Larry and Julius are playing a game, taking turns throwing a ball at a

Question

Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a player knocks the bottle off the ledge is 1/2, independently of what has happened before. What is the probability that Larry wins the game?

(A) 1/2
(B) 3/5
(C) 2/3
(D) 3/4
(E) 4/5

Show Answer

Official Answer

C

(A) 1/2
(B) 3/5
(C) 2/3
(D) 3/4
(E) 4/5

gaurav2m · Accepted Answer

If Larry wins, he either wins on the first move, or the third move, or the fifth move, etc. Let represent "player wins", and represent "player loses". Then the events corresponding to Larry winning are W,LLW,LLLLW.....

Thus the probability of Larry winning is
1/2,1/8,1/32......

This is a infinite geometric series with ratio: 1/4

Formula to solve this is  (1/1-r)  where r is the common ratio

hence the answer is :2/3

leminkhoa · Answer

This question is a decision-tree problem.

Obviously, because Larry goes first, Larry clearly has the advantage if she luckily wins in the first turn. This game is an unfair game for Julius -> The probability that Larry wins must be higher than 1/2.

We have the following scenarios: (LW as Larry wins, LL as Larry loses, JW as Julius wins, JL as Julius loses)

LW  
LL -> JW 
--------JL ->  LW 
--------------- LL -> JW
-----------------------JL ->  LW 
------------------------------LL ->

This will continue until there is one person wins the game or we can say it continues to infinitive.

The chance for Larry to win lies at 1st, 3rd, 5th, 7th turns -> odd turns

=> The probability now is: (0.5)+ (0.5)^3 + (0.5)^5 + (0.5)^7 + ... + (0.5)^n with n is an odd number

The question now asks us to count the sum of a sequence.

Applying the  Geometric Progression , we have b1=0.5, r = (0.5)^2, the sum is 0.5/  = 0.5/0.75 = 2/3

Shobhit7 · Answer

Larry can win in following scenarios: W, LLW, LLLLW, so on.

GP series of Probabilities:
1/2, 1/8, 1/32 and so on

Final Probability= Sum of all Probabilities in above series

Sum= a1(1-r^n) / (1-r)

Common ratio r= (1/8)/(1/2)= 1/4 
(i.e ratio of two adjacent terms with a2 as numerator and a1 as denominator)

No of terms n= infinite
So, r^n= 1/2^infinity tends to 0 as 1/infinity= 0

So, Sum= a1(1-0) / 1-0.25
= 0.5/0.75 = 2/3

Hence, Ans C

HarshavardhanR · Answer

How can Larry win?

W  - Win - 1/2 probability
 L  - Loss - 1/2 probability

W  OR  LL  W  OR  LLLL  W  OR  LLLLLL  W  OR ... so on till infinity.

P(Larry wins) =  1/2  + ( 1/2 x 1/2  x  1/2 ) + ( 1/2 x 1/2  x  1/2 x 1/2  x  1/2 ) + so on till infinity.

P(Larry wins) = 1/2 + 1/2^3 + 1/2^5 + 1/2^7 + so on till infinity.

This is an infinite GP with first term a = 1/2 and common ratio r = 1/2^2 = 1/4 (r <1).

Sum = a/1-r = (1/2) / (1-1/4) = 2/3. Choice C.

___
Harsha
Enthu about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com

Regor60 · Answer

Call the infinite series S, which begins with a win for Larry of 1/2 probability.

After adding a win with a probability of 1/2 for the initial toss there follows a loss for Larry and a loss for Julius, each probability of 1/2, multiplied by the same initial infinite series that began with Larry's winning, correct ?

So:

S = 1/2 + (1/2)(1/2)S and

3S/4 = 1/2 and

S = 4/6 = 2/3

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