View Full Version : Odds Question - Bingo
Mustang
04-14-2010, 10:20 AM
Probably should be Ping: Quiksand.
Someone at work was talking about playing bingo at the casino and how someone won $111K filling up their sheet in the first 50 numbers called.
For standard bingo (75 numbers that can be called) and 24 numbers on your card (and assuming random distribution for both.. no shenanigans), what are the odds of filling up your card with the first 50 numbers called?
50 - 50 obviously
you fill it up in the first 50
or
you don't
Lathum
04-14-2010, 10:30 AM
wait?
Hitting 50 numbers IN A ROW?
there is no way. I have no idea what the odds are but they have to be so beyond astronomical there is no way it could happen.
I think he means hitting 24 in the first 50. Your card has 24 numbers on it.
edit: should be trivial to calculate if you know how many numbers exist that aren't on your card, which I don't.
Lathum
04-14-2010, 10:34 AM
ahh, still, has to be pretty astronomical odds.
For a frame of reference the top prize in Keno is if you hit ten of the 20 numbers called. The odds od that happening are 1/8,911,711.2
lerriuqs
04-14-2010, 10:37 AM
I think he means hitting 24 in the first 50. Your card has 24 numbers on it.
edit: should be trivial to calculate if you know how many numbers exist that aren't on your card, which I don't.
Every number on the card is different so you'd have 26 wrong numbers come up in the first 50 and 51 numbers out of the 75 that aren't on the card at all...Don't remember my probability stuff well enough to do the calculation though...
Toddzilla
04-14-2010, 12:39 PM
(1) The odds of hitting the first number are 24 in 75.
(2)If you hit that, the odds of hitting the second number are 23 in 74 - if you miss the first number, the odds are 24 in 74.
(3) ? ? ?
(4) Profit!
Pumpy Tudors
04-14-2010, 01:11 PM
Why doesn't one of our FOFC programmers just use the brute force method on this?
1) Select 24 numbers from 1-75 at random.
2) Select 50 numbers from 1-75 at random.
3) If all the numbers from set 1 are in set 2, call it a "yes". If not, call it "no".
4) Run it a billion times and see how many times you get "yes".
Get to work!
Marmel
04-14-2010, 01:15 PM
The answer is obvious.
2/3
Rizon
04-14-2010, 01:21 PM
I'm only 33, can I view this thread? Or do I have to be 75?
Ran it 1 million times with no successes, so far.
path12
04-14-2010, 01:39 PM
I heard that if you double the number of cards you buy each round you can't lose.
Mustang
04-14-2010, 02:00 PM
1) Select 24 numbers from 1-75 at random.
Actually, you would need 5 numbers from each block of 15 (1-15, 16-30, etc)
DanGarion
04-14-2010, 02:04 PM
The odds are that you will lose more often than you will win.
Rizon
04-14-2010, 02:07 PM
The possibility of successfully navigating an asteroid field is approximately 3,720 to 1.
Mizzou B-ball fan
04-14-2010, 02:18 PM
Yahoo! Canada Answers - What are the odds of getting a blackout bingo (all numbers) before 51 balls are drawn? (http://ca.answers.yahoo.com/question/index?qid=20100405001750AA6wCHi)
JediKooter
04-14-2010, 05:13 PM
Was his name-O...
Greyroofoo
04-14-2010, 05:22 PM
I'm guessing
3.8791716274735488810706332426736e-20 percent
Pumpy Tudors
04-14-2010, 05:25 PM
Actually, you would need 5 numbers from each block of 15 (1-15, 16-30, etc)
I considered this at first, but for the purposes of figuring out the probability, does it really matter where the numbers are? Someone can (and will) correct me if I'm wrong, but I'd think that probability neither knows nor cares what the numbers are. We're just taking a set of numbers and finding out how often they come up in a bigger set. Whether that set is 5 numbers from blocks of 15 or just the numbers 1-24 in order, I don't think it makes a difference.
Pumpy Tudors
04-14-2010, 05:26 PM
Ran it 1 million times with no successes, so far.
ALSO I SAID A BILLION
Greyroofoo
04-14-2010, 05:33 PM
I think the odds can be figured by doing (24/75)*(23/74)*(22/73) all the way down to (1/52)
Pumpy Tudors
04-14-2010, 05:39 PM
I think the odds can be figured by doing (24/75)*(23/74)*(22/73) all the way down to (1/52)
But doesn't this assume that it's matching your 24 numbers on the first 24 pulls? What if only the first pull matches and then you have 26 misses in a row? Doesn't that change this to (24/75)*(23/74)*(23/73)*(23/72)...?
Greyroofoo
04-14-2010, 05:40 PM
But doesn't this assume that it's matching your 24 numbers on the first 24 pulls? What if only the first pull matches and then you have 26 misses in a row? Doesn't that change this to (24/75)*(23/74)*(23/73)*(23/72)...?
whoops! didn't fully read the question...
EagleFan
04-14-2010, 05:48 PM
Why not work it backwards to see what the odds are of having 25 drawings without getting a number? That is basically what those final 25 numbers would represent.
EagleFan
04-14-2010, 05:57 PM
I come up with...
1 in 212,085.4297...
Pumpy Tudors
04-23-2010, 04:43 PM
WHAT IS THE ANSWER
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