how can I win at greyhound racing using mathematical formula?
349693
Why don't you spend more time working an honest job than betting on dog races? You'd be surprised how good it feels to have worked hard to earn your money.
750282
X= total number of miles barry will drive\n\nx - 125 > 75
679486
Take your ticket to collect your winnings.\n\nOr, if you want to work it out yourself - take price and multiply by stake e.g. a £2.00 stake on a 4/1 winner would give you £10.00 return (£8.00 win + your £2.00 stake back)
84771
In pursuit, two teams start on the opposite sides of the track and "pursue" each other. You can win your heat by either having a better time over the designated distance, or by catching the other team.\n\nIf you catch the other team, you are 1/2 lap ahead of them so they just end it. Sort of like a slaughter rule in softball.\n\nIf you're get bold, you can go out way too fast early in hopes of catching the other team. If you fail, you end up getting killed by your opponents when you crash and burn later in the race.
1152243
I think it's 16 because of the math involved, you cut 16 in half and get 8, cut that in half you get 4, cut that in half and get 2, then a final winner.  Try doing that with 15 or 18
772558
Now there's some strange questions. I'll help as best I can. I hope I've understood the questions properly.\n\nQuestion 1) Bob will always win, assuming they play perfectly.\nIf Ann takes 1 matchstick, Bob should take 2. That leaves 3 left. Now whether Ann takes 1 or 2 from the remaining 3, Bob is left with 1 or 2, so he can take what's left and wins.\nIf Ann takes 2 matchsticks,  Bob should take 1. That also leaves 3 left, so again regardless of how many Ann takes from the 3 remaining ones, Bob can take what's left and win again.\n\nQuestion 2) The answer is 5.52pm.\nLets say the slow train moves at speed X. That means the fast train moves at speed 2.75X. This is because the fast one moves 2.75 times faster than the slow one (since it travels the same distance in 8 hours compared to the slow time of 22 hours, so it must be going 22/8 times faster).\nIn the time taken for the trains to meet, between them they have travelled the full distance. Therefore you can add their speeds together to calculate the relative approaching speed of the trains, which would be 2.75X + X = 3.75X.\nNow, the slow train travels at speed X and takes 22 hours to go the same distance. Therefore we can say that the time taken for the two trains to meet is 22X/3.75X, or 22/3.75, which is 5.8666.\nTo convert 0.8666 of an hour to minutes is simple - just multiply by 60. This gives 52.\n\nSo the total time taken is 5 hours 52 mins. \n\nTo put it another way:\nSpeed of train 1 = 2.75 times faster than train 2. So the speeds of the two trains together = 3.75 times faster than the slow train.\nThe slow train takes 22 hours to go the distance. Therefore the two trains together take 22/3.75 to do the same distance, which is 5.8666.\n\nI hope this helps :)
655518
a) the number of possible tickets is C(42,6) = 5245786\n1/5245786 = .000000190629\nb) the number of tickets with exactly 4 is \nC(6,4)C(36,2)\n= 15*630 = 9450\n9450/5245786 = .00180\nc) You don't win if you have less than 3 winning numbers, so either you can have 0, 1, or 2 winning numbers and not win\nP(0 numbers) = C(36,6)/5245786 = .37130\nP(1 number) = C(6,1)C(36,5)/5245786 = .43119\nP(2 numbers) = C(6,2)C(36,4)/5245786 = .16843\nP(no prize) = .37130 + .43119 + .16843\n= .97092
384498
How to win state and national lotteries.\nDo you want to win a BIG, BIG prize? Do you want to have all your financial worries sorted out? Do you want to win the lottery? The secret is very, very simple. Anybody can manage it. All you have to do is buy a ticket with the numbers that are going to be drawn at the next draw. If you do this I can GUARANTEE you will win the lottery.\n\n That's not exactly new information, is it?\n\nNeither is the idea that nobody can predict what those numbers will be. However, when a lottery machine selects the numbers this is in no sense a random action. It is 100% predictable. There is nothing stopping anybody calculating exactly which balls not only will, but MUST be drawn. Nothing stopping you. It will simply take a measuring and timing tolerance at the edge of possibility combined with a huge series of calculations.\n\nYou can measure the balls and know their properties. You can time the intervals between starting and stopping the contra-rotating tumble dryers. You can project the trajectories of every single ball and their interactions with each other. It can be done. But it would probably cost more than the size of the lottery jackpot to fund the project and nobody is going to be very happy with people standing around and timing the balls to that degree of accuracy, because one microsecond will totally disrupt all the calculations. The lottery ball machine is totally predictable and not in any way random, but exactly the same thing can be said about a woman. Both are determined and predictable systems, but they simulate randomness far more elegantly than true randomness would. True randomness is a willo-the-wisp entity, like infinity, the highest prime number and the function of half of the roundabouts in Warrington. Lottery machines are not random, but they are chaotic, and therefore, like the minds of women, very difficult to probe with the crude analytical tools we have at our disposal.