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Harold plays a game in which he starts with $2. [#permalink]
04 Sep 2012, 17:18

Expert's post

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

35% (02:39) correct
65% (02:02) wrong based on 82 sessions

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

I completely do not understandig how to start, attack or even figure out this problem without read the explanation

Also: is real to encounter such a problem during the exam. lost

Re: Harold plays a game in which he starts with $2. [#permalink]
04 Sep 2012, 22:38

1

This post received KUDOS

Expert's post

Hi Carcass,

This does seem like an exceptionally tough problem--but it's solvable!

Consider the possible outcomes. We could end with 4 dollars, but to do so, we need out outcomes to be (add, 1, add, 1). That's 1/2 * 1/2 * 1/2 * 1/2 = 1/16 odds.

To end with 3 dollar, we can go (add, 1, add, 0), (add 1, multiply 1), (multiply 1, add 1) or (add, 0, add, 1). The probabilities of this game are uniform, so each possible outcome has the same odds. There are four mutually exclusive outcomes that total 3 dollars. We should add the odds of mutually exclusive outcomes, so 1/16 + 1/16 + 1/16 + 1/16 = 1/4 is the odds of 3 dollars.

2 dollars can be: (Add, 0, Add, 0), (Multiply, 1, Multiply, 1), (Add, 0, Multiply, 1), or (Multiply, 1, Add, 0). That's 4 option, so 4/16 = 1/4 is the odds of breaking even.

There is exactly one way to end with 1 dollar, (multiply, 0, add, 1). That's 1/16

And everything else is 0.

So, (1/16)*4 + (1/4) * 3 + (1/4) * 2 + (1/16) * 1 is the expected outcome.

That's 0.25 + 0.75 + .5 + ~0.05, which is just slightly greater than 1.5. (D) is our answer! _________________

Re: Harold plays a game in which he starts with $2. [#permalink]
04 Sep 2012, 22:57

1

This post received KUDOS

carcass wrote:

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

I completely do not understandig how to start, attack or even figure out this problem without read the explanation

Also: is real to encounter such a problem during the exam. lost

No, this cannot be a real test question. First of all, the long-run average - on the GMAT, they will never leave an undefined term. Second, this is a more advanced subject in statistics (expectance), definitely not tested on the GMAT. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Harold plays a game in which he starts with $2. [#permalink]
04 Sep 2012, 22:58

1

This post received KUDOS

KapTeacherEli wrote:

Hi Carcass,

This does seem like an exceptionally tough problem--but it's solvable!

Consider the possible outcomes. We could end with 4 dollars, but to do so, we need out outcomes to be (add, 1, add, 1). That's 1/2 * 1/2 * 1/2 * 1/2 = 1/16 odds.

To end with 3 dollar, we can go (add, 1, add, 0), (add 1, multiply 1), (multiply 1, add 1) or (add, 0, add, 1). The probabilities of this game are uniform, so each possible outcome has the same odds. There are four mutually exclusive outcomes that total 3 dollars. We should add the odds of mutually exclusive outcomes, so 1/16 + 1/16 + 1/16 + 1/16 = 1/4 is the odds of 3 dollars.

2 dollars can be: (Add, 0, Add, 0), (Multiply, 1, Multiply, 1), (Add, 0, Multiply, 1), or (Multiply, 1, Add, 0). That's 4 option, so 4/16 = 1/4 is the odds of breaking even.

There is exactly one way to end with 1 dollar, (multiply, 0, add, 1). That's 1/16

And everything else is 0.

So, (1/16)*4 + (1/4) * 3 + (1/4) * 2 + (1/16) * 1 is the expected outcome.

That's 0.25 + 0.75 + .5 + ~0.05, which is just slightly greater than 1.5. (D) is our answer!

I did it bit differently. I caluculated all the possible outcomes and then added the $ values and took the average. Answer came out to be 25/16 which is option D. _________________

Re: Harold plays a game in which he starts with $2. [#permalink]
05 Sep 2012, 02:11

3

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

carcass wrote:

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

I completely do not understandig how to start, attack or even figure out this problem without read the explanation

Also: is real to encounter such a problem during the exam. lost

@sanki779: Your method is fine too.

The question involves the same concepts of probability and weighted average that you use in and out. The question here may be long and tedious but GMAT excels in making it sound tough (so that you wonder how to start) even if the concepts used are simple.

You start with $2 - There are 4 possibilities: add/multiply; 1/0

............................................................2 ..........3...............................2.............................2...............................0 (four possibilities after 1st round) 4....3....3....0.............3....2....2....0.............3....2....2....0.............1....0....0....0 (sixteen possibilities after 2nd round)

After 2 rounds, you will have a total of 16 possible outcomes - 4 in one case, 3 in four cases, 2 in four cases, 1 in one case and 0 in six cases.

Now you just have to find their weighted average to find the average amount of money he is left with at the end of the game.

Re: Harold plays a game in which he starts with $2. [#permalink]
25 Sep 2012, 04:41

This is the way i solved it (not sure if it is the correct approach but i did get the correct answer)

Harold Starts off with - 2 $

He has to play a game of Two SETS , Let the SETS be A and B ...

