Betting - why is 3/1 known as carpet and 33/1 double carpet?
481021
This is beacuse the first man ever to win at 3/1 ( Oliver Corn - Cheltenham 1876 ) was actually betting in order to get enough money to buy a carpet for his lounge . It turned out however that he had been misquoted by some , frankly , mischevious carpet salesmen and had to return the next day as they had only supplied half the carpet - the rest being offered at an extortionate price - You've guessed it - at 33 times the price - hence the double carpet addage .
134133
Suppose there were three horses running in a race. If you put money on each one of them to win, place or show, you are guarenteed to win three times because each of them will win, place or show. That is hedging a bet. I'm not sure why they use the word 'hedging', but that is the general idea.
451034
We play $3, $1 for front side, $1 for back side and $1 for over all score.  individual holes dont matter.
754346
Investments with a doubling would be a geometric sequence.  Terms depend on a *multiplication* of previous terms, so it is geometric.\n\nc) geometric sequence.
97799
i don't believe the person above me got the second part right.  he is right about the beginning - the first probability is 2/5:\n\npossible numbers to draw: 2 4 6 11 13 15\n\nso you can choose any of these six out of the fifteen given (6/15 = 2/5).  the next number you draw must be also a "winning" number.  BUT, you have to have an even number less than 7 AND an odd number greater than 10.  this gives us two scenarios:\n\nIF the first number picked was even and <7, then the possible numbers you can pick from are ONLY: 11 13 15, since picking another even number <7 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\n\nIF the first number picked was odd and >10, then the possible numbers you can pick from are ONLY: 2 4 6, since picking another odd number >10 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\nSO, finally, suppose you go with the first scenario.  the probability would be: (6/15)*(3/14) = 3/35  going with the second scenario will give you the same thing, 3/35.  thus adding these two, since you can have either scenario occur and still meet your requirements, gives you: 3/35+3/35 = 6/35.\n\nAnd that would be the odds: 6/35.
77575
because math is weird.and 3 is one less than 4.
1119147
If you are referring to the 1:0 (0:0) format, the first set of numbers is the final score and the second set is the half time score.  This was driving me crazy until I found the answer.
895594
win,place.and show .that's1.2,3,then exotic bets
403923
I don't think it's so much a method of calculation as it is a follow the leader thing.  SI did an article a year or two ago in which they talked about the guy who sets the line for one of Vegas' big sportsbooks.  He researched the teams, etc,, but ultimately sort of set it by feel.  An interesting note: his concern was to put the line where the money would flow evenly to each side rather than to actually predict the game's outcome.  After he set his, other sportsbooks followed his lead and came out with similar/identical numbers.  Then, of course, they move it according to which way the action is going so that their odds of losing money are reduced.
435663
The first card you draw has a 54/55 chance it's not Saddam Hussein (I'm assuming he's the only Hussein in the deck). The second card has a 53/54 chance; the third 52/53 and so on.  When 14 of these are multiplied, everything cancels out to 41/55, or about 74.5%.
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.
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.
381005
if the team is a (+) like cleveland that means they are the under dogs and the (-) means they are the favorite.  You can pretty much pick that out by looking at the records.  Majority of the time the team with the best record will be the favorite.  The 183 is the over/under for the TOTAL score of the 2 teams.  By simply adding the score for both teams.  If you choose under 183 and the score is 100-82 than you would win.  The -110 is what you would recieve off a bet.  Its complicated. For instance, on a $100 bet if you win on a -110 you will recieve 90 off that $100 bet.  If it was a +110 you would recieve $110.  Just like the (-) means the favorite team it also mean the same on bets meaning its the favor bet so your not going to win as much as a (+) bet or underdog bet.  Another example, in hockey you have lot of +200, +150 ect.  That just tells you what you will recieve on a $100 bet.  Out of a $100 bet and you win on a +200 odds.  You will recieve $200 dollars on top of the $100 you get back for better for total of $300.
328998
Well, I doubt any sixth grader knows of the floor function :)\n\nHow about 8/(3-(8/3))?\n\n3 - 8/3 = 1/3\n8/(1/3) is the same as 8*3 = 24\n\nP.S. Anyone ever played 24? It's a pretty similar game, where you get four numbers and have to make them equal 24 with these rules. Very fun :)
112490
Because the numbers after the decimal (to the right of) represent less than one. One times a number is that number. Less than one times a number is only a fraction of that number.\n\nMake sense?
799460
Figure the probability of each outcome first:\nP(red marble) = 4/18 = 2/9\nP(green marble) = 3/18 = 1/6\nP(white marble) = 5/18\nP(black marble) = 6/18 = 1/3\n\nThe expected winnings would just be:\n2/9 * $5 + 1/6 * $4 + 5/18 * (-$3) + 6/18 * (-$2)\n\nThis works out to an expected winning of $0.277778.\n\nAn alternate way to figure this is:\n[(4 * $5) + (3 * $4) + (5 * -$3) + (6 * -$2) ] / 18\n= (32 - 27) / 18\n= $5/18 = $0.277778\n\nThe game is better than fair, since you don't have to pay anything but you are expected to win more than $0.  (Technically it is unfair since the expected return is not equivalent to what you pay, but I'm going to keep playing it because it is unfair to the house, not to me!)
804992
Let us call the first 2 "2 sub 1", the second "2 sub 2" ... same thing for the 6's.\n\nLet us say that "2 sub 1" is distinct from "2 sub 2", and so on.\n\nWe have 8 choices for the first number, seven choices for the second number, etc. This means that there are 8*7*6*5*4*3*2*1 choices, or 8!. \n\nNow if you look at any number, take all of the 2's out. You see that you can rearrange the 2's in 3! different ways (including the original arrangement) using the procedure in the paragraph above. If we take out the 2's rearrange them, and put them back in we have created the a new arrangement. So for each arrangement of numbers, we have counted it 3! times. So we divide by 3!. The same thing for the sixes, we divide by 2!.\n\nThe result it (8!)/((3!)(2!)) = 3360 choices.