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Math/Statistics Help (Page 2)
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 2014-01-10 12:56 PM in reply to: McFuzz |
Elite  4564    Boise |  Subject: RE: Math/Statistics HelpOriginally posted by McFuzz Originally posted by Kido I'll disclose!
A coworker had some notion that he could bet every possible outcome of the 4 games in a parlay and no matter the outcome, win money. That's because he could only think of 4 combinations. I told him that Vegas is not that dumb AND there were many more outcomes than he thought. He challenged me to come up with others. It took me 30 seconds to figure out 4 more than he had before I quit because I proved my point. I then wanted to know the mathematical way to figure it and narrowed it down to 4! or 2^4. I ruled out 4! right away leaving 2^4. I used BT to confirm. 16 possible outcomes. You can't bet all 16 because the winning ticket would not pay 16:1 to make up for the 15 outcomes that you lose.
Your friend is right...he would in fact win money betting on every outcome. What your friend fails to realize is that the expected payout will be lower than the cost to play because the house always takes a cut. If Denver has a 5% chance of winning and pays 15:1, the expected payout is 0.05*15=0.75. Your friend can repeat this calculation for each team and he's unlikely to find any team where the expected payout is greater than 1.0 (which is the cost to play). Now your friend might still collect more money than he pays, but only if the winning team pays out greater than 16:1, but he can't know this is going to happen when he buys in (if he knew this when he buys in, why would he purchase the non-winning tickets?) The question you originally posed doesn't really capture what your friend is proposing. The question you posed (4 events each with two possible outcomes) only describes the first round. You can look at it as there are only 8 outcomes (one of the 8 teams will win 3 times and the other 7 will lose once and it doesn't matter when they lose). Alternatively, you can look at it as (2^4)*(2^2)*(2^1)=2^7=128 outcomes if you care when a team loses (but does it really matter when a team loses?). Off topic, there was an article awhile back about a state lotto in the northeast somewhere. Whenever the jackpot got up to a certain amount this group of people would buy lots of tickets. They expected payout for the drawing had eclipsed the cost to enter and they ended up winning a nice sum of money over the course of several years. |
| 2014-01-10 1:48 PM in reply to: Kido |
Sensei Sin City | Subject: RE: Math/Statistics Help Cranked out a spreadsheet using current money lines and $10 bet on each possible 4 teamer. Go figure, 8 combinations you win money, 8 combinations you lose. What I found interesting though, if the worst possible combination (based on odds - which is the four favorites winning), your net loss is $112.11. All the other losing combinations are less than that. However, the best ticket (four underdogs winning), nets $1059.60. Many of the other winners are pretty big two. It looks like you need at least two underdogs to prevail to be in the money. Granted, the faves winning is statistically more apt to happen so that's why the disparage between the money lost and money won. Next I'm going to figure out if I adjust the betting (more on the favorites and less on the underdogs) to see if there is an exact amount to bet on each outcome to find a "can't lose" situation. |
| 2014-01-10 1:58 PM in reply to: 0 |
Champion 7558 Albuquerque, New Mexico | Subject: RE: Math/Statistics Help Originally posted by Kido Originally posted by McFuzz Originally posted by Kido Your friend is right...he would in fact win money betting on every outcome. What your friend fails to realize is that the expected payout will be lower than the cost to play because the house always takes a cut. If Denver has a 5% chance of winning and pays 15:1, the expected payout is 0.05*15=0.75. Your friend can repeat this calculation for each team and he's unlikely to find any team where the expected payout is greater than 1.0 (which is the cost to play). Now your friend might still collect more money than he pays, but only if the winning team pays out greater than 16:1, but he can't know this is going to happen when he buys in (if he knew this when he buys in, why would he purchase the non-winning tickets?) The question you originally posed doesn't really capture what your friend is proposing. The question you posed (4 events each with two possible outcomes) only describes the first round. You can look at it as there are only 8 outcomes (one of the 8 teams will win 3 times and the other 7 will lose once and it doesn't matter when they lose). Alternatively, you can look at it as (2^4)*(2^2)*(2^1)=2^7=128 outcomes if you care when a team loses (but does it really matter when a team loses?). I'll disclose!
