I'm trying to do what the title says, and I've figured it out for 4 and 5 terms, checking my answers with this database but when I try 6 I get over double the correct answer. The rules are standard american rules, 75 numbers, free space in the middle, and the B column can have numbers 1-15, I can have 16-30, and so on.
For 4 terms the formula I derived was $$ \frac{{16*3*2*1}}{\frac{75!}{71!}} $$ The 16 came from the 16 possible squares the first number could be in, 3 is the squares the second number can be in, because once you know the first number, you know what has to be your bingo, and then 2 and 1 for the last two numbers. This gives the correct answer of 1.69 x 10^-6.
Getting bingo on the 5th term I used the formula $$ (\frac{{16*3*2*1}}{\frac{75!}{71!}})*4 + (\frac{{16*8*3*2*1}}{\frac{75!}{70!}}) + (\frac{{8*4*3*2*1}}{\frac{75!}{70!}}) $$ The first term is the same as the previous, except I multiplied it by 4, because since there is 5 numbers total now, one number could be anything not including the 4 in the final solution. Therefore there would be 4 different ways to put in the extra number. The extra number cannot go at the end, or else that would mean that they won in 4 turns. The second term represents the probability of getting a 5 in a row, not including when the BINGO starts on a value in the center cross-section. There are 16 possible initial squares, after selecting a cell, there are now 2 rows that can be picked, the horizontal row, or the vertical row. Therefore there are now 8 new cells that will still give possibility to getting a BINGO. After the second number is selected, there are now 3 possible numbers left to choose, so 3 x 2 x 1. The third term represents the probability of getting a 5 in a row when the first value is in the center cross-section. There are 8 cells that fit this criteria, after 1 is selected, there are 4 cells left, then 3 x 2 x 1. This gives the correct answer, which is 1.69 x 10^-5.
Now the problem appears. My logic for the next step is the same formula, except now since there are 6 numbers I multiply all the terms by 5, since there is now the same ways of getting 5 in a row, but now there is an extra number. The extra number can go in any order but the last, giving 5 slots it can go into. However, this answer is wrong, but I don't understand what's wrong with my reasoning.
If something didn't make sense please let me know, Thanks
Edit: Here's a picture of a bingo board. By center cross section I mean the "N" column and the third row.
