mid-Ohio United States Member #9 March 24, 2001 19898 Posts Offline

Posted: January 10, 2013, 11:39 am - IP Logged

Quote: Originally posted by Ronnie316 on January 10, 2013

If I give up my chances of winning on some draws to get BETTER ODDS on others, is it still BETTER ODDS?

If you've been tracking your results then the sums of your wins compared with the amounts you spent should tell you even if you haven't won a jackpot yet.

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

Kentucky United States Member #32652 February 14, 2006 7342 Posts Offline

Posted: January 10, 2013, 4:20 pm - IP Logged

Quote: Originally posted by Boney526 on January 10, 2013

If you bet on 5 numbers, then you can win on any of those 5.

The odds of you winning are 5/38 or 33 to lose and 5 to win.

For example, betting individually on the numbers 1,2,3,4,5 and 6 is 32 to 6 and pays out 5 to 1 (which is the same as 30 to 6.) This gives an edge of 5.26%. The math is simple, and sound. There is no difference between placing 1 dollar on each of these individually or betting on the 6 by placing your bet on the outer edge of the field of numbers. That should be common sense.

I didn't use 5 numbers for the example because the 5 number bet on a roulette wheel gets shortpaid due to rounding. The principle is the same. Sure, only one of your bets can win, but you've placed five. So the correct way to calculate the odds are 33 lose and 5 win for you. So 33:5 against you winning, which will be paid at 31:5 if you win.

"Sure, only one of your bets can win, but you've placed five."

First you said it's impossible to change the odds but after I demonstrated how making four additional bets lowers the number of chances against and does change the odds against winning, you're now saying if I make two bets, the odds are lowered to 18 to 1 as if 18 of the possible losing outcomes magically disappeared. It's an accumulative effect because the odds against any of the five numbers winning is still 37 to 1. The reduction can only be on the number of losing outcomes because only one number can win. Five players individually won't get better odds making the same five $1 bets.

By making that type of bet, four of those bets will always lose. The odds against the first bet winning are 37 to 1 and the odds against the second bet becomes 36 to 1 because even though the first bet eliminated one of the losing outcomes, there is still only one way the second bet can win. Because the second bet eliminated one of the losing outcomes against the first bet, those odds against winning are reduced to 36 to 1 too. By making five bets, the odds against any one of the five bets winning is 33 to 1.

"So 33:5 against you winning, which will be paid at 31:5 if you win."

The house determines the payoff odds by the accumulative effect whether it's by five players each making five different $1 bets each having odds of 37 to 1 or one player making five different $1 bets. The probability of house winning that bet is the number of chances to win (33) compared to the total number of losing chances (5) and a 84.8% chance of winning. The house pays off at 6.2 to 1 when the payoff should be 6.6 to 1 and the difference becomes the house edge.

If one player bets $5 on the first five numbers (0, 00, 1, 2, and 3) and wins, they are paid off at 6 to 1 or a net win of $30. If another player puts $1 on each of those 5 numbers and wins, they are paid off at 35 to 1 or a net win of $31. The first five numbers is the one bet on a roulette layout the house will gladly accept all day long because it raises the house edge from 5.26% to 7.89%.

I'm not disputing your numbers, just pointing out while the accumulative bets changes the overall effect, it doesn't make any of the 33 losing chances magically disappear. I'm sure Jimmy can't wait to jump in with his presentation of the Gambler's Fallacy.

United States Member #124493 March 14, 2012 7023 Posts Offline

Posted: January 10, 2013, 5:08 pm - IP Logged

Quote: Originally posted by Stack47 on January 10, 2013

"Sure, only one of your bets can win, but you've placed five."

First you said it's impossible to change the odds but after I demonstrated how making four additional bets lowers the number of chances against and does change the odds against winning, you're now saying if I make two bets, the odds are lowered to 18 to 1 as if 18 of the possible losing outcomes magically disappeared. It's an accumulative effect because the odds against any of the five numbers winning is still 37 to 1. The reduction can only be on the number of losing outcomes because only one number can win. Five players individually won't get better odds making the same five $1 bets.

By making that type of bet, four of those bets will always lose. The odds against the first bet winning are 37 to 1 and the odds against the second bet becomes 36 to 1 because even though the first bet eliminated one of the losing outcomes, there is still only one way the second bet can win. Because the second bet eliminated one of the losing outcomes against the first bet, those odds against winning are reduced to 36 to 1 too. By making five bets, the odds against any one of the five bets winning is 33 to 1.

"So 33:5 against you winning, which will be paid at 31:5 if you win."

