i want a free programe that can help choose lotto results, i want to get all the combinations of 1 to 49.
127465
If you throw dice and get a 6 three times in a row, the odds of it getting another 6 the fourth time are still 1/6, as good or as bad as for any other number. The history has nothing to do with it. Each throw of a dice or lottery draw is a series of random events, the outcome of which is not influenced by previous outcomes.\n\nAs the British magazine The Economist once wrote, lotteries are a voluntary tax on being bad at math ;-)
196605
There is also the superfecta, pick4, pick3 (best value) & the trifecta key ( bad bet though, you're better off boxing the horses). Hope this was helpful.
137686
its easy. figure out in this way there are 3 digits and since its an odd number the one's place or the last digit can be 1,3,7,9(5 is excluded).the no. of chances are 4. for the second digit the digits can be 0 to 9 leaving 5 so no. of chances are 9. for the first digit 0 and 5 are not possible and the no. of chances are 8.\n\nmultiply the no. of chances for every digit it ll give u the answer i.e 8*9*4 = 288. so u have 288 different numbers.
142595
Are you sure you want them *completely* random? Is it acceptable to have, say, 941 + 1 + 1 +... ?\n\nIf so, do something like (in pseudolanguage, my c++ is rusty ;-) :\n\n\nint sum = 1000; //or whatever\nArray allNumbers = EMPTY;\n\nfor (int i = 59; i > 0; i++) {\n  number = Random-between(1, sum - i);\n  push allNumbers, number;\n  sum -= number;\n}\n\n\nSo the first time it gets a result between 1 and 941, say 100.\nNext round, a number between 1 and 842...\nUntil it gets a number between 1 and at-least-1 minus 0.\n\nIf it picked 941 on the first round, say, (sum - i) would be (59 - 58), so a number between 1 and 1 all the way.\n\nThat should do the trick.
56164
Here is some code. Most of it is from the Microsoft help file:\n\n\n#include <stdlib.h>\n#include <stdio.h>\n#include <time.h>\n\nvoid main( void )\n{\n   int i;\n   int random1or2\n   /* Seed the random-number generator with current time so that\n    * the numbers will be different every time we run.\n    */\n   srand( (unsigned)time( NULL ) );\n\n   /* Display 10 numbers. */\n   for( i = 0;   i < 1000; i++ )\n   {\n      random1or2 = (rand() % 2) + 1;\n      printf( "%d&#92;n", random1or2);\n   }\n   return;\n}
40275
http://www.mycardmaker.com/
126071
if the order you draw them is important its a permutation.\n\nthe answer is basically; a 1 in 55 chance for the first card, a 1 in 54 chance for the second card (you've taken the first card out) and a 1 in 53 chance of getting the final card. all three events are independant of eachother. so the answer is 55 x 54 x 53 = 157410\nso 1 in 157410 chance.\n\nif the order you draw them is unimportant them its a combination.\nfor the first card there is a 3 in 55 chance\nthe second card has a 2 in 54 chance\nand the final card has a 1 in 53 chance\nthe answer is: (3/55)  x (2/54) x (1/53) = 1 in 8745
136914
You are dealing with probabilities here and your odds are going to be exponential. But here is what I have for you.\n\nOdds in rolling a:\n4 = 1/12\n5 = 1/9\n6 = 5/36\n7 or 11 = 2/9\n8 = 5/36\n9 = 1/9\n10 = 1/12\n\nNow for each roll after that you would take the odds and times it to the "n" power, where "n" is the number of the roll. So to roll any of these numbers 12 times in a row would mean taking the fraction to the 12th power. Below are those answers. (Hope you understand scientific notation.)\n\n(1/12)^12 = 1.13876838470415E+55\n(1/9)^12  = 1.13771030934265E+55\n(5/36)^12 = 5.1524055910299E-11\n(2/9)^12  = 1.14868734219891E+55\n\nIn other words you would be more likely to win the lottery several times before this would happen. Hope this helps.
462126
how to registar to vote\nhow to complete a job application or financial aid application\nhow to play tic-tac-toe, checkers, poker...\nhow to use google\nhow to parallel park
521965
There are 76,275,360 different possible tickets. Therefore if you buy one ticket your chances are 1 in 76275360. If you buy two tickets your chances are 2 /76275360 or 1/38137680.\n\nIf you buy 76275360 (different) tickets then your chances are 76275360/76275360 or 1! Hence you will definitely win.\n\n(Your odds are halved by doubling the amount of tickets you buy as long as each has different number combinations)
430044
Heres a program link. Its free too.
659562
This does not require hiring a professional, it's a rather simple & straightforward program. \n\nRaven put you on the right track, use the modulo (%) operator to break out the thousands from the hundreds, the hundreds from the tens, etc.  Define an array or map of strings with the words for each number up to twenty, then thirty, forty, fifty.. to ninety, then hundred, thousand.  Break the number apart, and get the appropriate words from the array & string them together.
4288
Easy! Just predict next week's lottery numbers...\nPS let me know too...
400825
http://www.funquizcards.com
474552
http://www.pyzam.com/countup.html\n\nhere ya go!!
31877
Here are some code snippets that might be useful.\nNote: I wrote these down, did _not_ compile them, or test them.\n\nThe deck of cards is a array of 52 unsigned integers,\nThey can take a value of 0-52.\n\n//Create an array to hold the deck:\n\nunsigned int Deck[52];\n\nfor(unsigned int i = 0; 1 < 52; i++)\n{\n   Deck[i] = i;\n}\n\n// Shuffle the deck\nsrand(); // Initialize the random number generator\n\nunsigned int temp;\n\n// The shuffle works this way. First round, every card gets a chance\n// to be in slot 51. Second round, each of the remaining cards gets a \n// chance to be in slot 50, third round etc.\n\n\n// Note: This method allows an array value to be swapped with itself, i.e. not moved, which is what you want to get all possible shufflings of the deck.\nfor(unsigned int j = 52; 0< j; j--)\n{\n   // Generate an index in the range 0-j\n   unsigned int k = rand( void ) % (j);\n\n   temp = Deck[k];\n   Deck[k] = Deck[j-1];\n   Deck[j-1] = temp;\n}\n\n///////////////////////////////////\n\nunsigned int GetFaceValueOfCard(unsigned int n)\n{\n  return Deck[n] % 13 + 1;    // Returns a value 1-13\n}\n\n// If you want to get fancy:\n\nchar GetSuitOfCard(unsigned int n)\n{\n   char Suit[4] = "HDCS";\n   return Suit[n/13];\n}
464574
The odds of getting all six are 13,983,816 to 1\n\nNote: the above answer forgets to divide by through the possible combinations of the six numbers. \n\nIf we multiply the numbers 49 x 48 x 47 x 46 x 45 x 44 we get 10,068,347,520. However, each possible group of six numbers (combination) can be drawn in different ways depending on which number in the group was drawn first, which was drawn second, and so on. There are 6 choices for the first, 5 for the second, 4 for the third, 3 for the fourth, 2 for the fifth, and 1 for the sixth. Multiply these numbers out to arrive at 6 x 5 x 4 x 3 x 2 x 1 = 720. We then need to divide 10,068,347,520 by 720 to arrive at the figure 13,983,816 as the number of different groups of six numbers.\n\nEdit: Yes, well the above answer has now changed their response. Well done and all that.