NY United States Member #23835 October 16, 2005 3501 Posts Offline

Posted: July 1, 2006, 11:00 pm - IP Logged

Quote: Originally posted by Coin Toss on July 1, 2006

This thread was triggered by another thread where the poster who put up the OP said that odds and combinations were two different things.

And you're perfectly correct about the house edge.

I figured that was what prompted this thread, but I didn't see anyone saying odds and combinations are different things. There was plenty of confusion about how to figure out the probability for any given number of combinations, and confusion about the technical difference between odds and probability. As near as I can tell, though, everybody but you was talking about the odds (or probability) of an event happening, not the odds somebody will give you for paying out if that event occurs. That's why I said I think you'll just confuse the issue if you talk about combinations and payouts at the same time. The only relation between probability and the odds the houses uses for payouts is that the payout has to lower than the probabilityfor the house to make money.

In the post you quoted I'm pretty sure she was referring to other people's confusion that the number of combinations not played determined the odds regardless of how many combinations were played. In her first statement the proper odds would be 1:999, but a lot of people don't understand the difference between expressing the likelihood of an event as a probability and as odds. I'm not looking to pick nits, especially with one of the few people who seems to understand the basic concept, but a minor technical error definitely doesn't mean her basic explanation is wrong.

Statement 2 is exactly right. With 10 tickets she would have 1 chance for every 1000 possible outcomes.

Statement 3 is a math error or a typo that she later acknowledges. She meant that there would be 9990 combinations not played, which is also correct.

She carried that error over into her conclusion, which is also correct (whether it's adjusted for the math error or not).

I see that she did say in a subsequent post that odds aren't the same as combinations, and she's correct They're cetainly related, but they are different things. You stated earlier that "combinations determine the odds." The number of possible comb9nations will determine the game's odds, but that's only half of the equation. A player can play any number of those possible combinations from one to all of them, and that will determine the player's odds of winning

In your response to the post you've quoted here it appears that you have the exact type of confusion she was addressing. You correctly noted that the game odds in pick 4 are 1:9999, but after that you got it all wrong:

you: If you play ten different combinations, your odds are now 9989:1.

That's wrong on two counts. Playing 10 combinations leaves 9990, not 9989, and there are 10 outcomes that would match your combinations, not 1. The proper odds with 10 tickets would be 9990:10 against. The probability is 1 in 1000, exactly as dvdiva stated.

you: In your example above each dollar played reduced the odds by 1,000, that's just not the way it works.

If she had said that each dollar reduced the odds against by 1000 you'd be right, but that isn't what she said. She said that playing 10 tickets reduced the odds by a factor of 10, and that's exactly right.

you: If it did, according to that methos, another ten dollars played would recude those odds to 100:1.

You're just perpetuating your error about what she said. She slipped up in her subsequent response and reverted to the odds of pick 3 instead of pick 4, but she still has the concept right. The second 10 tickets would give you twice as many chances to win as 10 tickets and therefore doubles the probability of winning with 10 tickets. For pick 4 that would mean 1:500, and for pick 3 it would be the 1:50 that she states.

In another response to the post you quoted above you said:

You're saying you buy ten straight tickets with ten different numbers and reduce the odds from 1:10,000 to 1: 1,000.

The usual payoff on a stright pick 4 is $5000. Lotteries aren't in business to pay out $5,000 on 1:000 to one odds. Think about that one.

As mentioned beofre, the odds bet behind the pass line bet on a crap table are the only true odds available anywhere.

Lotteries pay off based on the odds a game offers, not on the odds a player has, which are based on the number of tickets they buy. The game odds are determined by the number of combinations, and the player's odds are determined by how many of those combinations result in them having a winning ticket. Lotteries aren't in business to pay millions of dollars for a $1 bet, but they manage to do it all the time and still make a huge profit. When the lottery pays the winners they do it with the money from all of the losing tickets, and in the example above that would include 9 of the 10 tickets our winner bought.

The payout odds in craps have absolutely nothing to do with anything said by dvdiva. You may know a lot about craps or casino payouts, but as far as probability goes, you should be taking a class in odds 101 not trying to teach it.

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10392 Posts Offline

Posted: July 3, 2006, 6:29 pm - IP Logged

KY Floyd

You said

That's wrong on two counts. Playing 10 combinations leaves 9990, not 9989....

