Brain Teaser

[Shakone]

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

No. The thing you're not getting here, is that there is a big difference between

(i) a room full of 23 people and someone having the same birthday as you

(ii) a room full of 23 people and nobody having the same birthday.

In the second case, there's not just a chance someone has the same birthday as you, but also that they have the same birthday as one of the other 21 people there.

 

[MrGecko]

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

Shakone got it already - you need to buy at least 693 tickets in order to have a greater than 50% chance of having at least one winning ticket.

 

[counter_violent]

Hopefully The Bramble will come along and sort it out !

 

[MrGecko]

The birthday one explained:

There are 365 days in a year, and each is equally likely to be anyones birthday. So, if you get 2 people together, there are (365) * (365) = 133,225 combinations of birthdays - Person A could be born on the 1st Jan, Person B could be born on any of the 365 days of the year... Person A could be born on the 2nd Jan, and Person B could be born on any other of the 365 days in the year, and so on...

So, for a group of X people, there are (365)^ X combinations of birthdays possible.

Now, as we did last time, we are going to say "the probability of at least two people sharing a birthday" is;

100% - "Probability that everyones birthday is different".

So, what is the probability that everyones birthday is different? To do this we figure out how many combinations of different birthdays there are for X many people...

Lets start out with two people. Person A is born on one of the 365 days, and for Person B to have a different birthday, he needs to have been born on one of the other 364 days..

so there are (365) * (364) = 132,860 combinations where this can be true...

What if we introduce a third person? Well they need to be born on one of the other 363 days that are left, so there are

(365) * (364) * (363) = 48,228,180 combinations of three people having different Birthdays.

In fact, we can go on to say that there are (365) * (364) * (363) * ... * (365 - X + 1) = combinations of X many people each having different birthdays.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

So, we know that there are (365) ^ X possible combinations of birthdays, and...
(365) * (364) * (363) * ... * (365 - X + 1) possible combinations of different birthdays for everyone in a group of X many people.

The probability then that everyone in a group of X many people has a different birthday is:

Number of different birthday combinations / Number of birthday combinations

= (365) * (364) * (363) * ... * (365 - X + 1) / (365) ^X

Now we want to find the probability that at least two people share a birthday, which we expressed as 100% - P(nobody shares a birthday). For the chance of at least two people sharing a birthday to be > 50%, it should be unlikely (i.e. P < 50%) that everyone has a different birthday...

We solve this by trial and error, changing the value of X until the sum descrobed above is less than 0.5. Which happens to be when X = 23 - so if there are 23 people in a room, it is more likely than not that at least two of them share a birthday.


(apologies for the poor explanation, it's tricky to do this in text)

 

[sk1105]

In other words, the Dow will close up tomorrow.

 

[scose-no-doubt]

Scary how much stuff I forgot since I left school.
I fail to see how any of this can be applied to the markets.
Care to explicate, Gecko.

 

[TheBramble]

Hopefully The Bramble will come along and sort it out !

Most of my stuff gets done in camera these days CV.

The plane doesn't take off.

 

[TheBramble]

You have a 1 in 1000 chance of contracting a virus which makes you sign up to t2w and live the remainder of your days in misery.

However, there is a possible remedy in the taking of the blue (NOT the red) capsule.

If you take this capsule, although you are saved from a lifetime of t2w misery, you have a 50% chance of suffering the side effects of becoming an honorary premium member of crappy cows website. Seems like a good deal.

Or is it?

Mixed framing is the bane of professional statisticians and the cornerstone of professional marketers claims.

Good lesson from Mr. G.

 

[TheBramble]

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

Don't allow yourself to be groomed by Gecko and his wicked band of statisphiles. (Where's my t2w panic button?). He'll fiddle with your figures.

And don't be tricked by MG either - he was a statistical anomaly and a complete ******, not a trader. It's not 13/27. And it's not 1/3 - it's 1/2.

What's the probability your next (first?) child will be a boy? Makes NO difference what your last child was.