\n\nSo, I can't tell you how to win the lottery. However, I can offer you some interesting advice that you might find helpful.\n\nYou have to be in it to win it\nThis is a very common slogan used by lotteries and football pools companies alike. What this slogan is saying is you can't dream of winning the lottery or the pools unless you part with your beer tokens. Of course they are right. If you never buy a ticket you can never win. But how many times do you dream of winning? Do you have to win every week to be happy? Of course not. You only have to win once. So you only have to buy one ticket, at some time in the future to be able to legitimately dream of winning a huge jackpot.\n\nOne ticket will do\nYour chance of winning a share in a big jackpot prize is very low. In the UK national lottery (now named Lotto) your chances of having a share of a top jackpot, by getting 6 balls out of 6 representing the numbers 1 to 49 are 1 in 13,983,816. \n\nNow most weeks people win, often several people share the jackpot. So it does happen. But if you staked £5 every week on those odds you would not have a realistic expectation of winning a share in a jackpot in a normal lifetime. But you knew that anyway, didn't you?\n\nSo you might as well get used to the idea that winning the lottery is unlikely. But it's also nice to have a dream, it is good for your mental health. So here is the secret: Buy a ticket. One day. Maybe tomorrow. Maybe not.
226806
Go to Gamewinners (link below), or a similar "game cheat" web site for hints on winning your cases. \nEach case will probably be just as, if not more hard for you to solve. They tried to make the game as realistic as possible, that's why it seems so hard.
1029935
Always sing while you are winning\n\nRS
720147
Dutch win betting or "dutching" is a gambling term to mean structuring your bets (on two or more betting interests) in a manner that will realize an equal profit regardless of which one wins. Dutching is a powerful tool used to cover several horses in the same race and guarantee the maximum possible return. Using the calculator provided above all you need to do is type in the odds you expect to receive (example 5-1) in the odds converter and it will automatically convert the track odds into a priceline number. Copy the priceline number into the calculator and repeat this process for all of the horses that you are considering. Then determine the amount you wish to risk and click the button marked calculate. The calculator will then determine the exact amount needed to risk on each horse according to your desired risk to insure the same return no matter whom you choose to win. The advantage of dutch betting is it takes the guesswork out of what to do when you like more than one horse in the same race.\n\nhttp://www.oddsonracing.com/bettors_corner_tools.cfm
1154073
Heed the advice of "marnefirstinfantry" as it is very sound, and very true.  I apply my own rule of thumb with regards to success in any venture ... and it's not something I came up with on my own, but learned from those far more succesful than I've been in handicapping, in business and in life. I call it the 97% - 3% Rule, which in my observations has borne out to a greater than lessor extent, as I do believe approximately 3% of the MOST SUCCESSFUL persons in any venture, control approximately 97% of the assets in that venture ... which leaves  97% of the participants to split up the remaining 3% of the assets in that venture. \n\nSo, translating this to winning at handicapping would seem to indicate you've got your work cut out for you.   I know of no simple, fast and easy "SYSTEM" that consistently beats the ponies. What I've heard and observed from those that are successful at handicapping the horses, is that you've got to do your homework: study the Racing Form, Learn all you can about CLASS, PACE/ SPEED, TRAINERS & THEIR TRAINING METHODS, JOCKEYS, THE TRACK or TRACKS you'll be frequenting. And still  you might be a consistent loser if you're weak in one or more areas of handicapping. The "PROS" I've known treat their handicapping like a JOB!  And you've got to show up, and you've got to put in your 8 hours!!! Learn to watch each race closely for horses in trouble or who showed  good effort but maybe are still coming up to their best race.\n\nALSO, GET INFORMATION: Read Steve Davidowitz's and Andrew Beyer's books ... these are two Harvard graduates who claim to have "earned" their way through college by beating the ponies.  There are plenty of other books that will shed light on the game ... but, again, it's like a job to be a consistent winner ... So while you're "attending school" do a lot of "testing" before you plunge in with your wallet open ... when you've gained some consistency with your "play dough" then perhaps you can set a budget that is realistically comfortable for you, and play in earnest (money management is probably one of most overlooked aspects of the handicapping game, but it is critical to have any success).  Oh! and while  you're "in school" enjoy the classroom and your fellow students, there's a whole lot of fun to be had just being in the environment!  All I can say is, I've joined a whole lotta students, screaming our lungs out as the horses entered the stretch and "MY" longshot is in front by a length and  a half and I've got all of $2 bet on the nose.\n\nGood luck, have fun, and always use good judgement!
