would "guarantee" you would hit the next drawing. Never happen.

In searching the net because of this thread, it turns out some of the people promoting this "increase your chances with another ticket" theory are uing it to sell their own systems. Pretty weak. Almost equates to " 2 + 2 =5, would you like to buy my book on math?"

Findlay, Ohio United States Member #4855 May 28, 2004 400 Posts Offline

Posted: February 24, 2006, 1:19 pm - IP Logged

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 point of view of PROBABILITY, the two fractions are both accurate because:

2 ÷ 1000 = 0.002 as does 1 ÷ 500 = 0.002

But which should make more sense to use? Expressing odds the correct way can be just as deceiving as listing the odds as a “fractional probability”. If I play 250 of the total amount of pick 3 combinations (straight) my “odds” are 250 in 1000 or 250:750, which implies that there are 250 ways I can win and whopping 750 ways I can lose. Doesn’t it seem much more practical to take the odds of 250 in 1000 and express it as a reduced fraction of ¼ to indicated that the probability of winning is .25, ¼, 1 in 4 or 25%.

If I said my “odds” were 1 in 4 or ¼ someone may think that there is only four possible outcomes to the game, there is obviously 1000, but there may as well only be a total of four because I’m playing ¼ the total amount of outcomes and the effects of probability will perform the exact same way whether the game is a true 1 in 4 game or a 250 in 1000 game. Statistical analysis proves that to be true.

Playing two 6/49 combos DOES NOT reduce the total amount of combos from 13,983,816 down to 6,991,908, but expressing the odds of 2 in 13,983,816 as a reduced fraction of 1/6,991,908 is accurate if you look at it with the probability of actually winning in mind:

2 ÷ 13,983,816 = 0.00000014302247684

1 ÷ 6,991,908 = 0.00000014302247684

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.

Atlantic Mine, Michigan United States Member #416 June 23, 2002 1614 Posts Offline

Posted: February 24, 2006, 1:34 pm - IP Logged

"

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 inaacurate. Using that theory, you would reach a number of tickets that would "guarantee" you would hit the next drawing. Never happen."

It does happen when you buy all the tickets that are possible. But you got to remember the 2nd ticket you buy greatlly reduces your odds but the 1,000,000th ticket you buy does not have the same effect as buy a 2nd ticket. For instance you got from 1 in 175,000,000 with one ticket in your hand to 1 in 3,900,000 with 46 tickets but if you buy 87,500,000 you get down to 1 in 2 but you don't get down to 1 in 1 until you buy all 175,000,000 tickets.

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10344 Posts Offline

Posted: February 24, 2006, 2:40 pm - IP Logged

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 getting but now it's $100,000,000, pre- cash option, pre-tax. Do the math. This is where you do your fractions, splitting jackpots, not calulating odds.

(And no one who had the means to buy all the combinations possible would ever do so).

Another thing wrong with the assumptions here is that given 175, 711, 536 (for now, til it increases again) comboinations and two draws a week, people are assuming that they will all hit. Some will never hit. Read the threads on this site, there are Pick 4s that have never hit in some states!

And one more time, the second ticket you buy does not "freatly reduce your odds". It reduces them by one combination.

The second ticket takes 175,711,536 ro 175,711,535, that's all it can do.

You may find information to the contrary on the net but it is just hype to sell systems.

Somebody please show us one of these magic "second tickets" that has 87,000,000 combinations of numbers on one line of 5 numbers + a Mega or PB number.

Pennsylvania United States Member #1340 April 6, 2003 2450 Posts Offline

Posted: February 24, 2006, 7:19 pm - IP Logged

Quote: Originally posted by Bradly_60 on February 24, 2006

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 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

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 favorable outcomes are independent of eaach other.

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

United States Member #2338 September 17, 2003 2063 Posts Offline

Posted: February 24, 2006, 11:21 pm - IP Logged

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 will come out. 3.8 million is still an extreme long shot.