Posted: 1/27/2011 9:47:24 PM EDT
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More of a gambling question. How many combinations are there in an 11-leg parlay? Is it: 66 (11+10+9+8+7+6+5+4+3+2+1), or is it 39,916,800 (11X10X9X8X7X6X5X4X3X2X1)?
The casino has an 11-leg superbowl progressive parlay with a $14,000 payout on a $5 buy-in. Each option is either a straight-up bet, or to the half so there can't be any ties. How do you figure out the odds. |
| 11 either / or choices is the same as picking the outcome of 11 coin flips. You'd have 1 out of 2 chances for the first flip, then 1 out of 4 for the second, and so on. Ultimately on the 11th flip you'd have 1 chance out of 2048 to pick the correct combination. Since the payout ($14,000) divided by the cost per pick ($5) is 2800, this favors the house. |
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I have a stats minor...and I'm pretty dang good at it...but I don't 100% percent understand the question. If there are 11 possibilities and 11 possible outcomes (1234567891011) => From (00000000000) to (11,11,11,11,11,11,11,11,11,11,11) then the number of possible outcome is 11! (11 Factorial) in which is (1x2x3x4...etc). 11 Possibilities with 2 out comes, can only choose 1 then there are 22 possibilities (1, a) to (11, b) assuming 1-11 is the fist choice and A or B is the possible outcome of the first choice. 11 possibilities with 2 outcomes, in which case 1-11 could be on any choice followed by an A or B, then its ( 11! x 2 ). (11a*10a*9a*8a...) + (11b*10b*9b*8b...) Does any of that help you? |
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Quoted:
The way the bet is set up, you have 11 "either-or" choices (steelers/packers or over/under 48.5 points). So if you have 11 choices with two possibilities each, how many possibilities are there? 1/2^11 so, 1/(2*2*2*2*2*2*2*2*2*2*2) or 1 out of 2,048 ETA: Hmm... with a $14,000 pay out and $5 tickets, it seems like you could buy every combination and come out with a $3,760 profit. |
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I have a stats minor...and I'm pretty dang good at it...but I don't 100% percent understand the question. If there are 11 possibilities and 11 possible outcomes (1234567891011) => From (00000000000) to (11,11,11,11,11,11,11,11,11,11,11) then the number of possible outcome is 11! (11 Factorial) in which is (1x2x3x4...etc). 11 Possibilities with 2 out comes, can only choose 1 then there are 22 possibilities (1, a) to (11, b) assuming 1-11 is the fist choice and A or B is the possible outcome of the first choice. 11 possibilities with 2 outcomes, in which case 1-11 could be on any choice followed by an A or B, then its ( 11! x 2 ). (11a*10a*9a*8a...) + (11b*10b*9b*8b...) Does any of that help you? Getting closer. The choices are: 1.) Pitt/GB - Pick the winner 2.) over/under 2.5 inteceptions - pick over or under 3.) Who will have more tackles: Mathews or Polamolu - 4.) over/under 3.5 Field Goals - pick over or under 5.) over or under 51.5 points - pick over or under 6.). 7.)... 8.)... So there are 11 choices with two possibilities each. I think it is 11 factorial, right? |
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I have a stats minor...and I'm pretty dang good at it...but I don't 100% percent understand the question. If there are 11 possibilities and 11 possible outcomes (1234567891011) => From (00000000000) to (11,11,11,11,11,11,11,11,11,11,11) then the number of possible outcome is 11! (11 Factorial) in which is (1x2x3x4...etc). 11 Possibilities with 2 out comes, can only choose 1 then there are 22 possibilities (1, a) to (11, b) assuming 1-11 is the fist choice and A or B is the possible outcome of the first choice. 11 possibilities with 2 outcomes, in which case 1-11 could be on any choice followed by an A or B, then its ( 11! x 2 ). (11a*10a*9a*8a...) + (11b*10b*9b*8b...) Does any of that help you? Getting closer. The choices are: 1.) Pitt/GB - Pick the winner 2.) over/under 2.5 inteceptions - pick over or under 3.) Who will have more tackles: Mathews or Polamolu - 4.) over/under 3.5 Field Goals - pick over or under 5.) over or under 51.5 points - pick over or under 6.). 7.)... 8.)... So there are 11 choices with two possibilities each. I think it is 11 factorial, right? No, 11 factorial would be if you have 11 options and need to pick the order they occur in. For example, a password is 26 letters long and uses every letter of the alphabet once. The odds of guessing it correctly would be 1 in 26!. |
