i have already tried to find the number of bingo cards and have settled with infinite, definitely established that each column has 3003 combinations and 360360 permutations i will ad that when talking odds for a blackout then the combinations are necessary which leads me to think that then to really find all permutations where it concerns the possibility of the number of cards where a single line across is needed for a win then 360360 must also be calculated again with each column for permutations making much more greater possibility’s then a septillion ect well its just one massive bigg mother of a number . but in my curiosity i realised that although my numbers are infinite the odds of getting the right numbers from the ball drops is much cooler i am playing a game where 47 numbers are called there is 75 numbers 15 in each column this means that 9 should be called in each of them and two numbers extra that make two columns have 10 each you need 5 numbers from each of the columns to get a black out (this is every number on the card ) and from the nine numbers that will be called from your column this means that 4 can be wrong this way there is still 5 left to cover every number you need. i will add that you need to consider that there is 6 numbers that would probably not be called out of fifteen and the odds that one of your five numbers that needs to be called could be one of those 6 outa 15 numbers is fairly good as well. :rolleyes:
i guess i am saying that 15 P 5 where there is 3003 possible combinations of numbers and 360360 permutations , there is 9 P 5 or 10 P 5 numbers that would be called out
this then 9/5 = 9-5= 9/4 9x8x7x6x5 = 15120 n/k 1x2x3x4x5 =120 = 126 possible number combinations that will be called out or 252 if there is ten.
well it seems the odds have greatly increased seens that of my infinite amount of number possibility that there is only a different possibility of between 126 and 252 different combinations… if your allowed to have four errors in each column 4/126 or 4/252 is odds worth considering off course you still have to consider that the combinations could consist of a combination that consists of one or even more then 4 of the 6 possible number that would be left out as well… hmm i think my skills in maths are getting tested …
6/15 missing numbers are errors 4 errors outa 9 are tough chances the final odds are multiplied by the 5 columns CAN ANYONE HELP !!! 40 % then 40 % of that then probably divided by 5 like 3.2 % give or take… its heaps better then infinite…:dubious::smack:
the odds could also consider 15 P 9 and 15 P 10 i can see this truly i am curious about the 126 to 252 variability and how the variability changes the odds between 10 numbers and 9 numbers being called being that 10 represents 252 and 9 represents 126 right then 6 /15 per 9 numbers and 6/15 or 6/14 hence there is now ten numbers called odds become variable or 6/15 per 126 and 6/14 per 252 these odds as a percentage is say 40 % of 126 an 45% 252 gives 50 combinations per 9 numbers called and 120 per 10 numbers called noting this is for errors where for a win the odds are 9/15 per 126 and 10/15 per 252 this is 60% and 66% 72 and 165 combos for winning does this mean that winning is more probable isn’t that so cool that from a possible amount of infinite to odds that are higher for a hit on one of them then you are at missing at…
1 / infinite odds …
My head hurts.
If you are trying to determine the probability of having all 24 numbers on your card called in the first 47 drawn, you don’t have to bother with the columns. Think of it this way: color your 24 numbers’ balls red, and the other 51 white. Pull out 47 balls; you win if you pull out 24 red ones and 23 white ones. There are (75)C(47) combinations of 47 balls, (24)C(24) = 1 combination of 24 red balls, and (51)C(23) combinations of white balls, so the probability of winning is (51)C(23) / (75)C(47), or about 1 / 1,598,798. Note that this includes wins in 24 through 47 balls, and not just for 47.
If you are interested in the number of different cards possible, there are (15)P(5) ways to have the 5 B numbers, the same for the Is, Gs, and Os, and (15)P(4) for the Ns, so it’s (15)P(5)[sup]4[/sup] x (15)P(4) - but then you divide by 2 as every card has the same 12 winning results if you turn it upside-down (any other arrangement of the columns changes the numbers needed in the two diagonals), so the total is somewhere around 2.35946 x 10[sup]77[/sup].