Jump to content


Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account.
As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends.

Of course, you can also enjoy our collection of amazing optical illusions and cool math games.

If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top.
If you have a website, we would appreciate a little link to BrainDen.

Thanks and enjoy the Den :-)
Guest Message by DevFuse
 

Photo
- - - - -

Casino Game (big)


  • Please log in to reply
15 replies to this topic

#11 unreality

unreality

    Senior Member

  • Members
  • PipPipPipPip
  • 6370 posts

Posted 17 September 2007 - 01:12 AM

Hi UR,

I re-read your posts, and If I understand them,
you are saying that the equally likely cases are the numbers of reds and greens.
5 red 0 green = same likelihood as
4 red 1 green = same likelihood as
3 red 2 green = same likelihood as
2 red 3 green

Like that defines 4 equally likely cases.

Is that what you're saying?
Otherwise I don't understand where we differ.

Best,
BN



yep! same probability... cuz there is no random distribution. I'll post this in that topic so we can continue this discussion there not here


  • 0

#12 bonanova

bonanova

    bonanova

  • Moderator
  • PipPipPipPip
  • 5918 posts
  • Gender:Male
  • Location:New York

Posted 17 September 2007 - 05:47 AM


Hi UR,

I re-read your posts, and If I understand them,
you are saying that the equally likely cases are the numbers of reds and greens.
5 red 0 green = same likelihood as
4 red 1 green = same likelihood as
3 red 2 green = same likelihood as
2 red 3 green

Like that defines 4 equally likely cases.

Is that what you're saying?
Otherwise I don't understand where we differ.

Best,
BN



yep! same probability... cuz there is no random distribution. I'll post this in that topic so we can continue this discussion there not here <!-- s:D --><!-- s:D -->


Thank you, my friend -- I have seen the light.
You've taught me well.
If you'll excuse me I have a few things to do...

[1] Go tell sajow4 she was right.
Spoiler for ...

[2]Go to Morty's and flip some coins tonight.
Spoiler for ...

[3] Finally -and this is the big one - I'm going to buy a lotto ticket.
Spoiler for ...

  • 0
The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#13 Martini

Martini

    Senior Member

  • Members
  • PipPipPipPip
  • 770 posts

Posted 17 September 2007 - 05:49 PM

Right there ^ is where you are wrong. It is not randomly distributed. There would be 26 different options if and only if the chips under the cups were distributed randomly.


How does this make a difference?

There is a red marble and a black marble in a bag. You blindly put your hand in and pull one out (random). What is the probability that you pulled the black marble?

There is a red marble and a black marble in a bag. Someone grabs the one he wants (non-random). You have to guess what marble he picked and you guess black. What is the probability that you guessed correctly?
  • 0

#14 unreality

unreality

    Senior Member

  • Members
  • PipPipPipPip
  • 6370 posts

Posted 17 September 2007 - 08:40 PM

You guys aren't getting it... there are 2 random chances...

the first is where the chips are under the cups

the second is which ones you picked (easily determined)

The chances of the second do not change unless the first changes


Bonanova, i dont know where you got the 100-0-0-0 thing but think about it this way:

normally, there would be certain chances that certain combinations are more likely... like look at it with three cups of two different color chips:

red-red-red
red-red-blue
red-blue-red
blue-red-red
blue-blue-red
red-blue-blue
blue-red-blue
blue-blue-blue

but the chances for one, if they ARENT picked randomly, is just...

ZOMG NEVER MIND SORRY

wow i'm such an idiot.
  • 0

#15 unreality

unreality

    Senior Member

  • Members
  • PipPipPipPip
  • 6370 posts

Posted 17 September 2007 - 08:47 PM

ahhh actually I was right to an extent. We sort of met in the middle. This, copied from the front of my answer, was the base for the rest of my operations:

4 RED, 1 GREEN
3 RED, 2 GREEN
2 RED, 3 GREEN

Each are equally probable.


That statement is perfectly true... I just didnt elongate it. For example, 4 red and 1 green can be:

GRRRR
RGRRR
RRGRR
RRRGR
RRRRG

*slaps forehead*
  • 0

#16 bonanova

bonanova

    bonanova

  • Moderator
  • PipPipPipPip
  • 5918 posts
  • Gender:Male
  • Location:New York

Posted 17 September 2007 - 11:22 PM

ZOMG NEVER MIND SORRY

wow i'm such an ------.


Not true. Quit talking about my friends like that....
  • 0
The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users