Odds are based on to one....if the task is to pick one winning number out of a field of 10 numbers, the odds against your number are 9 TO 1, not 10 TOone....this if the odds are 9999 TO one (Pick 4) and you play ten striaght tickets, the odds against you are now 9989 TO one...

You're starting out with odds of 10,000 to one, which is not the case. It starts out 9999 to one.

Some Nevada casinos label their prop bets layouts like this, on the 11 it will say pays 15 TO 1. Then in other casinos (Reno) it might say PAYS 16 FOR 1. The 16 FOR 1 was a marketing ploy to let players think they were getting paid more for the same bet.

15 TO one and 16 FOR 1 are the same thing, the word FOR means that the payoff includes the amount of the wager.

Once again, dice, roulette, keno, lottery, whatever odds are odds, math is math. The formula being used to calculate the odds doesn't care what game it is calculating, some games just produce a lot bigger figures.

If I take a class it won't be from you. You're only verifying your opinionated lack of knowledge on this with such remarks. Where'd you learn this, backroom games in Newport, where the ace-deuce is barred on the come out. ? We used to watch those guys come up to the tables and run into a buzz saw everytime.Get off me.

If you really believe this:

The second 10 tickets would give you twice as many chances to win as 10 tickets and therefore doubles the probability of winning with 10 tickets.

For pick 4 that would mean 1:500, and for pick 3 it would be the 1:50 that she states.

you have absolutely no business telling anybody anything about odds.

As others have stated in these threads, only one straight wins and every single one of them has the same odds against them, no matter how many are played.

NY United States Member #23835 October 16, 2005 3501 Posts Offline

Posted: July 4, 2006, 1:40 am - IP Logged

Quote: Originally posted by Coin Toss on July 3, 2006

KY Floyd

You said

That's wrong on two counts. Playing 10 combinations leaves 9990, not 9989....

Odds are based on to one....if the task is to pick one winning number out of a field of 10 numbers, the odds against your number are 9 TO 1, not 10 TOone....this if the odds are 9999 TO one (Pick 4) and you play ten striaght tickets, the odds against you are now 9989 TO one...

You're starting out with odds of 10,000 to one, which is not the case. It starts out 9999 to one.

Some Nevada casinos label their prop bets layouts like this, on the 11 it will say pays 15 TO 1. Then in other casinos (Reno) it might say PAYS 16 FOR 1. The 16 FOR 1 was a marketing ploy to let players think they were getting paid more for the same bet.

15 TO one and 16 FOR 1 are the same thing, the word FOR means that the payoff includes the amount of the wager.

Once again, dice, roulette, keno, lottery, whatever odds are odds, math is math. The formula being used to calculate the odds doesn't care what game it is calculating, some games just produce a lot bigger figures.

If I take a class it won't be from you. You're only verifying your opinionated lack of knowledge on this with such remarks. Where'd you learn this, backroom games in Newport, where the ace-deuce is barred on the come out. ? We used to watch those guys come up to the tables and run into a buzz saw everytime.Get off me.

If you really believe this:

The second 10 tickets would give you twice as many chances to win as 10 tickets and therefore doubles the probability of winning with 10 tickets.

For pick 4 that would mean 1:500, and for pick 3 it would be the 1:50 that she states.

you have absolutely no business telling anybody anything about odds.

As others have stated in these threads, only one straight wins and every single one of them has the same odds against them, no matter how many are played.

Buono fortuna.

I guess I'll just start at the top.

you: Odds are based on to one

you (elsewhere, about odds in craps): 6:5 odds against

Maybe it's just my poor math skills, but that doesn't look like you expressed it as 1 to something. Does that mean that odds can be accurate without being expressed as n:1, or has your own argument shown that you aren't good at figuring odds? Odds are based on the ratio of possible outcomes favoring one result to possible outcomes favoring another result. If there are only two possible results (winning and losing in this case) then the two numbers, if left in their raw form, have to add up to the number of possible outcomes.

Let's start from a point where we agree. In pick 4 there are 10,000 possible outcomes, and if you have one ticket the odds are 9999:1 against winning.

you: if the odds are 9999 TO one (Pick 4) and you play ten striaght tickets, the odds against you are now 9989 TO one...

If you had 1 ticket there are 9999 unplayed combinations. If you wanted 10 tickets you'd need 9 more. Did I get that right? When I subtract 9 from the 9999 on one side of the equation I get 9990 combinations that aren't played, not 9989.

