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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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11 Feb 2008, 09:50
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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
[Reveal] Spoiler: OA
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11 Feb 2008, 10:10
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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.
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11 Feb 2008, 20:35
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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
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11 Feb 2008, 21:24
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...

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13 Jun 2012, 23:34
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...
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14 Jun 2012, 00:02
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...
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14 Jun 2012, 00:07
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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.

Hope it's clear.
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14 Jun 2012, 01:00
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!

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14 Jun 2012, 01:03
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!

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.
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14 Jun 2012, 01:09
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!

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

Regards,

Last edited by cyberjadugar on 14 Jun 2012, 01:14, edited 2 times in total.
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14 Jun 2012, 01:12
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?
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14 Jun 2012, 01:17
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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

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?
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14 Jun 2012, 01:24
omidsa 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

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,
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14 Jun 2012, 01:38
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.
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22 Nov 2012, 01:22
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.

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?
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22 Nov 2012, 04:57
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.

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.
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22 Nov 2012, 11:56
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!
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13 Sep 2013, 23:31
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?
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Re: Jim and Renee will play one game of Rock, Paper, Scissors. [#permalink]

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22 Jul 2014, 23:41
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?
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Re: Jim and Renee will play one game of Rock, Paper, Scissors. [#permalink]

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23 Jul 2014, 02:29
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.
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Re: Jim and Renee will play one game of Rock, Paper, Scissors.   [#permalink] 23 Jul 2014, 02:29

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