Manhattan GMAT Challenge Problem of the Week – 3 September 2012

by on September 3rd, 2012

Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people that enter our challenge, the better the prizes!

Question

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

Answer

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.

Special Announcement: If you want to win prizes for answering our Challenge Problems, try entering our Challenge Problem Showdown. Each week, we draw a winner from all the correct answers. The winner receives a number of our our Strategy Guides. The more people enter, the better the prize. Provided the winner gives consent, we will post his or her name on our Facebook page.

1 comment

  • hi,could you tell me de result for this 41+1x0 I said is 41
    (41+1)x0 I said is 0
    Thank you

Ask a Question or Leave a Reply

The author Manhattan Prep gets email notifications for all questions or replies to this post.

Some HTML allowed. Keep your comments above the belt or risk having them deleted. Signup for a Gravatar to have your pictures show up by your comment.