There are 36 numbers 6 different num form 1 bet, what's the min bet to win 3 num at least?
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.
629548
3. Each get one third of the prize. Not fair
88852
yes ther has to be a minumum number of 3
901297
1/54 (for 6-sided dice of course)\nThere are 6*6*6 = 216 possible outcomes, and\n1,1,1\n1,1,2\n1,2,1\n2,1,1\nare the only outcomes with sum<5, so 4/216 = 1/54
598521
What were the odds?\n\n4 to 1?\n\n8 to 1?\n\n20 to 1?\n\n**********************************\nEDIT\n\nIf the odds were 6 to 1, then you won $120.
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.
276877
What are the next winning lotto numbers?
263402
i dont know but that is only 6 numbers!
681817
The first dice can show any one of the six numbers: 1,2,3,4,5,6. So the total numbers of outcomes: 6\nThe second dice can show any one of the six numbers: 1,2,3,4,5,6. So the total numbers of outcomes: 6.\n\nSince both the dice are thrown together we can have 6*6 number of outcomes : 36.\n<<i.e 1 on dice A and any one of six on dice B = 6 possibilities, repeated for the numbers on dice A:: 6*6=36.)
864738
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.
682138
The one about the 3 bets is correct
660290
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56.\n\nThe winning combo is amongst the nice hot lucky winning numbers I listed above. Good luck!!!
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
600596
There are 6 possible outcomes for the roll. There is only one outcome when the number is 4. Therfore the chances of getting a 4 are 1/6.
239442
You're picking the wrong numbers. listen carefully, i'll tell you the winning numbers for last week 4-23-26-31-40-51.\nNo just joking, its a good idea to have three of the numbers to be even numbers and three to be odd numbers, these hit almost all the time, other than that it's all luck.
276262
You just have to list all the possible answers until you get 6 (your lowest number) in a row without skipping one.  After that you could get the next six by adding the 6 to those six.\n\nYou can get:  6, 11, 17, 18, 22, 23, 24, 28, 29, 30, 32, 33, 34, 35, 36, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55.  There are 6 in a row.  So the highest you CAN'T get was 49.\n\nIt only works if the two numbers are relatively prime (they have no common factors besides 1).  The formula for finding the number is:\n\n(higher number) times (lower number minus 1) minus lower number\n\nHere that is 11(6-1) - 6 = 11(5) - 6 = 55-6 = 49
745507
diidy, how could the odds get worse when you buy more tickets?\n\nI think there are actually two answers, depending on whether each ticket is independent of the others or not.\n\nIf each ticket is independent, and there's no guarantee as to how many tickets there are or how many winners there will be, then think of it this way.  The odds of any ticket NOT winning are 299:300.  Therefore the odds of NO ticket winning are 299^n:300^n, where n is the number of tickets.  The odds of at least one ticket winning are 1 - the odds of no ticket winning, or 1 - 299^n:300^n, or (300^n - 299^n):300^n.\n\nIf there are 300 tickets of which 1 is a winner, then the odds of winning are n:300.