in betting what doea it means 'odds are 3:5?
97109
it means if you bet five of some thing and you win you would win three. You Bet $5 and win the bet, you would win $3
1140710
a win for any game is 3 points\na tie is 1\nand a loss is 0\n\ndid you notice how many answers are the same after i had this answer heh
744146
OK... so the question isn't completely clear, but here's my best interpretation... The winner needs best of seven games, so the first team to rack up four wins will win the whole thing.  If the Flames win in exactly 6 games, then that means that game 6 was their fourth win.  In the previous 5 games, they must have won three and lost two.  There are 10 different ways they could have gotten these three wins and two losses.  It might have been Win, Win, Win, Lose, Lose.  Or it might have been Win, Win, Lose, Win, Lose.  And so on and so forth.  The probabilty of the first pattern is 0.8*0.8*0.8*0.2*0.2.  (20% chance of a loss).  The probability of the second pattern is the same: 0.8*0.8*0.2*0.8*0.2 (although the order of the numbers is different the product is the same).  There are ten patterns.  So, the chance of winning 3 out of five is 10*0.8^3*0.2^2.  Then, the chance of winning game 6 is 0.8.  So the odds are 0.16384 or about 16%.
933407
If the "spread" is Dallas over Washington by 3, and during the game Dallas is ahead by more than 3 points, then they are "covering the spread" and making good on that bet.\n\nPoints are a racket designed to make money for the house...
277670
probability is the # of favorable outcomes over the # of possible outcomes:\n\nprobability=\n# of favorable outcomes\n------------------------------\n# of possible outcomes\n\nfavorable outcomes would be the # of possible resultis in your favor\n\nexample:\ni have 5 cards- a joker, a jack, a queen, a king, and a ace. there is a 1 in 5 chance (1/5) that you will pull a jack...there is a 3 in 5 chance (3/5) that you will pull a royal card (jack, queen, king)\n\nin both examples, the first number (1 and 3) are favorable outcomes
1133393
If the odds are 5-1, and you win, you get back $5 for every dollar bet,  plus your original dollar back, which is $6.\n\n$5 X 5 + 5 (original bet) = $30.00.\n\nIn the question as it stands, a 5-1 with a payoff of $14.40 is not a 5-1 odds, but 6.20 -1 odds.. A two dollar bet returning $14.40 is 6.20 x 1, and for a return on a two dollar bet would render 6.20  x 1 (odds) and the return of you original dollar, or $7.20 x 2 dollar bet would come to $14.40.\n\nSo, a $14.40 return on a $5 winning ticket would be $6.20 x 5 + your original $5 or $36.00.
390663
basically you bet on a variety of things. the winning team most importantly, then the score, maybe the first scorers, top try/pointscorer etc.
8884
It would be the third set - (0,-9) (2,0) (1,-4.5)
177047
Save your money. Either a) the website is true and you really can break a physical law known to be true and tested for centuries (ie. that you cannot move anything around with just your eyeballs), or b) the website is lying. Now if you were a betting person what is more likely to be true, a or b?
559040
hedging as in hedging your bet... putting smaller amounts on more possibilities.. the winnings are smaller but you increase your odds of winning. its the same as diversifying
1112053
It means (X + 5)/3 = 25 I guess.
553417
if there are only 10 numbers to chose from, say 0-9 then the odds are 10*10*10, if only nine number say 1-9 then it is 9*9*9. that help ya?
913338
To win 1 number, you have 36!/35! = 36 different single numbers.  Since you are picking 6 numbers in one bet, you can cover 6!/5! = 6 numbers per bet; therefore, you get full coverage of those 36 numbers in 36/6 = 6 bets.\n\nTo win 2 numbers, you have 36!/(2!34!) = 18*35 = 630 different pairs of numbers.  Since you are picking 6 numbers in one bet, you can cover 6!/(2!4!) = 15 pairs per bet; therefore, you get full coverage of those 630 pairs in 630/15 = 42 bets.\n\nTo win 3 numbers, you have 36!(3!33!) = 6*35*34 = 7140 different sets of three numbers.  Since you are picking 6 numbers in one bet, you can cover 6!/(3!3!) = 20 sets of three numbers per bet; therefore, you get full coverage of those 7140 sets in 7140/20 = 357 bets.\n\nIn general, when choosing c numbers from a set of n numbers, to win x numbers at least, you need:\n\nn!(c-x)! / (c!(n-x!))   bets.
232212
If a coin is flipped several times, you expect that it will turn heads half (.5) of the times and tails half (.5) of  the times. This expectation, is the same if you flip it only once, that is, there is a 50% chance it will be tails and 50% chance it will be heads (even though you'll either end up with 100% of the single flip being heads/tails and 0% being tails/heads.\n \nSo, supposing the bet is of $1.00 (i.e., you get to keep the coin if you win), you receive $1.00 if you call the out come correcetly and zero otherwise. The expected pay off of a flip is\n \nE[pay off] = $1.00*.5 + $0.00*.5 = $0.50\n \nNow, the "how much would you pay" question should be changed to "what is the most you can pay and still expect a profit". The expected profit is given by:\n \nE[profit] = expected pay off - amount payed to enter the bet\nE[profit] = E[pay off] - amount payed to enter the bet\nE[profit] = $0.50 - X\n \nIf you set the expected profit to be the least possible but still more than zero, E[profit] becomes $0.01 (one cent). Now you solve for X.\n \n0.01 <= 0.50 - X or X <= 0.49 (where <= should be read as less than or equal).\n\nNow it should be observed that this answers the question: "what is the most you can pay and still expect a profit?". The question "How much would you pay to enter this bet?", is far more comlex and depends on something called "risk preference".\n \nIf you are "indeferent to risk", the solution above is still correct but, if you are "averse to risk", you may prefer not to enter a game even if it has a positive expected profit. For examle, you may not be willing to pay 0.20 to enter this bet even though its expeceted profit is 0.30, because there is still a chance that you'll lose all of your 0.20 and have a profit of -0.20.\n \nOn the other hand, if you have an attraction to risk, you might pay 0.60 to enter the game and have an expected profit of -0.10 because you might make a profit of 0.40. (When you enter a bet at a cassino or lottery, the odds are always against you, i.e., your E[profit] < 0.)\n \nHope this helps.
978546
The expected value you get for rolling a normal die is 3.5, but there is no 3.5 on a die.
1026334
well 1-20 is not very good odds. That would be the favor. To win $1 dollar you would have to lay(bet) $20. 5-4 isnt that good either, to win 5, you would bet 4. Horse races have long shots like 25-1. Were a you would win $25 for every $1 you bet. So if you bet $10 you would win $250. Other sports are different though baskatball is based on $100. Like the Mavs are favored. You $150. To win a hundred you would bet $150, and so on.
1107637
the odds stay the same i think. u cant bet wen the match has started xxx