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Need some help calculating the statistical odds of something happening (1 Viewer)

Horses Mouth

Horses Mouth

Footballguy
At work we play credit card roulette when we eat out.

Whoever's card is pulled out of the hat pays.

There are 8 of us.

I have paid 3 out of the last 4 times. 

My understanding is that the odds would be: 1/8 x 1/8 x 1/8 but then I get lost. 

How does not hitting on the fourth time affect the odds?

I'd like to figure out how lucky I am!

ETA: or another way of putting it, what are the odds of hitting 3 out of 4 times. 

 
Last edited by a moderator:
Horses Mouth said:
At work we play credit card roulette when we eat out.

Whoever's card is pulled out of the hat pays.

There are 8 of us.

I have paid 3 out of the last 4 times. 

My understanding is that the odds would be: 1/8 x 1/8 x 1/8 but then I get lost. 

How does not hitting on the fourth time affect the odds?

I'd like to figure out how lucky I am!
Click to expand...
The odds are 12.5% (or 1/8) every time you go out to eat.  The odds don't change and are not affected by previous outcomes.   Just consider yourself unlucky early because you'll be much closer to paying 12 times after you've gone out a total of 100.  Unless your friends are cheating.

 
There are four ways to get three pays, one no pay:  the no payncan happen first, second, third or fourth.

Each of these four ways has a (1/8)^3 * (7/8) probability, as described by others.  So multiply that by four.

This gives you a 7/1024 chance, or if you prefer your odds based off of ones, about 1 in 146 such cases of four independent visits.

Not absurdly rare.  The odds of any one of the eight (not just you specifically) paying three times out of four aren't that odd.  Just sucks to be the one.

 
Arodin said:
There are four ways to get three pays, one no pay:  the no payncan happen first, second, third or fourth.

Each of these four ways has a (1/8)^3 * (7/8) probability, as described by others.  So multiply that by four.

This gives you a 7/1024 chance, or if you prefer your odds based off of ones, about 1 in 146 such cases of four independent visits.

Not absurdly rare.  The odds of any one of the eight (not just you specifically) paying three times out of four aren't that odd.  Just sucks to be the one.
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agreed. it's not super rare. a hair under 2% chance. 

 
There was a guy here that would respond to questions like this with a "50/50 chance.  Either it happens or it doesn't".  That dude was a brilliant statistician.

 
Arodin said:
There are four ways to get three pays, one no pay:  the no payncan happen first, second, third or fourth.

Each of these four ways has a (1/8)^3 * (7/8) probability, as described by others.  So multiply that by four.

This gives you a 7/1024 chance, or if you prefer your odds based off of ones, about 1 in 146 such cases of four independent visits.

Not absurdly rare.  The odds of any one of the eight (not just you specifically) paying three times out of four aren't that odd.  Just sucks to be the one.
Click to expand...


Horses Mouth said:
agreed. it's not super rare. a hair under 2% chance. 
Click to expand...
Where in the world are you getting just under 2%?

And how are you guys both agreeing on something that actually a less than 1% chance of happening, not being "super rare"?  That's the definition of rare.  If that isn't a serious long shot, what is?

 
Rodrigo Duterte said:
Where in the world are you getting just under 2%?

And how are you guys both agreeing on something that actually a less than 1% chance of happening, not being "super rare"?  That's the definition of rare.  If that isn't a serious long shot, what is?
Click to expand...
Paying for lunch 5 out of 5 days? :shrug:

 
Isn't this the same as betting on red or black in roulette? Just because red comes up 5 times in a row, the odds of it coming up red again are the same. The laws of probability are what make people bet the opposite. It would be the odds of red hitting 6 times in a row instead of just the 50/50 chance of red or black. (simplifying even though there is also the option of zero hitting)

 
KCitons said:
Isn't this the same as betting on red or black in roulette? Just because red comes up 5 times in a row, the odds of it coming up red again are the same. The laws of probability are what make people bet the opposite. It would be the odds of red hitting 6 times in a row instead of just the 50/50 chance of red or black. (simplifying even though there is also the option of zero hitting)
Click to expand...
He's asking what the odds were against something that already occurred. His odds of getting stuck with the check again tomorrow are 1 in 8, even though he's already lost 3 times in 4 days and even though the odds of losing 4 times in 5 days are slim. That's no longer relevant. 

 
1/32

Basing this on 100% probability that someone's card is pulled at the first event.  The chance that same person goes on to pay 2 of the next 3 times is 1/8*(1/8+1/8) or 1/8*1/4.

The odds of YOU getting the short straw 3 of the next 4 times are 1/256.  (1/8)*(1/8)*(1/4).

 
Last edited by a moderator:
Is the same person always pulling the card out? Maybe there is something unique about your card that they can tell its your card.  :shrug:  

 
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