the minimum no. of cards to be dealt fro an arbitrarily shuffled deck of 52 cards to guarantee that 3 cards are from the same kind i.e.(diamond,spade,clove and heart.) is
(a) 3
(b)8
(c)9
(d)12
2007-06-10 14:40:03 · 4 answers · asked by honey 1 in Science & Mathematics ➔ Mathematics
"Guarantee" means 100% sure. No ifs or buts, no "with high probability".
The worst case scenario is you get 8, with 2 of each suite (2x4=8). To get the third, you need one more, so the minimum is 9.
2007-06-10 14:55:08 · answer #1 · answered by TV guy 7 · 0⤊ 1⤋
(c) 9.
In order to figure out what is necessary to guarantee having three of one suit, you have to figure out the longest you can go without getting three of any one suit.
You can have two of each suit (eight cards) and you will not have three of any one suit. However, when you receive the ninth card, no matter which suit it is, it will be third card its suit.
2007-06-10 14:45:55 · answer #2 · answered by McFate 7 · 2⤊ 0⤋
a) 3
The minimum is 3, if you are looking for three from the same suit.
There is nothing saying that you need to receive all the cards in one persons hand, or how many people are at the "table" so I would have to presume that you are only saying to deal three of one suit, how many cards to I need to deal.
9, would be the maximum amount required.
2007-06-10 14:49:43 · answer #3 · answered by Erik D 2 · 0⤊ 3⤋
its simply 9 dam sure
2007-06-10 15:14:58 · answer #4 · answered by nobody_rahul 3 · 0⤊ 1⤋