Brain Teaser

[MrGecko]

A sort of trading-related (via probability) brain teaser for you

T2W are feeling festive and have a Christmas lottery; T2W members can buy as many tickets as they like, and 1 in 1000 tickets are winners (each winner gets... it doesn't matter what each winner gets, it's an imaginary competition).

How many tickets do you need to buy in order to have more than a 50% chance of holding at least one winning ticket?

EDIT: Obviously the answer isn't 501

 

[scose-no-doubt]

As I don't really care whether people think I'm smart or not I'll save everyone the embarrassment...

What the f***?

 

[redart]

EDIT: Obviously the answer isn't 501

 

[MrGecko]

Well it's obviously not 501 because otherwise I wouldn't have asked in the first place...

 

[Shakone]

693

You haven't exactly posed the problem in a clear way to be honest.

 

[MrGecko]

How could it have been clearer?

 

[Atilla]

To be honest at first thought I thought

The Obviouse answer is 501.


Why is it not the obvious answer?

 

[MrGecko]

The reason it is not 501...

Well, lets start with the idea that it is 501... what needs to be true for 501 to be accurate?

501 is accurate if the probability of (getting at least one winner | you've bought 501 tickets) > 0.5 AND P(getting at least one winner | you've bought 500 tickets) < 0.5

(where | => Given that) - conditional probability,innit.

The mistake that most people make is that they count each and every ticket being a winner as disjoint in the sample space - that is, they don't fully consider the probability that both could be winners... or more accuratly, they consider this twice!

EDIT: The natural reaction is to do P(winner) + P(winner) + ...+ until they get to something > 50%.
Thats how you get to 501.

The problem is that the probability P(either ticket is a winner) = P(1st is a winner) + P(2nd is a winner) - P(1st AND 2nd are winners)... and it is this P(1st n 2nd) that trips people up

 

[MrGecko]

OK, so how do you get to 693?

Well, the probability of getting at least one winning ticket is the same as:

100% - no winning tickets at all!

Now, if 1/1000 are winners, it follows that 999/1000 are losers...

So, How many losers do I need to make it unlikely that I lose?
(unlikely to lose = likely to win!)

Then we do (999\1000) ^ X < 0.5
(how many tickets do I need to make the probability of them all being losers less than a half?)

That is, lets find out how many tickets we need to make it UNLIKELY that all are losers

(take the Logs of both sides gives

X = In(0.5) / In(999/1000) = 693 (rounded to integers)

 

[MrGecko]

If you want to check this, do:

(999/1000) ^ 693 - this is the probability that ALL 693 tickets are losers... which is Just lss than 50%!

 

[wackypete2]

err... what language is that?

Makes trading look easy.

Peter

 

[MrGecko]

err... what language is that?

Makes trading look easy.

Peter

Q: how many people do you neeed at a party to make it likely that at least 2 of them share a birthday?

A: 23

 

[Splitlink]

a sort of trading-related (via probability) brain teaser for you

t2w are feeling festive and have a christmas lottery; t2w members can buy as many tickets as they like, and 1 in 1000 tickets are winners (each winner gets... It doesn't matter what each winner gets, it's an imaginary competition).

How many tickets do you need to buy in order to have more than a 50% chance of holding at least one winning ticket?

Edit: Obviously the answer isn't 501

501

 

[scose-no-doubt]

Q: how many people do you neeed at a party to make it likely that at least 2 of them share a birthday?

A: 23

ok i understand the last one though I never would have though to approach it that way- but this one?

do you mean 253?

takes me back to my alevel days lol

 

[redart]

the plane takes off.

and 501

 

[MrGecko]

do you mean 253?

 

[scose-no-doubt]

f*ck off then

what've I missed?

 

[forker]

Q: how many people do you neeed at a party to make it likely that at least 2 of them share a birthday?

A: 23

4 lesbians and a lot of rum

 

[dick_dastardly]

After reading this thread I have arrived at two conclusions:

1) MrGecko is clever enuff to explain the answer to the Two Boy Puzzle

2) I would be too thick to understand it.



dd

 

[Splitlink]

I don't want the explanation, i just want the right answer to the first one, the T2W comp.