My kid's school PTA had a bingo event to raise money Friday night. I volunteered to call numbers. In my first game, I went 16 numbers without getting an I (for bingo novices, there are 75 letters, divided into 5 letter groups; 1-15 ar B, 16-30 are I, etc.) So I wondered what the odds of that are.
Mathematically inclined readers may want to pause here to work this out for themselves, if they haven't already figured it out.
You might think the odds would be (60/75)^n where n is the number of balls drawn. This would be correct if balls were replaced during the game, but once a ball is drawn it is out for the remainder of the game.
So the odds are (60/75)(59/74)...depending on how many numbers are drawn.
This can be expressed using the factorial function where 5!=5x4x3x2x1 as
odds = (60!/(60-n)!)/(75!/(75-n)!). Plugging in 16 (or setting up a spreadsheet so you can see how the odds change with varying n) I get about a 1.75% chance, which is unusual but not exactly shocking; even at 20 draws you have a >0.5% or 1 in 200 chance of this happening (and this doesn't take into account that there are 5 groupings of numbers)
I also calculated the odds while replacing balls and at lower numbers it doesn't change the odds as much as I'd have guessed: 2.8% and 1.15% at 16 and 20 draws, respectively.
More info on this kind of problem here. Nerdy readers may wish to work out the problem of what the odds of completely filling there bingo cards (24 spaces+free space) for a given number of draws.
One funny note. My second game calling "four corners" I mistakenly called out the I, N, G numbers not realizing those were of no use. I was quickly reprimanded by a grandmother looking type who obviously had a lot more bingo experience than I.
UPDATE: fixed formulas (removing errant 1- at start) per Ollies comments. Note to self: you know math professor is regular reader; proofread future math posts closely