Jim and Renee will play one game of Rock, Paper, Scissors. : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 23 Jan 2017, 06:17

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Jim and Renee will play one game of Rock, Paper, Scissors.

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Director
Joined: 01 May 2007
Posts: 792
Followers: 1

Kudos [?]: 287 [1] , given: 0

Jim and Renee will play one game of Rock, Paper, Scissors. [#permalink]

### Show Tags

11 Feb 2008, 08:50
1
This post received
KUDOS
4
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

58% (01:59) correct 42% (00:56) wrong based on 320 sessions

### HideShow timer Statistics

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?

A. 5/6
B. 2/3
C. 1/2
D. 5/12
E. 1/3
[Reveal] Spoiler: OA
Director
Joined: 01 Jan 2008
Posts: 629
Followers: 5

Kudos [?]: 176 [3] , given: 1

Re: Probability of Winning [#permalink]

### Show Tags

11 Feb 2008, 09:10
3
This post received
KUDOS
jimmyjamesdonkey 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

1/3 - equal probability of Jim winning, Renee winning and a tie.
SVP
Joined: 28 Dec 2005
Posts: 1575
Followers: 3

Kudos [?]: 148 [0], given: 2

Re: Probability of Winning [#permalink]

### Show Tags

11 Feb 2008, 19:35
1
This post was
BOOKMARKED
1/3

probability of jim winning is basically the sum of three probabilities:

probability of jim picking rock and renee picking scissors = 1/3*1/3 = 1/9

probability of jim picking scissors and renee picking paper = 1/3*1/3=1/9

probability of jim picking paper and renee picking rock = 1/3*1/3=1/9

1/9 + 1/9 + 1/9 = 3/9 = 1/3
Senior Manager
Joined: 20 Dec 2004
Posts: 255
Followers: 6

Kudos [?]: 106 [0], given: 0

Re: Probability of Winning [#permalink]

### Show Tags

11 Feb 2008, 20:24
maratikus wrote:
1/3 - equal probability of Jim winning, Renee winning and a tie.

Awesome...There cannot be any simpler solution to this problem than this one...

_________________

Stay Hungry, Stay Foolish

Intern
Joined: 03 Aug 2011
Posts: 46
Followers: 1

Kudos [?]: 17 [0], given: 17

Re: Probability of Winning [#permalink]

### Show Tags

13 Jun 2012, 22:34
You are wrong I think!.

In the stimulus there is no sign of definition of "TIE" situation! this comes from your experience in real world not from stimulus.
So the probability of losing is not 1/3 it becomes 2/3.(in the stimulus it had mentioned that just 1 situation of them game is wining situation so the other situations must be loosing situation as there is no explanation of tie situation)!

This means the probability becomes: 1C3 * 1/3*2/3=2/3...
_________________

Keep your eyes on the prize: 750

Kellogg MMM ThreadMaster
Joined: 28 Mar 2012
Posts: 314
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 29

Kudos [?]: 399 [0], given: 23

Re: Probability of Winning [#permalink]

### Show Tags

13 Jun 2012, 23:02
maratikus wrote:
jimmyjamesdonkey 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

1/3 - equal probability of Jim winning, Renee winning and a tie.

Pretty simple and straight forward apprach...
Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93516 [1] , given: 10568

Re: Probability of Winning [#permalink]

### Show Tags

13 Jun 2012, 23:07
1
This post received
KUDOS
Expert's post
omidsa wrote:
You are wrong I think!.

In the stimulus there is no sign of definition of "TIE" situation! this comes from your experience in real world not from stimulus.
So the probability of losing is not 1/3 it becomes 2/3.(in the stimulus it had mentioned that just 1 situation of them game is wining situation so the other situations must be loosing situation as there is no explanation of tie situation)!

This means the probability becomes: 1C3 * 1/3*2/3=2/3...

OA for this question is E (1/3) not 2/3. Refer to the solution below:

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?

A. 5/6
B. 2/3
C. 1/2
D. 5/12
E. 1/3

There are 3*3=9 combinations (cases) possible. In 3 cases there will be a tie (rock-rock, scissors/scissors, paper/paper). Now, out of 6 cases left Jim and Renee have equal chances of winning, so in 3 cases Jim will win and in other 3 Renee will win. So the probability Jim winning the game is (favorable outcomes) / (total # of outcomes) = 3/9 = 1/3.

