The general consensus (and the way the lottery explains your odds) seems to be that if you are playing a lottery with say 1,000,000 to 1 odds, is that you've lowered your odds by 50% by buying a second ticket.

In actuality, haven't you lowered your odds to just 999,999 to 1 odds?

I'd like to hear your opinions.

Thanks!

Oct 1, 2003, 11:11 am

(from http://www.saliu.com/probability.html)

~~~~~Probability starts with *logic*. There is a set of N elements. We can define a sub-set of n *favorable* elements, where n is less than or equal to N. *Probability* is defined as the rapport of the favorable cases over total cases, or calculated as:

Oct 1, 2003, 1:45 pm

in reality, buying 2 tickets does not magically eliminate 1/2 of all possible results...odds are not a fraction meant to be reduced. (but the states would LOVE for you to believe so). Million to one odds with 2 tickets mean you have 2 possibilities out of the 1,000,000 that could come up... there are still 999,998 left uncovered.

Oct 1, 2003, 8:28 pm

*Way I see it, Rick, not only are the odds not lowered in your favor in any capacity but they have increased. I say this because you are playing the stacked odds of any number of players plus whatever the state has setup! Something like an domino effect.*

*Teufellj*

Oct 3, 2003, 11:15 am

Well the Equation for the probabilty would be Number of Combos / number of tickets played. So if you had 1 million combos to chose from and bought one ticket 1,000,000/1 = 1 in 1,000,000 chances. If you bought 2 tickets 1,000,000/2 = 1 in 500,000 chance. 1,000,000/3 = 333,333. I mean it makes sense, if you played 3 tickets each day then it would take you 333,333 days to get that combo.

-Chet

Chet,

That is how Maryland explains it for the 6/49. It has the odds per game board as 1 in 13,983,816, but since there is 2 game boards per $1 wager; the odds get reduced to 1 in 6,991,908. Yet there are still the 13 million plus distinct combo's that I miss by every week. It seems to be a game of semantics.

Oct 3, 2003, 3:05 pm

Quote:Originally posted by Loops on October 01, 2003(from http://www.saliu.com/probability.html)

~~~~~Probability starts with

logic. There is a set of N elements. We can define a sub-set of nfavorableelements, where n is less than or equal to N.Probabilityis defined as the rapport of the favorable cases over total cases, or calculated as:np = -----N~~~~~~~~~~~~~ would equal your tickets purchased, provided they're of different combinations, andnwould equal the total number of combinations for the drawing (eg 120,526,770 for powerball). So yes, going from one ticket to two tickets essentially doubles your chances to win.N

There is a set of N elements. We can define a sub-set of n *favorable* elements, where n is less than or equal to N. **There is only ONE favorable outcome... winning, they only draw once.**

*Probability* is defined as the rapport of the favorable cases over total cases, or calculated as:

Oct 3, 2003, 7:22 pm

~~Probability starts with *logic~~*

This is the most important step, without logic, you can't deduce the probability. The problem with the lottery is the numbers are so large that when you say odds go from 120million:1 to 60million:1 with the purchase of an additional ticket, the mind can't grasp it. So let's compare it to something smaller.

The lottery is nothing more than a huge roulette wheel with incredibly crappy payouts. Each have one possible outcome per spin/drawing, right? So, in roulette, when you place bets on two numbers rather than one, your odds go from 38:1 to 19:1. We can easily prove this by checking payouts over multiple bets.

Let's take 38 spins where every number happened to come up once. For 1 bet: (36[amt won + bet for that win] - 37[number of losing spins] = -1). For 2 bets ((36 - 1)*2 - 2*36 = -2) Same percentage, only player two is betting twice as much, therefore loses twice as much . Seems logical, no?

How can this possibly change merely by increasing the number of choices on the wheel? Answer: It doesn't. N elements = total combinations, n = YOUR favorable elements (in other words, your tickets) P = n/N holds.

Loops, I know where you're coming from and I'm a big fan of Ion Saliu.. But using logic as our guide, I think that if you have only 10 possible single outcomes of an dvent and you pick one of them there are still nine possible outcomes available...your odds would be 9-1 on your first pick, 8-1 on your second pick because you have two picks, but only one can come in so one pick is eliminated from the drawing pool. The probability theory says that our odds should be 5 -1 with our second pick. But with 8 remaining outcomes, how can our odds possibly be 5-1?

Does that make any sense? Thanks for your input in this discussion.

It is really a problem of perspective. When they half the odds as in the Md. 6/49 from 1 in 13,983,816 per game board to 2 game boards per dollar dropping the odds to 1 in 6,991,908. It is really looking at the least amount of dollars to cover all the possible numbers, so 6,991,908 dollars covers all 13,983,816 positions. I don't know if the information that everyone is looking at is different, but where do these jackpot games give out odds just because an extra dollar is played? I don't see it, and that is quite different.

Oct 4, 2003, 4:03 pm

"What are the odds?" We hear that word tossed around, "Odds." Well, odds is defined as the ratio of success to failure. It's usually written as:

probability of success : probability of failure

This means if you place a bet on 1 combination in a 6/49 lottery with 13,983,816 combinations, your odds are:

1 : 13,983,815

If you place a bet on 2 combinations in a 6/49 lottery, your odds are:

2 : 13,983,814

However, the lottery doesn't boast about that, they'd show the odds as:

1 : 6,991,907

Simply dividing both sides by the success number.

If you placed a bet on 100 combinations in a 6/49 lottery, your odds are:

100 : 13,983,716

However, the lottery would boast that your odds are:

1 : 139,837.16

Nothing has really changed, your odds are still:

100 : 13,983,716

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