Akshay and Rahul on train
Rahul and Akshay are traveling in a train. The journey is really boring so they decided to play a little game. They take 10 balls and numbered them from 1 to 10 and then put them in a bag. The rules of the game are simple. First Rahul will take out 4 balls randomly. If none of the taken out numbers were consecutive than he wins otherwise Akshay gets to draw 4 balls. The game will continue till one of them draws a lot (4 balls) in which none of the balls were consecutively numbered.
Let the probability that Rahul wins be b a , where a and b are coprime positive integers, find a + b .
Assume that the balls are drawn with replacement.
The answer is 17.
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2 solutions
Why number of non-conscutive balls is ( 4 7 )
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That's a good question... I didn't have a particularly good way to explain it so decided not to. Essentially it is because if you are picking the numbers one after another in increasing order of size, you can't pick the number one bigger than your first choice, you can't pick the number one bigger than your second choice and you can't pick the number one bigger than your third choice. This effectively means you have only 10 - 3 = 7 numbers to choose from... To be honest, I only realised this after I started listing possibilities:
1,3,5,7 1,3,5,8 etc...
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You can use the stars and bars approach. Choose 4 of the balls at random and then order them as N 1 < N 2 < N 3 < N 4 . Now let a be the number of positive integers less than N 1 , b the number of integers between N 1 and N 2 , c the number of integers between N 2 and N 3 , d the number of integers between N 3 and N 4 and e the number of integers greater than N 4 and ≤ 1 0 .
We then have in general that a + b + c + d + e = 6 such that a , b , c , d , e are all non-negative integers. But if we require that none of the chosen numbers are consecutive then we will require that each of b , c , d be ≥ 1 . So define b ′ = b − 1 , c ′ = c − 1 and d ′ = d − 1 . Then our equation becomes a + b ′ + c ′ + d ′ + e = 3 such that a , b ′ , c ′ , d ′ , e are all non-negative integers. By the stars and bars method the number of solutions to this last equation is ( 3 3 + 5 − 1 ) = ( 3 7 ) = 3 5 .
Edit: After posting this comment I realized that Rishi Sharma's solution outlines the same idea as I have presented. I'll leave my comment up, however, as two explanations are better than one.
In this solution I shall only be explaining why the number of ways of picking non-consecutive balls is ( 4 n − 3 ) For rest of solution you can refer to Paul Hindess' solution.
Assume N dots on a line. Each dot will represent a number. Now randomly choose 4 numbers and put a line across each. These 4 lines will be dividing the assumed line in 5 portions. Now let the number of dots present in portion i be x i . The assumed variables will always follow the equation i = 1 ∑ 5 x i = n − 4 having the constraints x 1 , x 5 ∈ [ 0 , n − 4 ] and x 2 , 3 , 4 ∈ [ 1 , ( n − 4 ) ] Now number of integer solution can be found using stars and bars which comes out to be ( 4 n − 3 )
The probability of picking 4 all-non-consecutive balls from ten balls numbered 1 to 10 is ( 4 1 0 ) ( 4 7 ) = 6 1 .
[Note that ( r n ) refers to n C r = r ! ( n − r ) ! n ! .]
The probability of Rahul winning is 6 1 + ( 6 5 × 6 5 × 6 1 ) + ( 6 5 × 6 5 × 6 5 × 6 5 × 6 1 ) + . . . = 6 1 + 3 6 2 5 × 6 1 + ( 3 6 2 5 ) 2 × 6 1 + . . .
This is the sum to infinity S ∞ of a Geometric Progression with first term a = 6 1 and common ratio r = 3 6 2 5 .
S ∞ = 1 − r a = 1 − 3 6 2 5 6 1 = 1 1 6 .
So the answer to the problem is 6 + 11 = 17.