The house determines the payoff odds by the accumulative effect whether it's by five players each making five different $1 bets each having odds of 37 to 1 or one player making five different $1 bets. The probability of house winning that bet is the number of chances to win (33) compared to the total number of losing chances (5) and a 84.8% chance of winning. The house pays off at 6.2 to 1 when the payoff should be 6.6 to 1 and the difference becomes the house edge.

If one player bets $5 on the first five numbers (0, 00, 1, 2, and 3) and wins, they are paid off at 6 to 1 or a net win of $30. If another player puts $1 on each of those 5 numbers and wins, they are paid off at 35 to 1 or a net win of $31. The first five numbers is the one bet on a roulette layout the house will gladly accept all day long because it raises the house edge from 5.26% to 7.89%.

I'm not disputing your numbers, just pointing out while the accumulative bets changes the overall effect, it doesn't make any of the 33 losing chances magically disappear. I'm sure Jimmy can't wait to jump in with his presentation of the Gambler's Fallacy.

If you can win a jackpot following something fallacious, then so be it.

There is no measuring the intuitive mind. You either know fully, partially, or you dont know at all.

Lotto Flag formation=gamblers fallacy.

But statistically speaking 100 independent quick picks, have less of a chance of trapping the 5 correct numbers, then do

two 17 number wheels, that wheels a total of 22 numbers.

New Jersey United States Member #99032 October 18, 2010 1439 Posts Offline

Posted: January 10, 2013, 8:22 pm - IP Logged

Quote: Originally posted by Stack47 on January 10, 2013

"Sure, only one of your bets can win, but you've placed five."

First you said it's impossible to change the odds but after I demonstrated how making four additional bets lowers the number of chances against and does change the odds against winning, you're now saying if I make two bets, the odds are lowered to 18 to 1 as if 18 of the possible losing outcomes magically disappeared. It's an accumulative effect because the odds against any of the five numbers winning is still 37 to 1. The reduction can only be on the number of losing outcomes because only one number can win. Five players individually won't get better odds making the same five $1 bets.

By making that type of bet, four of those bets will always lose. The odds against the first bet winning are 37 to 1 and the odds against the second bet becomes 36 to 1 because even though the first bet eliminated one of the losing outcomes, there is still only one way the second bet can win. Because the second bet eliminated one of the losing outcomes against the first bet, those odds against winning are reduced to 36 to 1 too. By making five bets, the odds against any one of the five bets winning is 33 to 1.

"So 33:5 against you winning, which will be paid at 31:5 if you win."

The house determines the payoff odds by the accumulative effect whether it's by five players each making five different $1 bets each having odds of 37 to 1 or one player making five different $1 bets. The probability of house winning that bet is the number of chances to win (33) compared to the total number of losing chances (5) and a 84.8% chance of winning. The house pays off at 6.2 to 1 when the payoff should be 6.6 to 1 and the difference becomes the house edge.

If one player bets $5 on the first five numbers (0, 00, 1, 2, and 3) and wins, they are paid off at 6 to 1 or a net win of $30. If another player puts $1 on each of those 5 numbers and wins, they are paid off at 35 to 1 or a net win of $31. The first five numbers is the one bet on a roulette layout the house will gladly accept all day long because it raises the house edge from 5.26% to 7.89%.

I'm not disputing your numbers, just pointing out while the accumulative bets changes the overall effect, it doesn't make any of the 33 losing chances magically disappear. I'm sure Jimmy can't wait to jump in with his presentation of the Gambler's Fallacy.

The logic in the first two paragraphs is very, very, very flawed. You can't assume that your odds reduced (per bet) b/c only one will win. That's simply wrong because the ball only lands once, and then stops. If it was such that there were multiple, non repeatable results, then your logic would be fine.

I really don't feel like I can explain it any clearer. The odds of you winning are 5/38 with your example. That's the same as 33 to 5, not 33 to 1. There is no difference between putting a chip on 1,2,3,4,5 and 6 or putting 6 chips on the 6way. No difference at all.

There's the execption for 0,00,1,2 and 3 getting shor paid, so then it'd be better off for you to bet them individually, but that's the only exception.

How can you possibly claim that a 3 dollar bet on 1st Street is any different than 3 1 dollar bets on 1,2 and 3? (I realize you didn't use those specific details, but mathemtically it's all the same, just smaller numers)

But I'll repeat the point one more time. It's 33 to 5 because there are 33 ways to lose and 5 to win. In my example, it is 35 to 3. I don't see how you could possibly dispute that at all. And I don't see how you can tell me that I misunderstand the game, when quite clearly 5/38 is 33 to 5 not 33 to 1.

