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
- - - - -

Simple money probability


Best Answer Prime, 07 November 2013 - 07:19 AM

Not only I forgot about half dollar coin in my previous post, I also made few arithmetic errors in calculating probabilities.

Although, the answer who has a better probability is the same.

Spoiler for new improved probability including half-dollar

Go to the full post


  • Please log in to reply
6 replies to this topic

#1 BMAD

BMAD

    Senior Member

  • Members
  • PipPipPipPip
  • 1586 posts
  • Gender:Female

Posted 04 November 2013 - 01:31 AM

I have 19 coins in my pocket which total $1.00.  You have 15 coins in yours that also total $1.00.  Who has a greater probability of randomly pulling a dime out of their pocket?


  • 0

#2 Joyandwarmfuzzies

Joyandwarmfuzzies

    Newbie

  • Members
  • Pip
  • 13 posts

Posted 04 November 2013 - 01:41 AM

Spoiler for my two cents

  • 0

#3 radio1

radio1

    Newbie

  • Members
  • Pip
  • 3 posts

Posted 04 November 2013 - 02:47 PM

Spoiler for my two cents

It's not a complete answer, as the 15 coins could also be 9 dimes, 1 nickel, and 5 pennies.

I think you're on the right track, but the final solution must factor in all the possible combinations of coins that meet the 19 and 15-coin criteria.


  • 0

#4 Prime

Prime

    Senior Member

  • Members
  • PipPipPipPip
  • 872 posts
  • Gender:Male
  • Location:Illinois, US

Posted 07 November 2013 - 05:47 AM

“Simple” is a relative term. To me it would be a simple programming code, if it was feasible to traverse all variations. Alas, I am not going to live long enough to see my laptop running through 419 possible arrangements of coins. (My laptop is old and slow.)

Nonetheless, I’ll go beyond the OP question and give the exact probability.

Spoiler for the answer is

 

Oops, I forgot about 50-cent coin. Back to the drawing board.


Edited by Prime, 07 November 2013 - 05:56 AM.

  • 0

Past prime, actually.


#5 Prime

Prime

    Senior Member

  • Members
  • PipPipPipPip
  • 872 posts
  • Gender:Male
  • Location:Illinois, US

Posted 07 November 2013 - 07:19 AM   Best Answer

Not only I forgot about half dollar coin in my previous post, I also made few arithmetic errors in calculating probabilities.

Although, the answer who has a better probability is the same.

Spoiler for new improved probability including half-dollar


  • 0

Past prime, actually.


#6 Prime

Prime

    Senior Member

  • Members
  • PipPipPipPip
  • 872 posts
  • Gender:Male
  • Location:Illinois, US

Posted 14 November 2013 - 08:15 PM

Since my last post has not been recognized as the answer, there must be something else to this problem. I suppose, it could be that different coins are encountered with different frequency. E.g., a penny is a lot more common than a half-dollar coin.

I don’t have any data on how many of each coin type there are in circulation. However, I doubt an adjustment for each individual coin probability would change the final answer to the OP. I believe a 15-coin collection would still have a higher probability of spotting a dime.


  • 0

Past prime, actually.


#7 BMAD

BMAD

    Senior Member

  • Members
  • PipPipPipPip
  • 1586 posts
  • Gender:Female

Posted 14 November 2013 - 08:33 PM

My apologies
  • 0




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users