Or: no matter what sign Renee will select Jim can select 1 sign to win, 1 sign to get tie and 1 sign to loose, so the probability Jim winning the game is 1/3.

Answer: E.

Hope it's clear.
_________________
Intern
Joined: 03 Aug 2011
Posts: 46
Followers: 1

Kudos [?]: 17 [0], given: 17

Re: Probability of Winning [#permalink]

### Show Tags

14 Jun 2012, 00:00
Bunuel wrote:
In 3 cases there will be a tie (rock-rock, scissors/scissors, paper/paper).

please explain me where did you get this from the question stimulus? How do you decide that there is tie situation? Please read the Stimulus again!

The Official Answer is wrong!
_________________

Keep your eyes on the prize: 750

Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93516 [0], given: 10568

Re: Probability of Winning [#permalink]

### Show Tags

14 Jun 2012, 00:03
omidsa wrote:
Bunuel wrote:
In 3 cases there will be a tie (rock-rock, scissors/scissors, paper/paper).

please explain me where did you get this from the question stimulus? How do you decide that there is tie situation? Please read the Stimulus again!

The Official Answer is wrong!

We are told that Jim and Renee have an equal chance of choosing any one of the hand signs, so if they both choose Paper for example there will be a tie. How else?

So, the official answer is not wrong.
_________________
Kellogg MMM ThreadMaster
Joined: 28 Mar 2012
Posts: 314
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 29

Kudos [?]: 399 [0], given: 23

Re: Probability of Winning [#permalink]

### Show Tags

14 Jun 2012, 00:09
omidsa wrote:
Bunuel wrote:
In 3 cases there will be a tie (rock-rock, scissors/scissors, paper/paper).

please explain me where did you get this from the question stimulus? How do you decide that there is tie situation? Please read the Stimulus again!

The Official Answer is wrong!

Don't assume it as tie, but the result would be unpredictable when rock/rock, scissors/scissors or paper/paper combination happens. Also, according to question it is clearly mentioned about the rules for winning.

So, lets say there are two situations:
1. Win
2. Not win

Win - 3 cases for winning,
Not win - 6 cases

So, probability of winning Jim = 3/(3+6)=1/3
Thus, the answer is (E).

Regards,

Last edited by cyberjadugar on 14 Jun 2012, 00:14, edited 2 times in total.
Intern
Joined: 03 Aug 2011
Posts: 46
Followers: 1

Kudos [?]: 17 [0], given: 17

Re: Probability of Winning [#permalink]

### Show Tags

14 Jun 2012, 00:12
We are told that Jim and Renee have an equal chance of choosing any one of the hand signs, so if they both choose Paper for example there will be a tie. How else?

So, the official answer is not wrong.[/quote]

Choosing freely each hand does not means that choosing rock-rock means tie! . please just consider the statements of given problem and forget what you know about the real world. suppose somebody that in his real life has not played SPR game! then how could he finds there is a tie situation?
_________________

Keep your eyes on the prize: 750

Intern
Joined: 03 Aug 2011
Posts: 46
Followers: 1

Kudos [?]: 17 [1] , given: 17

Re: Probability of Winning [#permalink]

### Show Tags

14 Jun 2012, 00:17
1
This post received
KUDOS
cyberjadugar wrote:
Don't assume it as tie, but the result would be unpredictable when rock/rock, scissors/scissors or paper/paper combination happens. Also, according to question it is clearly mentioned about the rules for winning.

So, lets say there are two situations:
1. Win
2. Not win

Win - 3 cases for winning,
Not win - 6 cases

So, probability of winning Jim = 3/(3+6)=1/3
Thus, the answer is (E).

Regards,

Good Point!
The probability of wining is 1/3 and not wining is 2/3 then becomes 3C1* 1/3*2/3 =2/3
am i right?
_________________

Keep your eyes on the prize: 750

Kellogg MMM ThreadMaster
Joined: 28 Mar 2012
Posts: 314
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 29

Kudos [?]: 399 [0], given: 23

Re: Probability of Winning [#permalink]

### Show Tags

14 Jun 2012, 00:24
omidsa wrote:
cyberjadugar wrote:
Don't assume it as tie, but the result would be unpredictable when rock/rock, scissors/scissors or paper/paper combination happens. Also, according to question it is clearly mentioned about the rules for winning.