In Set A The following operations are equally possible ...

Scenario 1 * One dollar will be added to his 2 $ purse Scenario 2 * Zero dollars will be added to his 2 $ purse Scenario 3 * His purse will be multiplied by 1 (will remain at 2 $) Scenario 4 * His purse will be multiplied by 0 (will become zero)

So he has 1/4 chance of ending up with 3 $ (scenario 1) , 1/4 chance of ending up with 2 dollars (scenario 2) , 1/4 chance of ending up with 2 dollars (scenario 3) , and 1/4 chance of ending up with 0 dollars ...

Simplifying we get his odds @ - 1/2 for ending up with 2 dollars , 1/4 chance of ending up with 0 dollars and 1/4 chance of ending up with 3 dollars ...

Now we know that the same odds will be repeated again in set B of Game 1 , and again for all sets of all remaining games he plays .

After set B HAROLD will either be left with 0 $ , 3 $ or 2 $ ... Taking the average of the three we get 5/3 = 1.66 which falls into category (D) and does not overlap with any other answer choice ...

P.S. - How many of the Veteran gmatters here think that a possibility of such type of a question showing up on the GMAT is likely... _________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Re: Harold plays a game in which he starts with $2. [#permalink]
22 Mar 2014, 01:40

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Harold plays a game in which he starts with $2. Each game has 2 rounds [#permalink]
28 Apr 2015, 04:09

Expert's post

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

Harold plays a game in which he starts with $2. [#permalink]
28 Apr 2015, 05:04

1

This post received KUDOS

In each round, there is equal chance that one of the four operations is performed on the amount of money you start the round with x 1 + 1 x 0 + 0

Now Harold starts with $2. At the end of round 1, he'll be left with 1st case 2 x 1 = 2 2nd case 2 + 1 = 3 3rd case 2 x 0 = 0 4th case 2 + 0 = 2

Now he starts round 2 with either $2, 3, 0 or 2. So after playing round 2, he'll be left with 1st case (starting with $2) 2, 3, 0, 2 2nd case (starting with $3) 3, 4, 0, 3 3rd case (starting with $0) 0, 1, 0, 0 4th case (starting with $2) 2, 3, 0, 2

Each of these 16 outcomes has equal chance of occurring. So the probability that you end up with 0 is 6/16 (total 6 0s in the 16 outcomes) 1 is 1/16 2 is 4/16 = 1/4 3 is 4/16 = 1/4 4 is 1/16

Expected amount of money that Harold will be left with = (0*6/16) + (1*1/16) + (2*1/4) + (3*1/4) + (4*1/16) = 25/16, which is between $1.5 and $2

Re: Harold plays a game in which he starts with $2. Each game has 2 rounds [#permalink]
04 May 2015, 03:48

Expert's post

Bunuel wrote:

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

First, decode the game. In each round, you either add 1, multiply by 1, add 0, or multiply by 0. Each of those possibilities is equally likely. Now, notice that multiplying by 1 and adding 0 do the same thing: they leave the number unchanged. So, in effect, each round has 3 possible outcomes:

* Add 1 (25% or 1/4 chance)

* Leave the number unchanged (50% or 1/2 chance)

* Multiply by 0, turning the number to 0 (25% or 1/4 chance)

* Let’s now trace the game through. Harold starts with $2. After the first round, Harold either has $3 (1/4 chance), $2 (1/2 chance), or $0 (1/4 chance). Now figure out the second round from each starting point:

* Starting with $3: Harold finishes with $4 (1/4 chance), $3 (1/2 chance), or $0 (1/4 chance).

* Starting with $2: Harold finishes with $3 (1/4 chance), $2 (1/2 chance), or $0 (1/4 chance).

* Starting with $0: Harold finishes with $1 (1/4 chance) or $0 (1/2 + 1/4 = 3/4 chance).

* Add up the probabilities for each outcome, being sure to multiply by the probability of each starting point:

* Finish with $4: only one way, get to $3 in first round (1/4 chance) and then to $4 in second round (1/4 chance), for a total probability of (1/4)(1/4) = 1/16.

* Finish with $3: two ways (first $3 then $3, or first $2 then $3). The total probability is (1/4)(1/2) + (1/2)(1/4) = 1/4.

* Finish with $2: only one way (first $2 then $2). The probability is (1/2)(1/2) = 1/4.

* Finish with $1: only one way (first $0 then $1). The probability is (1/4)(1/4) = 1/16.

* Finish with $0: several ways (first $3 then $0, or first $2 then $0, or first $0 then $0). The probability is (1/4)(1/4) + (1/2)(1/4) + (1/4)(3/4) = 3/8.

This is a lot of computation! The key here is to be organized and fast. Also, check that the probabilities at the end of the second round add up to 1, as the ones above do.

Finally, to figure out the long-run average per game, you can simply multiply each outcome (in dollars) by its probability, then add all these products up.

Long-run average = ($4)(1/16) + ($3)(1/4) + ($2)(1/4) + ($1)(1/16) + ($0)(3/8) = 1/4 + 3/4 + 1/2 + 1/16 = 1 + 9/16, which is between $1.50 and $2.

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...