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# Harold plays a game in which he starts with $2.  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Moderator Joined: 01 Sep 2010 Posts: 2587 Followers: 424 Kudos [?]: 3371 [0], given: 720 Harold plays a game in which he starts with$2. [#permalink]  04 Sep 2012, 17:18
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Difficulty:

95% (hard)

Question Stats:

35% (02:43) correct 65% (02:01) wrong based on 75 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 [Reveal] Spoiler: OA _________________ Kaplan GMAT Instructor Joined: 25 Aug 2009 Posts: 644 Location: Cambridge, MA Followers: 72 Kudos [?]: 206 [1] , given: 2 Re: Harold plays a game in which he starts with$2. [#permalink]  04 Sep 2012, 22:38
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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!
_________________

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Prepare with Kaplan and save $150 on a course! Kaplan Reviews Director Joined: 22 Mar 2011 Posts: 612 WE: Science (Education) Followers: 78 Kudos [?]: 591 [1] , given: 43 Re: Harold plays a game in which he starts with$2. [#permalink]  04 Sep 2012, 22:57
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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. Intern Joined: 21 Aug 2012 Posts: 27 Followers: 0 Kudos [?]: 5 [1] , given: 4 Re: Harold plays a game in which he starts with$2. [#permalink]  04 Sep 2012, 22:58
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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. _________________ Please give kudos if you like my reply! Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5453 Location: Pune, India Followers: 1332 Kudos [?]: 6777 [3] , given: 177 Re: Harold plays a game in which he starts with$2. [#permalink]  05 Sep 2012, 02:11
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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.

Avg = (4*1 + 3*4 + 2*4 + 1*1 + 0*6)/16 = 25/16 = 1.56
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Re: Harold plays a game in which he starts with $2. [#permalink] 05 Sep 2012, 02:25 Expert's post Your explanation Karishma is very intuitive and clear. Also Eli gave a good explanation. Thanks _________________ Current Student Joined: 03 Sep 2012 Posts: 339 Location: United States Concentration: Healthcare, Strategy GMAT 1: 730 Q48 V42 GPA: 3.88 WE: Medicine and Health (Health Care) Followers: 11 Kudos [?]: 102 [0], given: 31 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...
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Re: Harold plays a game in which he starts with $2. [#permalink] 22 Mar 2014, 01:40 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Intern Joined: 16 Feb 2013 Posts: 8 Followers: 0 Kudos [?]: 0 [0], given: 9 Re: Harold plays a game in which he starts with$2. [#permalink]  22 Mar 2014, 04:13
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 Hi Bunuel, Could you please solve this problem for us? Thanks Math Expert Joined: 02 Sep 2009 Posts: 27169 Followers: 4226 Kudos [?]: 40974 [0], given: 5581 Re: Harold plays a game in which he starts with$2. [#permalink]  24 Mar 2014, 01:38
Expert's post
streamingline wrote:
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 Hi Bunuel, Could you please solve this problem for us? Thanks Can you please point out what didn't you understand here: harold-plays-a-game-in-which-he-starts-with-138374.html#p1118801 or here: harold-plays-a-game-in-which-he-starts-with-138374.html#p1118844. This way would be easier to address your doubts. Thank you. _________________ Re: Harold plays a game in which he starts with$2.   [#permalink] 24 Mar 2014, 01:38
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