Odds Question - Bingo

Mustang · 04-14-2010, 11:20 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?

MJ4H · 04-14-2010, 11:28 AM

50 - 50 obviously

you fill it up in the first 50

or

you don't

Lathum · 04-14-2010, 11:30 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.

MJ4H · 04-14-2010, 11:31 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.

Lathum · 04-14-2010, 11:34 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, 11:37 AM

Quote:

Originally Posted by MJ4H

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, 01:39 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, 02:11 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, 02:15 PM

The answer is obvious.

2/3

Rizon · 04-14-2010, 02:21 PM

I'm only 33, can I view this thread? Or do I have to be 75?

MJ4H · 04-14-2010, 02:29 PM

Ran it 1 million times with no successes, so far.

path12 · 04-14-2010, 02:39 PM

I heard that if you double the number of cards you buy each round you can't lose.

Mustang · 04-14-2010, 03:00 PM

Quote:

Originally Posted by Pumpy Tudors

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, 03:04 PM

The odds are that you will lose more often than you will win.

Rizon · 04-14-2010, 03:07 PM

The possibility of successfully navigating an asteroid field is approximately 3,720 to 1.

Mizzou B-ball fan · 04-14-2010, 03:18 PM

Yahoo! Canada Answers - What are the odds of getting a blackout bingo (all numbers) before 51 balls are drawn?

JediKooter · 04-14-2010, 06:13 PM

Was his name-O...

Greyroofoo · 04-14-2010, 06:22 PM

I'm guessing

3.8791716274735488810706332426736e-20 percent

Pumpy Tudors · 04-14-2010, 06:25 PM

Quote:

Originally Posted by Mustang

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, 06:26 PM

Quote:

Originally Posted by MJ4H

Ran it 1 million times with no successes, so far.

ALSO I SAID A BILLION

Greyroofoo · 04-14-2010, 06:33 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, 06:39 PM

Quote:

Originally Posted by Greyroofoo

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, 06:40 PM

Quote:

Originally Posted by Pumpy Tudors

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, 06:48 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, 06:57 PM

I come up with...

1 in 212,085.4297...

Pumpy Tudors · 04-23-2010, 05:43 PM

WHAT IS THE ANSWER

04-14-2010, 11:31 AM	#4
MJ4H Coordinator Join Date: Jan 2002 Location: Hog Country	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. Last edited by MJ4H : 04-14-2010 at 11:32 AM.

04-14-2010, 01:39 PM	#7
Toddzilla Pro Starter Join Date: Jan 2001 Location: Burke, VA	(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! Last edited by Toddzilla : 04-14-2010 at 01:40 PM.

04-14-2010, 02:11 PM	#8
Pumpy Tudors Bounty Hunter Join Date: Oct 2000 Location: Pittsburgh, PA	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! __________________ No, I am not Batman, and I will not repair your food processor.

04-14-2010, 02:15 PM	#9
Marmel Grizzled Veteran Join Date: Oct 2000 Location: Manchester, CT	The answer is obvious. 2/3 __________________ 81-78 Cincinnati basketball writer P. Daugherty, "Connor Barwin playing several minutes against Syracuse is like kids with slingshots taking down Caesar's legions."

04-14-2010, 02:39 PM	#12
path12 Coordinator Join Date: Feb 2003 Location: Seattle, WA	I heard that if you double the number of cards you buy each round you can't lose. __________________ We have always been at war with Eastasia.

04-14-2010, 11:28 AM	#2
MJ4H Coordinator Join Date: Jan 2002 Location: Hog Country	50 - 50 obviously you fill it up in the first 50 or you don't

04-14-2010, 11:30 AM	#3
Lathum Favored Bitch #1 Join Date: Dec 2001 Location: homeless in NJ	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.

04-14-2010, 11:34 AM	#5
Lathum Favored Bitch #1 Join Date: Dec 2001 Location: homeless in NJ	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

04-14-2010, 02:29 PM	#11
MJ4H Coordinator Join Date: Jan 2002 Location: Hog Country	Ran it 1 million times with no successes, so far.

04-14-2010, 03:04 PM	#14
DanGarion Coordinator Join Date: Nov 2003 Location: The Great Northwest	The odds are that you will lose more often than you will win. __________________ Los Angeles Dodgers Check out the FOFC Groups on Facebook! and Reddit! DON'T REPORT ME BRO!

04-14-2010, 03:18 PM	#16
Mizzou B-ball fan General Manager Join Date: Aug 2001 Location: Kansas City, MO	Yahoo! Canada Answers - What are the odds of getting a blackout bingo (all numbers) before 51 balls are drawn?

04-14-2010, 06:13 PM	#17
JediKooter Coordinator Join Date: Dec 2004 Location: San Diego via Sausalito via San Jose via San Diego	Was his name-O... __________________ I'm no longer a Chargers fan, they are dead to me Coming this summer to a movie theater near you: The Adventures of Jedikooter: Part 4

04-14-2010, 06:22 PM	#18
Greyroofoo Pro Starter Join Date: Nov 2003 Location: Alabama	I'm guessing 3.8791716274735488810706332426736e-20 percent

04-14-2010, 06:33 PM	#21
Greyroofoo Pro Starter Join Date: Nov 2003 Location: Alabama	I think the odds can be figured by doing (24/75)(23/74)(22/73) all the way down to (1/52)

04-14-2010, 06:48 PM	#24
EagleFan Hall Of Famer Join Date: Nov 2000 Location: Mays Landing, NJ USA	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.

04-14-2010, 06:57 PM	#25
EagleFan Hall Of Famer Join Date: Nov 2000 Location: Mays Landing, NJ USA	I come up with... 1 in 212,085.4297...

04-23-2010, 05:43 PM	#26
Pumpy Tudors Bounty Hunter Join Date: Oct 2000 Location: Pittsburgh, PA	WHAT IS THE ANSWER __________________ No, I am not Batman, and I will not repair your food processor.

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