After you've sold your excess 192 tickets at twice par to hedge your better than 50% chance of winning the t2w lottery to clueless newbs who think they are especially lucky tickets, you get on the plane...

...which does not take off.

 

[dick_dastardly]

Don't allow yourself to be groomed by Gecko and his wicked band of statisphiles. (Where's my t2w panic button?). He'll fiddle with your figures.

And don't be tricked by MG either - he was a statistical anomaly and a complete ******, not a trader. It's not 13/27. And it's not 1/3 - it's 1/2.

What's the probability your next (first?) child will be a boy? Makes NO difference what your last child was.

After you've sold your excess 192 tickets at twice par to hedge your better than 50% chance of winning the t2w lottery to clueless newbs who think they are especially lucky tickets, you get on the plane...

...which does not take off.

Thanks for explaining that in a way I understand

.... and the bit about my next child aswell.





another one....

My Uncle went to Thailand and when he came back she was me Aunt ( true strory BTW ).

What is the probability of this?



Additional info - my other uncles are/were straight afaik and the David Beckham allele is recessive ( I hope ).

dd

 

[MrGecko]

Care to explicate, Gecko.

Not really any application to the markets tbh, just thought it would get the old brain ticking.

As for the two boy puzzle (which I had never heard of)... I can see why it could be 1/2, and understand the basic logic of why it could be 13/27 (haven't gone through it on paper). I expect that both answers are correct in theory but the maths geeks all decided to agree on 13/27 in order that their ego's are massaged everytime some regular bod says "1/2".

 

[scose-no-doubt]

Mixed framing is the bane of professional statisticians and the cornerstone of professional marketers claims

what?

 

[FX Bandit]

Enough of this bullsh1t Gekko.

First off, it was a lot better round here when you were f***ing banned.

Second, have I won a prize in this f***ing T2W competition or not? If so, can I have a blender you b4stards, because mine packed in this morning?

Third of all, why the f*** are you called Gekko anyhow? This is a trading site, not the bloody notice board at the local lizard house. What the f*** has a gekko got to do with trading?

That's all I've got to say on the matter.

 

[FX Bandit]

Not really any application to the markets tbh, just thought it would get the old brain ticking.

As for the two boy puzzle (which I had never heard of)... I can see why it could be 1/2, and understand the basic logic of why it could be 13/27 (haven't gone through it on paper). I expect that both answers are correct in theory but the maths geeks all decided to agree on 13/27 in order that their ego's are massaged everytime some regular bod says "1/2".

Christ in a f***ing brothel, are you smoking crack?

There's an equal probabilityness that a child could be anything, right? So, one could be this or that, the next could be this or that, nobody f***ing knows. 50/50, at best.

 

[Scotty2Cues]

This one is just pure logic - no maths needed and has nothing to do with trading

In a room are 3 light bulbs. Outside the room are three switches connected to bulbs.

There is one closed door to the room and no windows (so you cant see in the room)

You are only allowed to enter the room once (or open door once)

You can turn the switches on/off as many times as you like until you open door.

How do you figure which switch is connected to which bulb?

Dont post the solution if you know it

A very nice solution to this!

 

[Scotty2Cues]

Dont post the solution if you know it

cant believe i put that

 

[Dommo]

Nice problem, nice solution.

 

[Dommo]

You need to heat up one of the bulbs first, then switch it off, and switch one of the others on before entering the room

Highlight here.

 

[MrGecko]

err...

so originally you can't see into the room, and you have 3 switched (with 3 lightbulbs?). You can turn them on or off as many times as you like, but you can only look in the room once - and can't reach to turn any swithces when you're in the room... then you can leave the room, do whatever you want with the switches, but aren;t allowed back into the room to see the changes?

Unless it's a "trick" question like "open the door, put a mate in and get him to shout", I can't see how it can be done. If you can't see the changes of turning the switches on and off, you can't deduce anything.

 

[scose-no-doubt]

look at the wiring