I expect that it's pretty obvious that the 1 and the 9999 for 1 ticket add up to 10,000. That's because all of the possible outcomes are represented by the odds. I'm cautiously optimistic that you'll agree that if you bought 5000 tickets they would cover half of the possible outcomes. Even if you're really bad at math it's probably intuitive that you would then have a 50% chance of winning. Personally, I think that a 50% chance of winning means there's also a 50% chance of losing and I'd have odds of 1:1. By your reckoning perhaps you think it's 5000 to 1. Even if you're bad at math I would think it's intuitive that 5000:1 is a lot worse than 50/50 or 1:1.

you: math is math

There we go. Another point of agreement. Let's review some math that I pointed out in the other thread. Fractions (which, just like odds, are ratios) can be expressed in different ways and still have the same value. 1/2 is the same as 2/4, or to get into pick 4 territory, 5000/10000. Watch closely, because this part is new. Math has a lot of rules and one of them is that the same thing can sometimes be expressed in multiple ways. 2+4 is the same thing as 4+2, for example. Another rule is that all ratios, whether it's a fraction, the gear ratio in a transmission, or odds can be expressed as many different numbers and still have the same value as long as both numbers have been changed in the same fashion. In the example of 1/2 and 2/4 both the top and bottom number have been multiplied by 2. For 5000/10000 both numbers have been multiplied by 5000.

Now let's get back to the player's odds in pick 4 when they have 10 tickets. There are 10,000 possible numbers that can be drawn and the player has tickets with 10 of those numbers. By my math (and math being math it should work the same for your math) that leaves 9990 numbers that aren't on any of the 10 tickets. That means there are 9990 numbers that, if drawn, won't match any of the 10 tickets. In turn that means that on the side of the odds equation that expresses the possible outcomes that result in the player losing we have 9990. There are also 10 numbers that , if drawn, will result in the player winning, because the number drawn will match one of the 10 tickets. That means that on the side of the odds equation that expresses the possible outcomes that result in the player wining we have 10. That makes the odds of losing 9990:10. (9990 outccomes that lose) : (10 outcomes that win), adding up to 10,000 possible outcomes.

If we take that 9990:10 and refer back to the math rules above we can divide both sides by 10 and we get odds of 999:1. Does it get any clearer now that the odds are based on "to one"?

you: You're only verifying your opinionated lack of knowledge

My lack of knowledge is a matter of fact, not opinion. There are all sorts of things I don't know, but unlike many people I usually know that I don't know, and I don't post my ignorance all over the internet. Fortunately I'm not the one displaying their lack of knowledge here, and I can be easily educated by those who do know the things I don't.

you: Where'd you learn this, backroom games in Newport

I'm not really sure, because it's been a very long time. I'm pretty sure that fractions were 3rd grade math when I was growing up, but maybe it was 4th grade. Either way, it's so elementary that it should really be called arithmetic. "Math" sort of implies something a bit more complicated. I don't think it was even called math class until I was at least in 6th grade.

you: only one straight wins and every single one of them has the same odds against them, no matter how many are played

That's true, and yet so thoroughly ignorant. If you buy 10,000 tickets with all 10,000 possible outcomes for pick 4 every single ticket has losing odds of 9999:1, and yet by playing multiple tickets the player has guaranteed that they will win. The odds against a single ticket winning a game with n possible outcomes will always be (n-1):1, but the odds of 1 of y tickets winning will always be (n-y):y.

you: Get off me.

There are two ways to make that happen. If you post about odds and get it right nobody else will need to post the correct information. If you don't post about odds while getting it wrong I can still help others who are confused without correcting you. Two choices, but I'm not offering odds.

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10392 Posts Offline

Posted: July 4, 2006, 1:57 am - IP Logged

Believe what you want. As long as people believe that an addittional ticket will "cut the odds in half" the lottery commissions will laugh at the players.

And those same people will tell people to go to casinos that chump people into thinking that 16 for 1 pays a lot better than 15 to 1.

When you say:

If you had 1 ticket there are 9999 unplayed combinations. If you wanted 10 tickets you'd need 9 more. Did I get that right? When I subtract 9 from the 9999 on one side of the equation I get 9990 combinations that aren't played, not 9989.

You still don't get it. Out of 10,000 possible combinations the odds against picking the right one are 9999 TO 1, not 10,000 to 1. This flaws the rest of your argument.

NY United States Member #23835 October 16, 2005 3501 Posts Offline

Posted: July 5, 2006, 2:52 pm - IP Logged

Quote: Originally posted by Coin Toss on July 4, 2006

Believe what you want. As long as people believe that an addittional ticket will "cut the odds in half" the lottery commissions will laugh at the players.

