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a deck of 14 cards  numbered1through 14 is dealt 7 each to two players. Each player score is the sum of their cards. lowest score wins. how many different winning scores are possible?

 Dec 14, 2018
 #1
avatar+94526 
+1

Here's my best attempt :

 

When each player has an even number of cards, there will only be 1 equal sum between the players....to see this....let's suppose that we have  cards 1 - 4

 

The  possible outcomes between two players are

 

(1, 2)  <    (3,4)

(1, 3)   <   (2,4)

(2, 3)   =   (1,4)      

 

Note that the number of "winning"   scores is     C(4,2)/ 2 - 1  =  3 - 1  =  2

 

When each player has an odd number, n,  of cards, the number of winning scores is just C(2n, n) / 2

 

Suppose we have cards 1- 6

 

The outcomes are

(1,2, 3)  <  (4, 5, 6)

(1,2,4)  <   (3, 5, 6)

(1, 2, 5) <  (3, 4, 6)

(1, 2, 6) <  (3, 4, 5)

(1, 3, 4) < ( 2, 5, 6)

(1,3, 5) < ( 2, 4, 6)

(1, 3, 6) < ( 2, 4, 5)

(1, 4, 5) > (2, 3, 6)

(1, 4, 6) >( 2, 3, 5)

(1, 5, 6) > ( 2, 3, 4)

 

Note that the number of winninig scores is just   C(6,3)/2  = 20/2 =  10

 

This implies....that.....for 14 cards....each player holds 7  and  the number of winning hands =

 

C(14, 7) / 2   =    1716 winning hands

 

 

cool cool cool

 Dec 14, 2018
edited by CPhill  Dec 14, 2018
edited by CPhill  Dec 15, 2018
edited by CPhill  Dec 15, 2018
 #2
avatar+95177 
+2

14C7    possible hands that i can have
half will win  (someone has to win)
3432/2 = 1716 winning hands

 Dec 15, 2018

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