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The way the bet is set up, you have 11 "either-or" choices (steelers/packers or over/under 48.5 points). So if you have 11 choices with two possibilities each, how many possibilities are there? 1/2^11 so, 1/(2*2*2*2*2*2*2*2*2*2*2) or 1 out of 2,048 ETA: Hmm... with a $14,000 pay out and $5 tickets, it seems like you could buy every combination and come out with a $3,760 profit. The math here is correct, there are 2048 possible combinations of 11 either-or choices. My understanding (having never done any significant gambling) is that something like this normally pays out at less than true odds. $14000 seems like a high payout based on ticket price. Also, I think each separate choice has attached odds usually too, which affects the payout amount. However, there's really not enough info in the OP for me to have a full picture. |
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I have a stats minor...and I'm pretty dang good at it...but I don't 100% percent understand the question. If there are 11 possibilities and 11 possible outcomes (1234567891011) => From (00000000000) to (11,11,11,11,11,11,11,11,11,11,11) then the number of possible outcome is 11! (11 Factorial) in which is (1x2x3x4...etc). 11 Possibilities with 2 out comes, can only choose 1 then there are 22 possibilities (1, a) to (11, b) assuming 1-11 is the fist choice and A or B is the possible outcome of the first choice. 11 possibilities with 2 outcomes, in which case 1-11 could be on any choice followed by an A or B, then its ( 11! x 2 ). (11a*10a*9a*8a...) + (11b*10b*9b*8b...) Does any of that help you? Getting closer. The choices are: 1.) Pitt/GB - Pick the winner 2.) over/under 2.5 inteceptions - pick over or under 3.) Who will have more tackles: Mathews or Polamolu - 4.) over/under 3.5 Field Goals - pick over or under 5.) over or under 51.5 points - pick over or under 6.). 7.)... 8.)... So there are 11 choices with two possibilities each. I think it is 11 factorial, right? There are 2^11 possibilities, so the odds are 1/2^11. Of course, since it sounds like this is a progressive parlay, it's you can lose some and still receive some payout, so it's a bit more complex than that. |
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Quoted: Quoted: I have a stats minor...and I'm pretty dang good at it...but I don't 100% percent understand the question. If there are 11 possibilities and 11 possible outcomes (1234567891011) => From (00000000000) to (11,11,11,11,11,11,11,11,11,11,11) then the number of possible outcome is 11! (11 Factorial) in which is (1x2x3x4...etc). 11 Possibilities with 2 out comes, can only choose 1 then there are 22 possibilities (1, a) to (11, b) assuming 1-11 is the fist choice and A or B is the possible outcome of the first choice. 11 possibilities with 2 outcomes, in which case 1-11 could be on any choice followed by an A or B, then its ( 11! x 2 ). (11a*10a*9a*8a...) + (11b*10b*9b*8b...) Does any of that help you? Getting closer. The choices are: 1.) Pitt/GB - Pick the winner 2.) over/under 2.5 inteceptions - pick over or under 3.) Who will have more tackles: Mathews or Polamolu - 4.) over/under 3.5 Field Goals - pick over or under 5.) over or under 51.5 points - pick over or under 6.). 7.)... 8.)... So there are 11 choices with two possibilities each. I think it is 11 factorial, right? Everyone saying 1/2^11 is correct. Now what you would want to do is look up some stats from the season and determine if there are any statistically significant stats that you can bet on. For example, "sucker bets" are the 50/50 chances like Packers or Steelers. But go look at average scores, average fg's, etc and determine if there's a good chance that a bet covering a 50% portion would be wise. Again, for example. 3.5 FGs over/under. Go look up the average number of FG's in the NFL in the last couple years. If more than...say 75% of the games had more than 3.5 fg's then bet over. If 75% of the games has less than 3.5fg's then bet under. If its anything close to 50%, then I would advise not betting if you intend to make money. Do that with every stat on the list except for, like I said, the A or B discrete outcomes. ETA: Never play cards with a statstitian. |