I recently vacationed on a cruise ship, and the ever-popular game Bingo was played. On one game they offered a $100,000 prize for achieving a cover-all by the 48th ball.
What are the odds of covering all the spaces on a Bingo card within the first 48 balls drawn? Show your work.
0.00000125093989261493
Or one in a hair under 800,000.
Can you show the explicit derivation, please.
Hmm, so if each card only costs 12 cents to play, the payoff isn’t too bad. Anything more than that, and it might be a sucker’s game…
(51 choose 24)/(75 choose 48) = 82861/66238993967
Essentially, there’s one way to choose the 24 numbers on your card and 51 choose 24 to pick 24 other balls versus 75 choose 48 ways to pick 48 balls.
Given that there were about 500 cards in play, the odds would be 1 in 1600 that the ship would have to cough up the $100K to someone.
Some of the cards were free while others were paid for by passengers, so this game was a givaway to the passengers - especially since there was still a smaller cash prize (~$500) for the eventual coverall. The $500 came from the passsengers who bought cards. I suppose the $100K would have come from the ship’s entertainment budget.
Each column is 5 out of 15 numbers. (In a blackout, order is irrelevant).
That’s 15!/(5! x 10!) where ! is factorial; or (15x14x13x12x11)/120 = 3003 … If I still remember my combinatorics. You divide by 120 because 1,2,3 is the same as 1,3,2, etc.
The number of ways to select 5 columns of 3003 -> 3003^5 = 2.442174 e17
The number of ways to select 48 out of 75 is (Again, order does not matter): 75!/(48! x 27!) = 183,535,527,771,815,939,300
According to my scientific calculator app.
Divide it out and I get 1 in 751, which can’t be right. Even worse of the middle is a freebie.
md2000, I think what you just calculated is the odds, given a sequence of numbers that could possibly be a coverall, that your card is one. The catch is that most sequences of numbers can’t be a coverall.