And those same people will tell people to go to casinos that chump people into thinking that 16 for 1 pays a lot better than 15 to 1.

When you say:

If you had 1 ticket there are 9999 unplayed combinations. If you wanted 10 tickets you'd need 9 more. Did I get that right? When I subtract 9 from the 9999 on one side of the equation I get 9990 combinations that aren't played, not 9989.

You still don't get it. Out of 10,000 possible combinations the odds against picking the right one are 9999 TO 1, not 10,000 to 1. This flaws the rest of your argument.

Nice talkin with ya. I give up. Happy 4th.

Bye.

My odds may be worse than they are with megamillions, but I'll try just once more

you: Out of 10,000 possible combinations the odds against picking the right one are 9999 TO 1, not 10,000 to 1.

I don't know why you don't get it, but we agree on that. I specifically said so in my previous post. Maybe you don't understand that those odds are based on already having 1 of the 10,000 possible numbers? By having 1 number there are 9999 numbers that result in losing and 1 that results in winning. That's where the numbers in "9999:1" come from. 9999 losing possibilities and 1 winning possibility. If you buy more tickets the number on the winning side goes up by the number of additional tickets and the number on the losing side goes down.

Maybe this approach will help. If having 1 number that will win against 9999 that will lose is what gives you 9999:1 odds, what would happen if you had 9999 tickets? Personally I find it incredibly obvious that it would now be 1:9999 in favor of winning. In case it isn't obvious to you, that's exactly the same math as for 1 ticket, except that we're now looking at it from the opposite side. Instead of the house having 9999 chances to win against your 1 chance the house now has only 1chance against your 9999. You and the house have simply traded places. Let's actually do the math:

9999 numbers that can lose : 1 number that result in winning - 9998 numbers that no longer lose + 9998 additional tickets 1 number that loses : 9999 numbers that result in winning

Do you see where it now says 1:9999? If you think that buying 10 tickest means subtraction 10 from 9999 and keeping the 1 on the right side for 9989:1 odds then by your method buying 9998 tickets would mean subtracting 9998 from 9999 and keeping the 1, which means you're claiming that having 9998 out of 10,000 possible combinations would give you 1:1 odds. Only 2 out of 10,000 numbers would result in losing but you think it's a 50/50 shot? Sorry, but that would be just plain stupid.

As I said before, the math itself is 3rd or 4th grade. It doesn't get any simpler and I don't know how to make it any easier.

In another thread you found it worthwhile to quote from a book that says that buying 100 tickets makes you 100 times more likely to win, but you've spent a lot of time claiming that 2 tickets doesn't cut the odds in half. Did you quote it because you believe it's true or because you wanted to post more incorrect information? The book author obviously gets it. Buying n times as many tickets makes you n times as likely to win, whether n is 2, 200 or 2000. If the odds are bad enough 2000 times as likely to win can still mean you'll almost certainly lose, but nobody has said otherwise.

Kentucky United States Member #32652 February 14, 2006 7344 Posts Offline

Posted: July 5, 2006, 5:30 pm - IP Logged

Floyd said:There are also 10 numbers that , if drawn, will result in the player winning, because the number drawn will match one of the 10 tickets. That means that on the side of the odds equation that expresses the possible outcomes that result in the player wining we have 10. That makes the odds of losing 9990:10. (9990 outccomes that lose) : (10 outcomes that win), adding up to 10,000 possible outcomes.

Whether you walk out of the store with 10 or even 1000 different pick-4 straight combinations, there are still 9999 outcomes that lose because only ONE straight combination is drawn.

Floyd also said:If we take that 9990:10 and refer back to the math rules above we can divide both sides by 10 and we get odds of 999:1. Does it get any clearer now that the odds are based on "to one"?

10 combinations reduce the losing outcomes by 1/1000 or .1% and 1000 combinations would reduce the losing outcomes by 10%. How does reducing the number of losing outcomes increase the number of winning combinations when only ONE combination is drawn?

Floyd said: I'm cautiously optimistic that you'll agree that if you bought 5000 tickets they would cover half of the possible outcomes. Even if you're really bad at math it's probably intuitive that you would then have a 50% chance of winning.

But, but, but, since the majority of state run Pick-4 lotteries pay $5000 for a dollar ticket, you would be out $5000 before the drawing. What you have is a 50% chance of losing the $5000 with a 0.00% chance of winning.