New Jersey United States Member #99032 October 18, 2010 1439 Posts Offline

Posted: January 10, 2013, 8:23 pm - IP Logged

Quote: Originally posted by Ronnie316 on January 10, 2013

If I give up my chances of winning on some draws to get BETTER ODDS on others, is it still BETTER ODDS?

No.

It's exactly the same. But something tells me you were looking to just post that same sentence for the 150th time in an effort the be the "boss" of Lottery Post.

United States Member #116268 September 7, 2011 20244 Posts Offline

Posted: January 10, 2013, 8:24 pm - IP Logged

Quote: Originally posted by Boney526 on January 10, 2013

The logic in the first two paragraphs is very, very, very flawed. You can't assume that your odds reduced (per bet) b/c only one will win. That's simply wrong because the ball only lands once, and then stops. If it was such that there were multiple, non repeatable results, then your logic would be fine.

I really don't feel like I can explain it any clearer. The odds of you winning are 5/38 with your example. That's the same as 33 to 5, not 33 to 1. There is no difference between putting a chip on 1,2,3,4,5 and 6 or putting 6 chips on the 6way. No difference at all.

There's the execption for 0,00,1,2 and 3 getting shor paid, so then it'd be better off for you to bet them individually, but that's the only exception.

How can you possibly claim that a 3 dollar bet on 1st Street is any different than 3 1 dollar bets on 1,2 and 3? (I realize you didn't use those specific details, but mathemtically it's all the same, just smaller numers)

But I'll repeat the point one more time. It's 33 to 5 because there are 33 ways to lose and 5 to win. In my example, it is 35 to 3. I don't see how you could possibly dispute that at all. And I don't see how you can tell me that I misunderstand the game, when quite clearly 5/38 is 33 to 5 not 33 to 1.

Wow, thats THREE very-s Stack. Someone here is quite FLAWED............

United States Member #116268 September 7, 2011 20244 Posts Offline

Posted: January 10, 2013, 8:26 pm - IP Logged

Quote: Originally posted by Boney526 on January 10, 2013

No.

It's exactly the same. But something tells me you were looking to just post that same sentence for the 150th time in an effort the be the "boss" of Lottery Post.

Is that what I am.... The boss of LP. I never thought of it that way but thanks........

United States Member #116268 September 7, 2011 20244 Posts Offline

Posted: January 10, 2013, 8:27 pm - IP Logged

Quote: Originally posted by Boney526 on January 10, 2013

No.

It's exactly the same. But something tells me you were looking to just post that same sentence for the 150th time in an effort the be the "boss" of Lottery Post.

Im sure somehow the odds are the same, but it sure seemed like lots of chances to hit 3 of 5 with my wheel.

United States Member #116268 September 7, 2011 20244 Posts Offline

Posted: January 10, 2013, 8:41 pm - IP Logged

Quote: Originally posted by Boney526 on January 10, 2013

No.

It's exactly the same. But something tells me you were looking to just post that same sentence for the 150th time in an effort the be the "boss" of Lottery Post.

I guess I have to take your word for it on this one boney........ 3 in 54 sound BETTER than 46 in 306.

United States Member #124493 March 14, 2012 7023 Posts Offline

Posted: January 10, 2013, 8:54 pm - IP Logged

I want to be boss too!

If I give up my chances of winning on some draws to get BETTER ODDS on others, is it still BETTER ODDS?

If I give up my chances of winning on some draws to get BETTER ODDS on others, is it still BETTER ODDS? hmm

Boney is right. You cant beat the lottery all the time. I thank him for pointing out the errors in my ways. I guess I will have better chances at the slot machines.

United States Member #124493 March 14, 2012 7023 Posts Offline

Posted: January 10, 2013, 9:29 pm - IP Logged

Quote: Originally posted by RJOh on January 10, 2013

"But statistically speaking 100 independent quick picks, have less of a chance of trapping the 5 correct numbers, then do

two 17 number wheels, that wheels a total of 22 numbers.

Someone do the math on that one please."

If you have the statistics then you have the proof. There's no way to prove it mathematically.

The Gambler's fallacy, also known as the Monte Carlo fallacy (because its most famous example happened in a Monte Carlo Casino in 1913),^{[1]}^{[2]} and also referred to as the fallacy of the maturity of chances, is the belief that if deviations from expected behaviour are observed in repeated independent trials of some random process, future deviations in the opposite direction are then more likely.