CTK Exchange
CTK Wiki Math
Front Page
Movie shortcuts
Personal info
Terms of use
Privacy Policy

Interactive Activities

Cut The Knot!
MSET99 Talk
Games & Puzzles
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Math as Language
Things Impossible
My Logo
Math Poll
Other Math sit's
Guest book
News sit's

Recommend this site

Manifesto: what CTK is about Search CTK Buying a book is a commitment to learning Table of content Products to download and subscription Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

CTK Exchange

Subject: "Probability problem"     Previous Topic | Next Topic
Printer-friendly copy     Email this topic to a friend    
Conferences The CTK Exchange This and that Topic #919
Reading Topic #919
Member since Jan-6-10
Jan-06-10, 02:20 PM (EST)
Click to EMail rabinca Click to send private message to rabinca Click to view user profileClick to add this user to your buddy list  
"Probability problem"

This is my first time here. I'm long ago graduated from school. My knowlwedge of probability theory is not terribly advanced and I've been wrestling with a probability problem and have no one else to turn to for help. I'm hoping someone here can help me. Here's the problem...

Imagine a game in which the object is to draw a specific card, say the Queen of Hearts, from a standard deck of playing cards. Each time a card is drawn it is removed from the deck if it is not the Queen of Hearts and set aside before the next card is drawn so the deck gets smaller with each unsuccessful draw.

My question is, "What is the probability that the Queen of Hearts will be the very last card drawn, i.e. it will not be drawn until all of the other 51 cards have been?" More generally, how would I go about computing the probability that the Queen of Hearts is not drawn until the nth trial?

I've been reading what I can find on probability theory on the internet but, given my level of math knowledge and skills, the stuff is either too basic or too advanced for me to understand well enough to extrapolate from it to solve the problem. I thank you in advance for any help you can provide.

Best regards,


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
Charter Member
2467 posts
Jan-06-10, 02:26 PM (EST)
Click to EMail alexb Click to send private message to alexb Click to view user profileClick to add this user to your buddy list  
1. "RE: Probability problem"
In response to message #0
   The probability that your card won't be drawn on the first try is 51/52. That on the second is 50/51; on the third 49/50, etc. Your card won't be drawn on the 50th draw with probability 2/3 and on the 51st with the probability 1/2.

The product of these probabilties is the probability that the card will be left over:

51/52·50/51·49/50·...·2/3·1/2 = 1/52.

Come to think of it. The cards are left over with equal probabilities and one is always there with probabilitiy 1. For a specific card then it is 1/52.

  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top

Conferences | Forums | Topics | Previous Topic | Next Topic

You may be curious to have a look at the old CTK Exchange archive.
Please do not post there.

Copyright © 1996-2018 Alexander Bogomolny