Bingo Blackout

[LondonBroil]

A friend asked me today what the odds of getting a bingo blackout (all 24 numbers + free space) covered when 48 numbers have been called out. I looked it up and found 0.00000125 which I assume is 800,000-1. What formula do they use to get that?

[BruceZ]

[ QUOTE ]
A friend asked me today what the odds of getting a bingo blackout (all 24 numbers + free space) covered when 48 numbers have been called out. I looked it up and found 0.00000125 which I assume is 800,000-1. What formula do they use to get that?

[/ QUOTE ]

You should really give us all the rules when you ask a question, especially about a game like bingo for which those smart enough to do the math are not likely to be very familiar with the game. Based on the info you provided, I would have to deduce that the rules of the game are that there are a total of 75 differently numbered balls, from which 48 are drawn without replacement, and you have 24 different numbers on your card in addition to one that you get for free. If this is right, then the probability of filling your card is:

C(75-24,24) / C(75,48) =~ 0.00000125 or 799,398-to-1.

That is, the total number of ways to draw the 48 numbers out of 75 is C(75,48). If your 24 are drawn, there would be 24 balls to be drawn out of the remaining 75-24, so there are C(75-24,24) ways to do this. As always, C(n,k) = n!/(n-k)!/k! = n*(n-1)*(n-2)*...*(n-k+1) / [k*(k-1)*(k-2)*...*1], where the numerator has k terms.

Let me know if I got the rules right. I suppose that deducing the rules from the odds made the problem more interesting. [img]/images/graemlins/wink.gif[/img]

[LondonBroil]

You got them exactly right. Thanks for explaining the math. I could never grasp the whole 'Choose' concept but you did a very nice job.

[Siegmund]

If we want to nitpick... p(N)= C(75-24,N-24) / C(75,N) is the chance of getting a blackout in N *or fewer* draws.

The chance of completing the blackout on the Nth draw can be written a few different ways, but p(N)-p(N-1) is the easiest.

[BruceZ]

[ QUOTE ]
If we want to nitpick... p(N)= C(75-24,N-24) / C(75,N) is the chance of getting a blackout in N *or fewer* draws.

The chance of completing the blackout on the Nth draw can be written a few different ways, but p(N)-p(N-1) is the easiest.

[/ QUOTE ]

I'll go you one easier (to compute). P(N)*24/N. In our case this is P(48)/2.

p(48) - p(47) = p(48)/2 exactly. That's because when all 24 of our numbers are drawn, exactly half of the 48 numbers drawn are on the card, so the probability that the last number drawn is on the card is 1/2, so we get our last number ON the 48th draw exactly half of the times that we get all our numbers BY the 48th draw. For general N > 24, the probability of getting it on the Nth draw is

p(N) - p(N-1) = p(N)*24/N.

The 800,000-to-1 (approx.) that he asked about corresponds to making it in 48 or fewer draws, so that is what I calculated. The probability of making it on exactly the 48th draw is exactly half of this or about 1.6 million-to-1.

[BogusPomp]

As a follow-up, let's say we are playing 3 different bingo cards during a "blackout" game.

What is the probability that 2 of these cards will BINGO on the same number?

(Assume we keep drawing numbers w/o replacement regardless if another BINGO has already been called)

[BogusPomp]

[ QUOTE ]
As a follow-up, let's say we are playing 3 different bingo cards during a "blackout" game.

What is the probability that 2 of these cards will BINGO on the same number?

(Assume we keep drawing numbers w/o replacement regardless if another BINGO has already been called)

[/ QUOTE ]


Apologies for bumping this ... but I want to take a stab at this one.

Can I use Excel BINOMDIST(2,3,p(n),false), where p(n) is the probability of getting a blackout on 'exactly' the nth ball, then take the sum of all n's....= 43.389% ???

Am I on the right track? Thanks in adv