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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
Harold plays a game in which he starts with $2. Each game has 2 rounds
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10 Jun 2021, 22:10
10 Jun 2021, 22:10
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Question Stats:
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58% (03:05) wrong based on 12 sessions
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Harold plays a game in which he starts with $2. Each game has 2 rounds; in each round, the amount of money he starts the round with is randomly either added to or multiplied by a number, which is randomly either 1 or 0. The choice of arithmetic operation and of number are independent of each other and from round to round. If Harold plays the two-round game repeatedly, the long-run average amount of money he is left with at the end of the game, per game, is between
A. $0 and $0.50
B. $0.50 and $1
C. $1 and $1.50
D. $1.50 and $2
E. $2 and $2.50
A. $0 and $0.50
B. $0.50 and $1
C. $1 and $1.50
D. $1.50 and $2
E. $2 and $2.50
Intern
Joined: 11 Jan 2021
Posts: 29
Given Kudos: 1275
GRE 1: Q167 V159
Re: Harold plays a game in which he starts with $2. Each game has 2 rounds
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12 Jun 2021, 01:20
12 Jun 2021, 01:20
1
The starting amount is 2$. At the beginning of round 1, there are 4 cases possible as the amount gets added or multiplied by 1 or 0.
Adding by 1: 3$
Adding by 0: 2$
Multiplying by 1: 2$
Multiplying by 0: 0$
The amount at the beginning of round 1 would be {3, 2, 2, 0}.
At the beginning of round 2, each of the above 4 cases would have its own 4 cases.
When the amount is 3$,
Adding by 1: 4$
Adding by 0: 3$
Multiplying by 1: 3$
Multiplying by 0: 0$
So, the array is {4, 3, 3, 0}
Repeating the steps for the other 3 cases, the respective arrays would be:
{3, 2, 2, 0}
{3, 2, 2, 0}
(1, 0, 0, 0}
Now, we take the average of all these cases. There would be 16 elements.
\(\frac{(4+3+3+0+3+2+2+0+3+2+2+0+1+0+0+0)}{16}\) = \(\frac{25}{16}\) = 1.5625
So, the answer is (D) 1.50$ and 2$
P.S. This solution is very lengthy and I can't think of any shortcut to solve this.
Adding by 1: 3$
Adding by 0: 2$
Multiplying by 1: 2$
Multiplying by 0: 0$
The amount at the beginning of round 1 would be {3, 2, 2, 0}.
At the beginning of round 2, each of the above 4 cases would have its own 4 cases.
When the amount is 3$,
Adding by 1: 4$
Adding by 0: 3$
Multiplying by 1: 3$
Multiplying by 0: 0$
So, the array is {4, 3, 3, 0}
Repeating the steps for the other 3 cases, the respective arrays would be:
{3, 2, 2, 0}
{3, 2, 2, 0}
(1, 0, 0, 0}
Now, we take the average of all these cases. There would be 16 elements.
\(\frac{(4+3+3+0+3+2+2+0+3+2+2+0+1+0+0+0)}{16}\) = \(\frac{25}{16}\) = 1.5625
So, the answer is (D) 1.50$ and 2$
P.S. This solution is very lengthy and I can't think of any shortcut to solve this.
gmatclubot
Re: Harold plays a game in which he starts with $2. Each game has 2 rounds [#permalink]
12 Jun 2021, 01:20
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