13 Sep 05
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?
14 Sep 05
Originally posted by jimslyp69OK:
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?
(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.
01 Oct 05
Originally posted by PBE6Please expand on the pigeonhole principle, never heard of it.
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. Afte ...[text shortened]... lems that can be solved using the pigeonhole principle, a simple but powerful mathematical tool.
19 Oct 05
1 edit
Originally posted by sonhouseGiven 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.
Please expand on the pigeonhole principle, never heard of it.
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' 😉).
19 Oct 05
Originally posted by royalchickenThat's funny, I thought it was called the Crab Nebula. 😕
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' 😉).