So, lets say there are two situations:
1. Win
2. Not win

Win - 3 cases for winning,
Not win - 6 cases

So, probability of winning Jim = 3/(3+6)=1/3
Thus, the answer is (E).

Regards,

Good Point!
The probability of wining is 1/3 and not wining is 2/3 then becomes 3C1* 1/3*2/3 =2/3
am i right?

Hi omidsa,

You have correctly stated that the probability of wining is 1/3 and not wining is 2/3.

The expresion - 3C1* 1/3*2/3 =2/3, can also be viewed as:
(probability of choosing first sign)*(probability of choosing sign which results to not winning)
(3/3)*(2/3)=2/3, and I hope this is what you wanted to convey.

Regards,
Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93516 [0], given: 10568

Re: Probability of Winning [#permalink]

### Show Tags

14 Jun 2012, 00:38
omidsa wrote:

Choosing freely each hand does not means that choosing rock-rock means tie! . please just consider the statements of given problem and forget what you know about the real world. suppose somebody that in his real life has not played SPR game! then how could he finds there is a tie situation?

I think you over-thinking this simple problem. If somebody has never played this game he/she could use common sense to decide that paper/paper is a tie.
_________________
Intern
Joined: 03 Oct 2012
Posts: 10
GMAT 1: 620 Q39 V38
GMAT 2: 700 Q49 V38
Followers: 0

Kudos [?]: 6 [0], given: 23

Re: Probability of Winning [#permalink]

### Show Tags

22 Nov 2012, 00:22
Bunuel wrote:
omidsa wrote:
You are wrong I think!.

In the stimulus there is no sign of definition of "TIE" situation! this comes from your experience in real world not from stimulus.
So the probability of losing is not 1/3 it becomes 2/3.(in the stimulus it had mentioned that just 1 situation of them game is wining situation so the other situations must be loosing situation as there is no explanation of tie situation)!

This means the probability becomes: 1C3 * 1/3*2/3=2/3...

OA for this question is E (1/3) not 2/3. Refer to the solution below:

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?

A. 5/6
B. 2/3
C. 1/2
D. 5/12
E. 1/3

There are 3*3=9 combinations (cases) possible. In 3 cases there will be a tie (rock-rock, scissors/scissors, paper/paper). Now, out of 6 cases left Jim and Renee have equal chances of winning, so in 3 cases Jim will win and in other 3 Renee will win. So the probability Jim winning the game is (favorable outcomes) / (total # of outcomes) = 3/9 = 1/3.

Or: no matter what sign Renee will select Jim can select 1 sign to win, 1 sign to get tie and 1 sign to loose, so the probability Jim winning the game is 1/3.

Answer: E.

Hope it's clear.

I'm not sure that I'm right but since you admit the chance of a tie, than there will be 9 ways for Jim to win. 3 ways with one win and 2 ties and 3 with 2 wins and 1 tie. There will be total of 27 combinations. Am I right?
Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93516 [0], given: 10568

Re: Probability of Winning [#permalink]

### Show Tags

22 Nov 2012, 03:57
felixjkz wrote:
Bunuel wrote:
omidsa wrote:
You are wrong I think!.

In the stimulus there is no sign of definition of "TIE" situation! this comes from your experience in real world not from stimulus.
So the probability of losing is not 1/3 it becomes 2/3.(in the stimulus it had mentioned that just 1 situation of them game is wining situation so the other situations must be loosing situation as there is no explanation of tie situation)!

This means the probability becomes: 1C3 * 1/3*2/3=2/3...

OA for this question is E (1/3) not 2/3. Refer to the solution below:

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?

A. 5/6
B. 2/3
C. 1/2
D. 5/12
E. 1/3

There are 3*3=9 combinations (cases) possible. In 3 cases there will be a tie (rock-rock, scissors/scissors, paper/paper). Now, out of 6 cases left Jim and Renee have equal chances of winning, so in 3 cases Jim will win and in other 3 Renee will win. So the probability Jim winning the game is (favorable outcomes) / (total # of outcomes) = 3/9 = 1/3.

