All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
Picking cards (Posted on 2006-08-17) Difficulty: 3 of 5
A player picks random cards from an ordinary card deck, without returning them to the deck. How many cards should he pick so as to get at least one ace with 70% probability?

No Solution Yet Submitted by atheron    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Puzzle Solution Verification Comment 8 of 8 |
I was reading the  computer program assisted respective "table of probability" and "simulated result" methodology by Charlie and Penny, and decided to perform a verification with the calculations done by the  aid of an online solver.

VERIFICATION 
Case 1: 14 cards
When we pick 14 cards, the probability of NOT having a single ace is:
comb(48,14)/comb(52,14)
= 2109/7735
So, the probability of having at least a single ace
= 5626/7735
 ~ 0.727343245

Case 2: 13 cards
When we pick 13 cards the probability of NOT having a single ace is:
comb(48,13)/comb(52,13)
= 6327/20825
So, the probability of having at least a single ace
= 14498/20825
~ 0.696182473
Since in the latter case, 100*(required probability) rounds to 70, while in the former case it rounds to 73, we must conclude that the player must pick precisely 13 cards. 

Edited on April 7, 2022, 11:35 pm
  Posted by K Sengupta on 2022-04-07 23:25:54

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information