Floyd said: As I said before, the math itself is 3rd or 4th grade. It doesn't get any simpler and I don't know how to make it any easier.

Do the new 3rd or 4th grade math text books include how to determine pot odds when playing Texas Hold'em too?

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10392 Posts Offline

Posted: July 5, 2006, 6:58 pm - IP Logged

KY FLOYD

Listen to Stack47

Also, consider the payout - the average $5000 straight payout for a Pick 4 is nowehre near the 9999 odds against. So if you were crazy enough to play 9999 straight tickets and had only one possible loser, even if you won you're still taking it in the shorts to the tune of 9999 minus 5000.So you "win" and the lottery makes $4999 off of you.

Michigan United States Member #22395 September 24, 2005 1583 Posts Offline

Posted: July 5, 2006, 7:43 pm - IP Logged

KY Floyd

Let's start from a point where we agree. In pick 4 there are 10,000 possible outcomes, and if you have one ticket the odds are 9999:1 against winning

Just a question.

Michigan has a small pamphlet to inform everyone about how to play The Daily 4 Game. In it, it says, "Bet Type Straight Odds 10,000-1"

So are you saying if I do buy a ticket it would be improper to say the odds are 10000:1? In a normal conversation, I never heard anyone state their exact odds based on how many tickets they purchased. If someone asks, the answer I have always heard is 10000:1. Also, if you never purchase a ticket aren't the odds much greater than 10000:1 - it is absolutely certain that you won't win, therefore infinity:1? Therefore printing the 10,000-1 information in the pamphlet is useless and inaccurate?

NY United States Member #23835 October 16, 2005 3501 Posts Offline

Posted: July 6, 2006, 3:29 pm - IP Logged

Quote: Originally posted by truecritic on July 5, 2006

KY Floyd

Let's start from a point where we agree. In pick 4 there are 10,000 possible outcomes, and if you have one ticket the odds are 9999:1 against winning

Just a question.

Michigan has a small pamphlet to inform everyone about how to play The Daily 4 Game. In it, it says, "Bet Type Straight Odds 10,000-1"

So are you saying if I do buy a ticket it would be improper to say the odds are 10000:1? In a normal conversation, I never heard anyone state their exact odds based on how many tickets they purchased. If someone asks, the answer I have always heard is 10000:1. Also, if you never purchase a ticket aren't the odds much greater than 10000:1 - it is absolutely certain that you won't win, therefore infinity:1? Therefore printing the 10,000-1 information in the pamphlet is useless and inaccurate?

The numbers in the Michigan pamphlet are for the probability, which is slightly different than odds. Probability is expressed as the possibilities that favor one outcome (in this case, winning) compared to the total number of possibilities. For pick 4 that means the probability of winning with a single ticket is one winning number out of 10,000 possible numbers, or 1:10,000. Probability usually refers to the chance of the event happening, so the smaller number comes first.

Properly speaking, odds are expressed as the possibilities that favor one outcome compared to the possibilities that favor another outcome. For one ticket there is only one number that can be drawn that results in winning compared to the 9999 numbers that will result in losing, thus 9999:1. Note that with odds the numbers will commonly refer to the odds against winning, which is why the bigger number comes first. It's not at all uncommon for people to use the term odds when the number they use is really the probability. It's technically incorrect, but it probably doesn't matter to the vast majority of people.

For flipping a coin there are two possibilities, and the probability of getting heads is 1 in 2. The odds of getting heads is 1 to 1.

For rolling a single die any given outcome it would be 1 in 6 or 5 to 1.

If you don't buy any tickest then there are 10,000 numbers that result in losing against 0 chances of winning. That would be 10,000: 0 if it was expressed that way. The odds might commonly be refereed to as "infinite" but infinity:1 woul dimply that there is 1 chnace to win against an infinite number of chances to lose. The pamhlet has really twisted things up since they refer to "odds" but use anumber that is cleary the probability, but as a practical matter the difference between 10,000:1 and 9999:1 would be insignificant. I doubt that anyone is actually confused about their chances.

NY United States Member #23835 October 16, 2005 3501 Posts Offline

Posted: July 6, 2006, 3:45 pm - IP Logged

Quote: Originally posted by Stack47 on July 5, 2006

Floyd said:There are also 10 numbers that , if drawn, will result in the player winning, because the number drawn will match one of the 10 tickets. That means that on the side of the odds equation that expresses the possible outcomes that result in the player wining we have 10. That makes the odds of losing 9990:10. (9990 outccomes that lose) : (10 outcomes that win), adding up to 10,000 possible outcomes.

