Harold plays a game in which he starts with $2.

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.

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.

Your explanation Karishma is very intuitive and clear.

Also Eli gave a good explanation. Thanks

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...

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

D is the correct answer here.

MANHATTAN GMAT OFFICIAL SOLUTION:

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.

The correct answer is D.

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