Question

In a game of chance, first 20 natural numbers are listed on a piece of paper and three numbers are chosen at random. If the numbers form an A.P. you will get Rs. 10000, If the numbers form a GP you will get Rs. 20000, otherwise you have to pay Rs. 8000. What is the mean expected gain and the corresponding risk, if you play the game?

Answer 1

ways of choosing 3 numbers out of 20 is C(20,3)=1140  [(a,b,c) with 1 <= a < b < c <= 20]

finding the series of GP,

Sets with common ratio 2/1: (1,2,4) , (2,4,8) , (3,6,12) , (4,8,16) , (5,10,20)

Sets with common ratio 3/1: (1,3,9) , (2,6,18)
Set with common ratio 4/1: (1,4,16)
Sets with common ratio 3/2: (4,6,9) , (8,12,18)
Set with common ratio 4/3: (9,12,16)

# of such sets = 11

Probability of GP= 11/1140

finding series of AP,

with common difference 1: (1,2,3).(2,3,4)............(18,19,20) so, # terms=18

with common difference 2:(1,3,5),(2,4,6).............(16,18,20 so, # terms=16

with common difference 3:(1,4,7),(2,5,8)............(14,17,20) so,#terms=14

with common difference 4 :(1,5,9),(2,6,10)..........(12,16,20) so # terms=12

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similarly this way,

last will be

with common difference 9:(1,10,19),(2,11,20) so, # terms=2

total terms of AP=18+16+14+12+10+8+6+4+2=90 terms

probability of three numbers being in AP=90/1140

probability of losing=1-90/1140-11/1140=1039/1140