A chauffer has 5 pairs of white gloves, 3 pairs of black gloves and 2 pairs of brown gloves in his draw. Without looking, how many gloves does he need to pull out of the draw to be sure he makes a complete pair?
ANd another one...
How many cards can be pulled out of a complete set of 52 playing cards, without having a full house combination?
Originally posted by jimslyp69 A chauffer has 5 pairs of white gloves, 3 pairs of black gloves and 2 pairs of brown gloves in his draw. Without looking, how many gloves does he need to pull out of the draw to be sure he makes a complete pair?
ANd another one...
How many cards can be pulled out of a complete set of 52 playing cards, without having a full house combination?
OK:
(1) The chauffeur can pull out 5 lefty white gloves, 3 lefty black gloves and 2 lefty brown gloves. After that, the next glove will complete some pair, so the chauffer needs to pull out 5+3+2+1 = 11 gloves to be sure.
(2) A full house, consisting of one pair and one three of a kind, can be put off the longest by drawing a pair from each rank. After that, any card will complete some three of a kind making a full house. So the most cards that can be drawn without making a full house is 13*2 = 26 cards. The 27th card will make the full house.
Interesting questions. These are examples of problems that can be solved using the pigeonhole principle, a simple but powerful mathematical tool.
Originally posted by Bowmann In the chauffeur's sock drawer, he has a ratio of 5 pairs of blue socks, 4 pairs of brown socks, and 6 pairs of black socks...
(1) The chauffeur can pull out 5 lefty white gloves, 3 lefty black gloves and 2 lefty brown gloves. After that, the next glove will complete some pair, so the chauffer needs to pull out 5+3+2+1 = 11 gloves to be sure.
(2) A full house, consisting of one pair and one three of a kind, can be put off the longest by drawing a pair from each rank. Afte ...[text shortened]... lems that can be solved using the pigeonhole principle, a simple but powerful mathematical tool.
Please expand on the pigeonhole principle, never heard of it.
Originally posted by sonhouse Please expand on the pigeonhole principle, never heard of it.
Given two finite sets S and T, f: S->T is bijective if and only if card S = card T. In particular, if card T < card S, there must exist some t in T such that t = f(s1) and f(s2) for distinct s1, s2 in S.
A shavixmirian example: The average person has, I would guess, on the order of 10^5 pubes, and certainly nobody has more than 10^6 (I should hope). Since there are about 7*10^9 people in the world, the pigeonhole principle guarantees that there exist two people with the same number of pubes (this is an example of what's known as 'fuzzy logic' 😉).
Originally posted by royalchicken A shavixmirian example: The average person has, I would guess, on the order of 10^5 pubes, and certainly nobody has more than 10^6 (I should hope). Since there are about 7*10^9 people in the world, the pigeonhole principle guarantees that there exist two people with the same number of pubes (this is an example of what's known as 'fuzzy logic' 😉).
That's funny, I thought it was called the Crab Nebula. 😕
13 Sep '05 23:36
Chauffer problem
Back to Top