A coworker had some notion that he could bet every possible outcome of the 4 games in a parlay and no matter the outcome, win money. That's because he could only think of 4 combinations. I told him that Vegas is not that dumb AND there were many more outcomes than he thought. He challenged me to come up with others. It took me 30 seconds to figure out 4 more than he had before I quit because I proved my point. I then wanted to know the mathematical way to figure it and narrowed it down to 4! or 2^4. I ruled out 4! right away leaving 2^4. I used BT to confirm. 16 possible outcomes. You can't bet all 16 because the winning ticket would not pay 16:1 to make up for the 15 outcomes that you lose.
I disagree. First of all, that's not how parlays actually work - Betting every outcome my HELP is changes to win, but it's far from a guarantee. In fact, betting the moneyline on every out come (say $100 a pop) costs you $1600 bucks. HOWEVER, if all 4 favorites win? You WIN $300. Net loss of $1300. However, if the four underdogs win, you get paid $12,700 and net $11,200 with the 15 losing tickets. Just playing around, you need at least TWO of the underdogs to win to get a winning parlay ticket that will cover the $1500 you lose on the other losing tickets. There is NO guarantee to win money. If I get time, maybe I will put all 16 possible outcomes into a parlay calculator assuming a $100 bet on each. The other wrinkle, maybe you can adjust you bet size based on the payout. For example, put $10 on the 4 teamer using the underdogs. That way, you invest less in the long shots so lose less when the favorites win... If you cover the table, you are guaranteed to win money, but you are not guaranteed to win enough money to cover the cost of the bets. In order to gain (win more than the cost of the bets) money covering the table, you need one of the underdogs paying at least N:1 where N is the number of bets you are making. Which underdog? I haven't worked out the math for a more optimal distribution of bets, but if *I* were putting money on the table, *I* would calculate the expected payout for each bet and concentrate my bets on the top 30% expected payouts. The best chance of getting *some* of your money back is to concentrate your bets on the favorites but as you indicated, you'll only get back a small part of your total bet. There's a reason I don't gamble (Vegas or lottery tickets). I'll occasionally buy a few lottery tickets when the jackpot grows past the odds and I'll sometimes participate in office pools for the social aspects. The real point I was trying to make in my post is that there are different ways to define "outcome" and the possible outcomes will change depending on what is chosen. The problem as stated and the solutions provided so far are appropriate if your friend is only betting on the results of the first round playoffs, not on an eventual superbowl winner. Edited by McFuzz 2014-01-10 2:03 PM |
| 2014-01-10 2:01 PM in reply to: McFuzz |
Pro 15655 | Subject: RE: Math/Statistics Help This thread is a corker......everybody saying the same thing a bit differently. |
| 2014-01-10 2:01 PM in reply to: McFuzz |
Sensei Sin City | Subject: RE: Math/Statistics Help Got it figured... I calculated how much you have to put down on each four teamer to cover the $10 you put on each game (win $160). Problem is, if you add up the minimum to cover $160, it costs you 188! Plug in enough to cover 188, you need 221! Basically it will always cost you %18 MORE to cover all your bets - which, is around what the house rakes... They got it doped out!... Cool. (for them)...
