Simple probablity question

[deftron]

Say you have a 1 in 1000 chance of something occuring

And you repeat it 300 times (Each time 1 in 1000 chance)

Is the overall chance (after try 300) of it happening 30%
or different ?


30% sounds right to me ...

but if the chance was 1 in 10 and you repeated it 10 times
you aren't going to have a 100% overall chance

 

[chuckywang]

No. The overall chance of it happening is 1-(999/1000)^300.

 

[chuckywang]

About 25.93%.

 

[deftron]

What would be the 10 tries (1 in 10 chance) percentage ?

 

[chuckywang]

Originally posted by: deftron
What would be the 10 tries (1 in 10 chance) percentage ?

1 - (9/10)^10

which is about 65.13%.

 

[deftron]

Thanks. Get it now.

 

[HomeBrewerDude]

you not specific enough.

the overall chance of it not happening would be .999^300
the overall chance of it happening just once would be 30*.001*.999^299
the overall chance of it happening more than once is a little more complicated.

 

[deftron]

Once or more is all I was looking for

 

[chuckywang]

Originally posted by: HomeBrewerDude
you not specific enough.

the overall chance of it not happening would be .999^300
the overall chance of it happening just once would be 300*.001*.999^299
the overall chance of it happening more than once is a little more complicated.

Fixed.

The three probabilities you wrote down should add up to 1. Therefore, the overall chance of it happening more than once is 1-.999^300-300*.001*.999^299.

 

[HomeBrewerDude]

Originally posted by: chuckywang

Originally posted by: HomeBrewerDude
you not specific enough.

the overall chance of it not happening would be .999^300
the overall chance of it happening just once would be 300*.001*.999^299
the overall chance of it happening more than once is a little more complicated.

Fixed.

The three probabilities you wrote down should add up to 1. Therefore, the overall chance of it happening once is 1-.999^300-300*.001*.999^299.

technically you only needed to bold ONE zero!

 

[chuckywang]

Originally posted by: HomeBrewerDude

Originally posted by: chuckywang

Originally posted by: HomeBrewerDude
you not specific enough.

the overall chance of it not happening would be .999^300
the overall chance of it happening just once would be 300*.001*.999^299
the overall chance of it happening more than once is a little more complicated.

Fixed.

The three probabilities you wrote down should add up to 1. Therefore, the overall chance of it happening more than once is 1-.999^300-300*.001*.999^299.

technically you only needed to bold ONE zero!

Depends if you quantize the digits or the number.

 

[ArmenK]

Originally posted by: HomeBrewerDude
you not specific enough.

the overall chance of it not happening would be .999^300
the overall chance of it happening just once would be 30*.001*.999^299
the overall chance of it happening more than once is a little more complicated.

Actually the probability of it happening more than once is not hard to find.
P(event happens more than once)=1-.999^300-300*.001*.999^299=.036858

 

[HomeBrewerDude]

Originally posted by: chuckywang

Originally posted by: HomeBrewerDude

Originally posted by: chuckywang

Originally posted by: HomeBrewerDude
you not specific enough.

the overall chance of it not happening would be .999^300
the overall chance of it happening just once would be 300*.001*.999^299
the overall chance of it happening more than once is a little more complicated.

Fixed.

The three probabilities you wrote down should add up to 1. Therefore, the overall chance of it happening once is 1-.999^300-300*.001*.999^299.

technically you only needed to bold ONE zero!

Depends if you quantize the digits or the number.

at 1 in the morning i don't do much but complain or make smart assed comments.

 

[chuckywang]

Originally posted by: HomeBrewerDude

Originally posted by: chuckywang

Originally posted by: HomeBrewerDude

Originally posted by: chuckywang

Originally posted by: HomeBrewerDude
you not specific enough.

the overall chance of it not happening would be .999^300
the overall chance of it happening just once would be 300*.001*.999^299
the overall chance of it happening more than once is a little more complicated.

Fixed.

The three probabilities you wrote down should add up to 1. Therefore, the overall chance of it happening once is 1-.999^300-300*.001*.999^299.

technically you only needed to bold ONE zero!

Depends if you quantize the digits or the number.

at 1 in the morning i don't do much but complain or make smart assed comments.

Same here. Procrastinating my paper.

 

[HomeBrewerDude]

Originally posted by: chuckywang

Originally posted by: HomeBrewerDude

Originally posted by: chuckywang

Originally posted by: HomeBrewerDude

Originally posted by: chuckywang

Originally posted by: HomeBrewerDude
you not specific enough.

the overall chance of it not happening would be .999^300
the overall chance of it happening just once would be 300*.001*.999^299
the overall chance of it happening more than once is a little more complicated.

Fixed.

The three probabilities you wrote down should add up to 1. Therefore, the overall chance of it happening once is 1-.999^300-300*.001*.999^299.

technically you only needed to bold ONE zero!

Depends if you quantize the digits or the number.

at 1 in the morning i don't do much but complain or make smart assed comments.

Same here. Procrastinating my paper.

i hear ya

 

[deftron]

Ok .. so if you play pick 3 every day for a year ($365), you only have a 31%
chance of getting an exact match at least once ($500 payout) ... No wonder the lottery makes so much money

Pick 4 is even worse with a 4% (?) chance of winning througout the year


I can't believe there's that many people in line at 7-11 to get tickets every day.
Someone needs to get them a calculator

 

[jagec]

Originally posted by: deftron
Ok .. so if you play pick 3 every day for a year ($365), you only have a 31%
chance of getting an exact match at least once ($500 payout) ... No wonder the lottery makes so much money

Pick 4 is even worse with a 4% (?) chance of winning througout the year


I can't believe there's that many people in line at 7-11 to get tickets every day.
Someone needs to get them a calculator

That's why some people refer to the lottery as a tax on people who are bad at math.

 

[ArmenK]

Originally posted by: deftron
Ok .. so if you play pick 3 every day for a year ($365), you only have a 31%
chance of getting an exact match at least once ($500 payout) ... No wonder the lottery makes so much money

Pick 4 is even worse with a 4% (?) chance of winning througout the year


I can't believe there's that many people in line at 7-11 to get tickets every day.
Someone needs to get them a calculator

Calculate expected value:
E[pick 3]=$500*(1/1000)-$1*(999/1000)=-$0.50
That is a pretty bad investment.