Whether you walk out of the store with 10 or even 1000 different pick-4 straight combinations, there are still 9999 outcomes that lose because only ONE straight combination is drawn.

Floyd also said:If we take that 9990:10 and refer back to the math rules above we can divide both sides by 10 and we get odds of 999:1. Does it get any clearer now that the odds are based on "to one"?

10 combinations reduce the losing outcomes by 1/1000 or .1% and 1000 combinations would reduce the losing outcomes by 10%. How does reducing the number of losing outcomes increase the number of winning combinations when only ONE combination is drawn?

Floyd said: I'm cautiously optimistic that you'll agree that if you bought 5000 tickets they would cover half of the possible outcomes. Even if you're really bad at math it's probably intuitive that you would then have a 50% chance of winning.

But, but, but, since the majority of state run Pick-4 lotteries pay $5000 for a dollar ticket, you would be out $5000 before the drawing. What you have is a 50% chance of losing the $5000 with a 0.00% chance of winning.

Floyd said: As I said before, the math itself is 3rd or 4th grade. It doesn't get any simpler and I don't know how to make it any easier.

Do the new 3rd or 4th grade math text books include how to determine pot odds when playing Texas Hold'em too?

Stack

Whether you walk out of the store with 10 or even 1000 different pick-4 straight combinations, there are still 9999 outcomes that lose because only ONE straight combination is drawn.

If I had tickets for 10% of the possible outcomes how could 99.99% of those outcomes result in losing? If I walk out of the store with 1000 tickets that all have different numbers on them there are 1000 combinations that result in one of my tickets winning. That means there are 9000 combinations where I lose and 1000 where I win: 9000:1000, or 9:1 If I have ten tickets it's 9990 combinations that don't match any of my tickets and 10 combinations that do. That's 9990:10 or 999:1.

How does reducing the number of losing outcomes increase the number of winning combinations when only ONE combination is drawn?

How many combinations are drawn determines the game's odds. The player's odds are determined by the game odds and the number of tickets the player has. For 10 tickets with different numbers only 1 ticket could match the winning number, but the player has 10 chances to do that. 10 times as many chances to win as with 1 ticket.

since the majority of state run Pick-4 lotteries pay $5000 for a dollar ticket, you would be out $5000 before the drawing. What you have is a 50% chance of losing the $5000 with a 0.00% chance of winning.

Through this entire thread I've only been talking about the odds, which are function of math, not finances. I'm not talking about how much you win or lose or how good a payout you get from a lottery or casino and I never have been. If you buy tickets with 5000 different numbers you have a 50% chance of having a winning ticket and a 50% chance of not having a winning ticket. If you have a winning ticket then you've won, period. Whether or not you turned a profit is an independent issue. If you have tickets for 50% of the possible combinations then your odds are 1:1, but I never said it was smart bet. The house always has an edge and anyone with sense always compares the risk and the reward before making their decision.

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10392 Posts Offline

Posted: July 6, 2006, 7:49 pm - IP Logged

KY Floyd

Through this entire thread I've only been talking about the odds, which are function of math, not finances. I'm not talking about how much you win or lose or how good a payout you get from a lottery or casino and I never have been

Well, I started the thread, you didn't, and it was going to lead to payoffs verses true odds, but kept getting sidetracked. Perhaps next time you can't start your own thread.

KY Floyd

For flipping a coin there are two possibilities, and the probability of getting heads is 1 in 2. The odds of getting heads is 1 to 1.

Baloney. the coin has two sides.

You know, I'd really love to see your concept of sports betting, laying 11 to win 10 and covering a pointspread. Want to tell us what percentage of winners you have to hit to overcome the vig?

New Mexico United States Member #12305 March 10, 2005 2984 Posts Offline

Posted: July 6, 2006, 7:54 pm - IP Logged

Baloney. the coin has two sides.

You know, I'd really love to see your concept of sports betting, laying 11 to win 10 and covering a pointspread. Want to tell us what percentage of winners you have to hit to overcome the vig?

Lawdylawdylawdy. We back on Boogie Street.

Coin Toss, if you won't think I'm one of them other kind of guys I'll tell you I'm getting one hell of a kick out of this thread. Lots of deep laughter going on here. Deep (voiced) laughter.