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

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

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New post 26 Nov 2010, 07:13
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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
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Re: Jim and Renee will play one game of Rock, Paper, Scissors.  [#permalink]

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New post 26 Nov 2010, 07:23
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anilnandyala 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 wi

5/6

2/3

1/2

5/12

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

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New post 17 Aug 2015, 09:41
Bunuel wrote:
anilnandyala 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 wi

5/6

2/3

1/2

5/12

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.


I got a bit confused reg tie. But now it is clear.
Thanks Bunuel for the solution.
Can u help me refer few more problems such as this?
What can be a general strategy to deal with this kind of problems?
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Re: Jim and Renee will play one game of Rock, Paper, Scissors.  [#permalink]

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New post 17 Aug 2015, 10:31
Mechmeera wrote:
Bunuel wrote:
anilnandyala 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 wi

5/6

2/3

1/2

5/12

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.


I got a bit confused reg tie. But now it is clear.
Thanks Bunuel for the solution.
Can u help me refer few more problems such as this?
What can be a general strategy to deal with this kind of problems?


Was not able to find similar questions to this one but you can practice more questions on probability and combinations from the links below:

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html


Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html


Also, check articles on these topics in our Important Topics Directory.

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: Jim and Renee will play one game of Rock, Paper, Scissors.  [#permalink]

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New post 27 Jul 2016, 06:44
Hello! First question here.

If for example they play 3 times and the question ask for the prob. that Jim will win the three times

Would it be 1/3*1/3*1/3=1/27?

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

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New post 27 Jul 2016, 06:58
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Re: Jim and Renee will play one game of Rock, Paper, Scissors.  [#permalink]

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New post 29 Aug 2016, 12:43
Top Contributor
anilnandyala 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?

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


Here's another way to look at it.
Let's say that Renee already knows what she is going to select. So, that part is done.

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

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

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New post 02 Sep 2018, 20:32
1
Probability of winning = (3C1*1C1) / (3C1*3C1) = 1/3
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Re: Jim and Renee will play one game of Rock, Paper, Scissors. &nbs [#permalink] 02 Sep 2018, 20:32
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