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Probability-Manhattan CAT

Expert replies
by rishijhawar » Tue Jul 19, 2011 11:58 am
Jim and Renee will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Jim and Renee have an equal chance of choosing any one of the hand signs, what is the probability that Jim will win?
5/6
2/3
1/2
5/12
1/3
OA to follow.
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Source: — Data Sufficiency |

by Brent@GMATPrepNow » Tue Jul 19, 2011 1:28 pm
rishijhawar wrote:Jim and Renee will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Jim and Renee have an equal chance of choosing any one of the hand signs, what is the probability that Jim will win?
5/6
2/3
1/2
5/12
1/3
OA to follow.
A quick way to solve this question is to first recognize that they will tie 1/3 of the time. For example, if Jim picks Rock, there's a 1/3 chance Renee will also picks Rock. Similarly, if Jim picks Paper, there's a 1/3 chance Renee will also picks Paper etc.

So, if there's a 1/3 chance they tie, there must be a 2/3 chance they don't tie (i.e., someone wins).

At this point, we should recognize that, if they do not tie, each person has an equal chance of winning (unless one of them is a mind-reader).

In other words, we can expect Jim to win 1/2 of the games that are not ties.
So, Jim should win 1/3 of all games (since 1/2 of 2/3 is 1/3)

The answer is E
Cheers,
Brent
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by Brent@GMATPrepNow » Tue Jul 19, 2011 1:42 pm
rishijhawar wrote:Jim and Renee will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Jim and Renee have an equal chance of choosing any one of the hand signs, what is the probability that Jim will win?
5/6
2/3
1/2
5/12
1/3
OA to follow.
Another option is to use the basic probability formula along with some counting.

So P(Jim wins) = [# of outcomes where Jim wins] / [total number of possible outcomes]

Let's begin with the denominator (it's always best to begin with the denominator - you'll see why shortly)

If Jim can choose a hand sign in 3 ways and Renee can choose a hand sign in 3 ways, the total number of possible outcomes equals 9 (3 x 3 = 9)

At this point, we know that P(Jim wins) = [# of outcomes where Jim wins] / 9
This means we can eliminate answer choices A, C, and D since it is impossible for ??/9 to simplify to equal 5/6, 1/2 or 5/12

We're now left with 2/3 and 1/3

From here, we could apply some logic. Since each person has an equal probability of winning, the answer cannot be 2/3 since 2/3 + 2/3 is greater than 1. So, the answer must be E

Alternatively, we could list all of the outcomes where Jim wins and then count them.

Aside: In many cases, listing and counting outcomes is a LOT of work. However, since the total number of outcomes is 9, I know that the number of outcomes where Jim wins will be less than 9, so it might be quite easy to just list them.

Here are the outcomes where Jim wins:
1. J=Rock, R=Scissors
2. J=Scissors, R=Paper
3. J=Paper, R=Rock

So, the [# of outcomes where Jim wins] = 3, which means P(Jim wins) = 3 / 9
= 1/3
= E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Brent@GMATPrepNow » Tue Jul 19, 2011 1:48 pm
rishijhawar wrote:Jim and Renee will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Jim and Renee have an equal chance of choosing any one of the hand signs, what is the probability that Jim will win?
5/6
2/3
1/2
5/12
1/3
OA to follow.
Here's another way to look at it.
Let's say that Renee already knows what she is going to select. It's a done deal.

At this point, what are the possible outcomes?
a) Jim selects the one sign that allows him to win
b) Jim selects the one sign that allows Renee to win
c) Jim selects the one sign that results in a tie.

Since each of these outcomes is equally likely, the probability of each outcome is 1/3
So, the probability that Jim wins is 1/3

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by rishijhawar » Wed Jul 20, 2011 11:30 pm
Brent, your explanations are great, as always. Never thought of so many ways. Thanks so much.
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by amit2k9 » Thu Jul 21, 2011 3:01 am
this question has been discussed before a lot.
3 options hence probability of each chance = 1/3

R-S,P-S and S-P are the only 3 outcomes possible.

hence in one chance the probability is 1/3 only.
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by Brent@GMATPrepNow » Fri Jul 22, 2011 6:07 am
amit2k9 wrote:this question has been discussed before a lot.
3 options hence probability of each chance = 1/3

R-S,P-S and S-P are the only 3 outcomes possible.

hence in one chance the probability is 1/3 only.
I think I should point out that there are actually 9 possible outcomes (organized here as Jim-Renee):
R-R
R-P
R-S
P-R
P-P
P-S
S-R
S-P
S-S

Since 3 of the 9 outcomes are favorable to Jim winning, the probability is 3/9 = 1/3

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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