105308
its easy to do\n\ntype your names down one column, and the teams across the top\n\nat the end where you want to total just put in the formula =sum(b2:E10)
912698
Do the game when the students can make one, two or three tally marks.  The object of the game is to make the last mark without going over.    The team gets a choice of going first, or choosing the number the game will go to.  \n\nThere is a pattern to the madness, but I want you to figure it out.
1060575
This should have what you need...nice drop down menu:\n\nhttp://racing.aqha.com/racing/handicapping/Stats.aspx
698606
Hmm, interesting question, actually, and a practical application of math.  Hey, all you folks who ask what this stuff is good for, here is a real-life situation where you can use it!\n\nGetting to the problem.  Obviously, the contestant to go last spins again if his score is lower than the best score.  If tied, there is a spin-off where his winning chance is 50%; so, if there is a tie, he should respin if he has 45 or less (giving him at least 11 out of 20 good numbers), not spin if he has 55 or more (giving him at most 9 out of 20 good numbers - better to go to the spin-off, where he has a 50% chance), while 50 is a "push".   \n\nNow, for the contestant to go 2nd to last: First, let's assume the first player busts (of course, the second player respins if his score is lower than the first player), which essentially makes this a two-player game.  Based on what number the second player stops on, this is the winnning percentage I calculate for the last player:\n\n100 - 4.875%\n95 - 14.375%\n90 - 23.375%\n85 - 31.875%\n80 - 39.875%\n75 - 47.375%\n70 - 54.375%\n65 - 60.875%  \n60 - 66.875%\n55 - 72.375%\n50 - 77.375%\n\nSo, you may think the second player should respin whenever his odds are under 50%, but that is actually incorrect.  For example, if the second player has 70, giving him a 45.625% chance of winning, respinning would give him a 70% chance of going bust - he is better off stopping with the 45.625% chance.  To determine whether or not respinning improves the second player's chances, a second calculation is needed that considers each of the 20 scores after the second spin - some of them bust the player but the others improve his chance of winning.  Based on that, here is a new table, giving the last player's chance of winning if the second player respins with the score given:   \n\n95 - 95.24375%\n90 - 90.9625%\n85 - 87.13125%\n80 - 83.725%\n75 - 80.71875%\n70 - 78.0875%\n65 - 75.80625%  \n60 - 73.85%\n55 - 72.19375%\n50 - 70.3125%\n45 - 69.68125%\n\nSo, comparing the two tables, the critical points are 55 and 60.  If the second player has 60, he should stop - the last player's chance of winning improves if the second player spins.  At 55, however, the second player should spin, improving his odds by about 0.2%.\n\n\nMoving on to the three player game... now it gets really messy.  The player to go first needs a score that can beat the other two players.  Does he stop on 55 again, hoping the other two players both bust?  He'll need a better score than that to have a 50% chance of winning, but, as was noted in the two-player analysis, stopping with a 30% chance can be better than making a second spin.  As for how to do a complete analysis, uh, beats me.  This may require a computer to do all the math.\n\n* * * * * *\n\nThinking about the three-player game, maybe it isn't as messy as I initially thought - the only additional consideration for the first player, as opposed to the second player, is that either player will stop if the first player's score is beaten.  This can be calculated from the first table.  For this table, listed is the chance of the first player losing (i.e. either player beats him) if he stands on the score given:\n\n100 - 9.51% \n95 - 26.68%\n90 - 41.29%\n85 - 53.59%\n80 - 63.85%\n75 - 72.31%\n70 - 79.18%\n65 - 84.69%\n\nSimilar to the two-player game, I can now calculate the odds if the first player decides to respin.  Since I’m getting a little tired crunching these numbers, for this last table I’ll only list the critical ones:\n\n70 - 83.4%\n65 - 82.3%\n\nIn other words, the first contestant should respin on 65, stop on 70.\n\nIn conclusion, I think the optimum strategies are.....\n\nFirst player: spin on 65, stop on 70\n\nSecond player: spin on 55, stop on 60 (assuming his score at least matches the first player)
1114869
Here's a formula on this webpage you can use