Or: no matter what sign Renee will select Jim can select 1 sign to win, 1 sign to get tie and 1 sign to loose, so the probability Jim winning the game is 1/3.

Answer: E.

Hope it's clear.

I'm not sure that I'm right but since you admit the chance of a tie, than there will be 9 ways for Jim to win. 3 ways with one win and 2 ties and 3 with 2 wins and 1 tie. There will be total of 27 combinations. Am I right?

No, that's not correct. Notice that they are playing one game, not three. So, there are total of 3*3=9 combinations, out of which there are 3 cases with at tie and 3 cases to win for each of the players.
_________________
Intern
Joined: 03 Oct 2012
Posts: 10
GMAT 1: 620 Q39 V38
GMAT 2: 700 Q49 V38
Followers: 0

Kudos [?]: 6 [0], given: 23

Re: Probability of Winning [#permalink]

### Show Tags

22 Nov 2012, 10:56
Hope it's clear.[/quote]
I'm not sure that I'm right but since you admit the chance of a tie, than there will be 9 ways for Jim to win. 3 ways with one win and 2 ties and 3 with 2 wins and 1 tie. There will be total of 27 combinations. Am I right?[/quote]

No, that's not correct. Notice that they are playing one game, not three. So, there are total of 3*3=9 combinations, out of which there are 3 cases with at tie and 3 cases to win for each of the players.[/quote]
My bad, I misinterpreted the question. Thank you!
Manager
Joined: 23 May 2013
Posts: 127
Followers: 1

Kudos [?]: 55 [0], given: 110

Re: Probability of Winning [#permalink]

### Show Tags

13 Sep 2013, 22:31
maratikus wrote:
jimmyjamesdonkey 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

1/3 - equal probability of Jim winning, Renee winning and a tie.

So if we go by this logic, if there are 3 players - regardless of the game - the probability of each one winning is 1/4, if 4 players then 1/5, if 5 players then 1/6....and so on.
Is my understanding correct?
_________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Intern
Joined: 03 Oct 2012
Posts: 26
Concentration: Strategy, General Management
GMAT 1: 750 Q49 V42
Followers: 0

Kudos [?]: 9 [0], given: 142

Re: Jim and Renee will play one game of Rock, Paper, Scissors. [#permalink]

### Show Tags

22 Jul 2014, 22:41
Here's my approach:

There are 3 signs each can pick and there are 3 people. Hence the Total Outcomes possible are 3^3 = 27

We want One sign to win which implies that the other two have to be the same (3C2 ways of picking this). And that one sign can be picked in 3 ways.

Therefore Answer = (3*3c2)/3^3 = 9/27 = 1/3.

Is this right?
Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93516 [0], given: 10568

Re: Jim and Renee will play one game of Rock, Paper, Scissors. [#permalink]

### Show Tags

23 Jul 2014, 01:29
SachinWordsmith wrote:
Here's my approach:

There are 3 signs each can pick and there are 3 people. Hence the Total Outcomes possible are 3^3 = 27

We want One sign to win which implies that the other two have to be the same (3C2 ways of picking this). And that one sign can be picked in 3 ways.

Therefore Answer = (3*3c2)/3^3 = 9/27 = 1/3.

Is this right?

No, that's not right. Even not to analyse the rest of your solution notice that there are TWO people playing Jim and Renee, not three people. Please refer to the discussion above for a correct approach.
_________________
Re: Jim and Renee will play one game of Rock, Paper, Scissors.   [#permalink] 23 Jul 2014, 01:29

Go to page    1   2    Next  [ 22 posts ]

Similar topics Replies Last post
Similar
Topics:
4 A game is played with a six sided, 1 28 Dec 2015, 22:45
5 The game of blackjack is played with a deck consisting of 7 27 Feb 2013, 03:54
A and B decide to play a game based on probabilities. They 3 07 Aug 2012, 06:17
45 Rita and Sam play the following game with n sticks on a 12 03 Apr 2012, 12:02
20 Jim and Renee will play one game of Rock, Paper, Scissors. 7 26 Nov 2010, 06:13
Display posts from previous: Sort by

# Jim and Renee will play one game of Rock, Paper, Scissors.

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.