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| 2014-01-10 3:07 PM in reply to: 0 |
Sensei Sin City | Subject: RE: Math/Statistics Help Originally posted by McFuzz Originally posted by Kido If you cover the table, you are guaranteed to win money, but you are not guaranteed to win enough money to cover the cost of the bets. In order to gain (win more than the cost of the bets) money covering the table, you need one of the underdogs paying at least N:1 where N is the number of bets you are making. Which underdog? I haven't worked out the math for a more optimal distribution of bets, but if *I* were putting money on the table, *I* would calculate the expected payout for each bet and concentrate my bets on the top 30% expected payouts. The best chance of getting *some* of your money back is to concentrate your bets on the favorites but as you indicated, you'll only get back a small part of your total bet. There's a reason I don't gamble (Vegas or lottery tickets). I'll occasionally buy a few lottery tickets when the jackpot grows past the odds and I'll sometimes participate in office pools for the social aspects. The real point I was trying to make in my post is that there are different ways to define "outcome" and the possible outcomes will change depending on what is chosen. The problem as stated and the solutions provided so far are appropriate if your friend is only betting on the results of the first round playoffs, not on an eventual superbowl winner. Originally posted by McFuzz Originally posted by Kido Your friend is right...he would in fact win money betting on every outcome. What your friend fails to realize is that the expected payout will be lower than the cost to play because the house always takes a cut. If Denver has a 5% chance of winning and pays 15:1, the expected payout is 0.05*15=0.75. Your friend can repeat this calculation for each team and he's unlikely to find any team where the expected payout is greater than 1.0 (which is the cost to play). Now your friend might still collect more money than he pays, but only if the winning team pays out greater than 16:1, but he can't know this is going to happen when he buys in (if he knew this when he buys in, why would he purchase the non-winning tickets?) The question you originally posed doesn't really capture what your friend is proposing. The question you posed (4 events each with two possible outcomes) only describes the first round. You can look at it as there are only 8 outcomes (one of the 8 teams will win 3 times and the other 7 will lose once and it doesn't matter when they lose). Alternatively, you can look at it as (2^4)*(2^2)*(2^1)=2^7=128 outcomes if you care when a team loses (but does it really matter when a team loses?). I'll disclose!
A coworker had some notion that he could bet every possible outcome of the 4 games in a parlay and no matter the outcome, win money. That's because he could only think of 4 combinations. I told him that Vegas is not that dumb AND there were many more outcomes than he thought. He challenged me to come up with others. It took me 30 seconds to figure out 4 more than he had before I quit because I proved my point. I then wanted to know the mathematical way to figure it and narrowed it down to 4! or 2^4. I ruled out 4! right away leaving 2^4. I used BT to confirm. 16 possible outcomes. You can't bet all 16 because the winning ticket would not pay 16:1 to make up for the 15 outcomes that you lose.
I disagree. First of all, that's not how parlays actually work - Betting every outcome my HELP is changes to win, but it's far from a guarantee. In fact, betting the moneyline on every out come (say $100 a pop) costs you $1600 bucks. HOWEVER, if all 4 favorites win? You WIN $300. Net loss of $1300. However, if the four underdogs win, you get paid $12,700 and net $11,200 with the 15 losing tickets. Just playing around, you need at least TWO of the underdogs to win to get a winning parlay ticket that will cover the $1500 you lose on the other losing tickets. There is NO guarantee to win money. If I get time, maybe I will put all 16 possible outcomes into a parlay calculator assuming a $100 bet on each. The other wrinkle, maybe you can adjust you bet size based on the payout. For example, put $10 on the 4 teamer using the underdogs. That way, you invest less in the long shots so lose less when the favorites win... Fine... ONE ticket "wins" money. I concede. But if the object is to cover all outcomes and walk away with more money than you started - AKA "WIN", I'm not sure how you can say you win if you are 130 dollars poorer than when you started even if you "won" $30 bucks on one of the tickets. I think you lost $130. But if you want to pick nits. Yes, you "won" money. I have done the distribution. I found there are probably two games you don't want to bet, resulting in a $140 investment. Those two games cover so little of your loses on the other bets, they are not worth it. They happen to be all four faves winning and all faves except Carolina (which is basically a pick-em) The other 6 actually hedge your bet pretty well, and reduce the "risk" from 160 down to about 40. I'm not even going to get taking points into it... Too much work. Edited by Kido 2014-01-10 3:12 PM |
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