Blackout Probability in 48 numbers
 

What is the probability of getting a blackout in bingo in 48 numbers or less? Can you then multiply that percentage by the number of cards playing to find out the likelihood of blackout when there are 300 cards playing that game?

Re: Blackout Probability in 48 numbers
 

Hiya,

I'm not familiar with bingo, so I can't answer your first question. However, the answer to the second question is 'no'. Think of it this way...the odds of rolling a '6' on a die is 1/6. If you roll 6 dice, are the odds of getting at least one '6' 6*1/6?

PP

Re: Blackout Probability in 48 numbers
 

Bingo has a 90 number board.
Each player is given a card with 15 numbers on it.
New numbers are drawn at random from 1-90 (with no replacement, so the longest possible is 90 draws) until a player has ticked off all their numbers, the first player to do so is the winner.
all I know is that 48 is hard.
A room with 200 or so people tends to get a winner at somewhere around 52-55 I believe (totally from old memories)

Re: Blackout Probability in 48 numbers
 

Not to say that you're wrong... but the Bingo I played years ago had a board with 25 numbers on it (5 columns with 5 numbers each). The columns were labeled B, I, N, G, and O. And there were 75 possible numbers, not 90.

Another restriction is that each column on your board was restricted to a set of 15 numbers... So, in the B column, you could only have numbers 1-15, I column, 16-30, etc. Therefore, the number of possible bingo cards is not 75c25, but rather 15c5 * 5 (for blackout purposes)... for non-blackout purposes, you'd have 15p5 * 5.

This is what I remember, anyway... I'm no bingo expert, and the game undoubtedly has tons of variations.

CORRECTION: Re: Blackout Probability in 48 numbers
 

Not 15c5 * 5, but (15c5)^5... and (15p5)^5. My mistake.