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Three children, John, Paul, and Ringo, are playing a game. Each child

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Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

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New post 18 May 2017, 13:05
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A
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C
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Three children, John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number different from those of the other two children, he is declared the winner. If all of the children choose the same number, the process repeats until one child is declared the winner. If Ringo always chooses 2 and the other children select numbers randomly, what is the probability that Ringo is declared the winner?

A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3

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Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

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New post 22 Jan 2019, 09:01
Official Explanation

The best way to think about this question is to focus on the final round, the one in which two children choose one number and one child chooses the other, so that there is a winner. It doesn’t really matter whether this is the first round or the hundredth, the probabilities will come out the same. In this final round, there are three possibilities: Ringo chooses 2 while both John and Paul choose 1; Ringo and John choose 2 while Paul chooses 1; and Ringo and Paul choose 2 while John chooses 1. Each of the three possibilities is equally likely, and Ringo wins in only one of them, so the probability that he wins is \(\frac{1}{3}\). Choose (C).


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Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

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New post 18 May 2017, 13:10
Is the answer 1/2?

I'm thinking that Ringo selects 2. So the other two have to select 1.

so the probability that the first one can selects 1 is 1/2. the second has to select 1. so answer 1/2?
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

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New post 19 May 2017, 11:05
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SajjadAhmad wrote:
Three children, John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number different from those of the other two children, he is declared the winner. If all of the children choose the same number, the process repeats until one child is declared the winner. If Ringo always chooses 2 and the other children select numbers randomly, what is the probability that Ringo is declared the winner?

A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3


This is one of my all-time favorite questions!!

The main concept here is that all 3 children are equally likely to win this game (unless one of them possesses supernatural powers that allow him to know what numbers the other two boys will choose :-))

Also note that, if everything is random, the probability of winning by choosing the number 2 is the same as the probability of winning by choosing the number 1.

So, regardless of what number Ringo chooses, his probability of winning is exactly the same as each of the other boys winning.

Since all 3 boys have the same probability of winning, P(Ringo wins) = 1/3

Likewise, P(John wins) = 1/3 and P(Paul wins) = 1/3

Answer:

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Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

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New post 19 May 2017, 11:13
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This is one of my all-time favorite questions!!

The main concept here is that all 3 children are equally likely to win this game (unless one of them possesses supernatural powers that allow him to know what numbers the other two boys will choose :-))

Also note that, if everything is random, the probability of winning by choosing the number 2 is the same as the probability of winning by choosing the number 1.

So, regardless of what number Ringo chooses, his probability of winning is exactly the same as each of the other boys winning.

Since all 3 boys have the same probability of winning, P(Ringo wins) = 1/3

Likewise, P(John wins) = 1/3 and P(Paul wins) = 1/3

Answer:

Cheers,
Brent[/quote]

I'm having trouble with this one.

I wrote down the other two outcomes that are possible:

1,2
2,1
2,2
1,1

He only wins if they get 1,1. So how is it not 1/4?

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Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

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New post 19 May 2017, 12:45
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gmathopeful19 wrote:
I'm having trouble with this one.

I wrote down the other two outcomes that are possible:

1,2
2,1
2,2
1,1

He only wins if they get 1,1. So how is it not 1/4?

Posted from my mobile device


Great question!
You are answering a slightly different question.
The question asks us to find P(Ringo is declared the winner)
Your solution is the correct answer to P(Ringo wins the FIRST round of the game).

Keep in mind that it's possible that nobody wins in the first round, in which case they need to play again, until someone is declared the winner.

In fact, P(Ringo wins the FIRST round of the game) = 1/4
P(John wins the FIRST round of the game) = 1/4
P(Paul wins the FIRST round of the game) = 1/4
P(NOBODY wins the FIRST round of the game) = 1/4

Cheers,
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

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New post 06 Jul 2017, 01:45
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GMATPrepNow wrote:
gmathopeful19 wrote:
I'm having trouble with this one.

I wrote down the other two outcomes that are possible:

1,2
2,1
2,2
1,1

He only wins if they get 1,1. So how is it not 1/4?

Posted from my mobile device


Great question!
You are answering a slightly different question.
The question asks us to find P(Ringo is declared the winner)
Your solution is the correct answer to P(Ringo wins the FIRST round of the game).

Keep in mind that it's possible that nobody wins in the first round, in which case they need to play again, until someone is declared the winner.

In fact, P(Ringo wins the FIRST round of the game) = 1/4
P(John wins the FIRST round of the game) = 1/4
P(Paul wins the FIRST round of the game) = 1/4
P(NOBODY wins the FIRST round of the game) = 1/4

Cheers,
Brent

This is a question that I never get right. I actually got confused with two possible ways of finding the answers. The first one of course led to a 1/4, by similar reasoning used by others. The other is that Ringo does win any round of the game.
Let me elucidate.

\(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}(\frac{1}{4}+\frac{1}{4}+.....)\)

I can solve this one algebraically but this also wouldn't result in the answer.

Now please also help me with this - How do you define the event?
Is it

Ringo wins the first round?
or
Ringo wins any round?
or
In how many ways can Ringo win the game?
or something else?

VeritasPrepKarishma Bunuel, I request other experts to also chip in.

Many thanks GMATPrepNow Brent
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

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New post 06 Jul 2017, 03:15
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Probability of declaring Ringo a winner = No. of ways John and Paul chooses 1/No. of ways John and Paul chooses a different number

No. of ways John and Paul chooses 1 = 1.
No. of ways John and Paul chooses a different number = [J-1,P2], [P-1,J-2], [J-1,P-1] = 3.

Probability = 1/3. Ans - C.
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

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New post 17 Jan 2019, 05:39
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SajjadAhmad wrote:
Three children, John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number different from those of the other two children, he is declared the winner. If all of the children choose the same number, the process repeats until one child is declared the winner. If Ringo always chooses 2 and the other children select numbers randomly, what is the probability that Ringo is declared the winner?

A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3

\(? = P\left( {{\rm{Ringo}}\,\,{\rm{wins}}} \right)\)

Brent´s solution explores symmetries and it is probably the GMAT-scope way of dealing with this good question.

In the same "vibe", I prefer the following argument:

1. Ringo´s fixed preference is irrelevant once it is known. (I mean, if he always chooses 1 is exactly the same problem, hence the answer cannot be different.)

2. John and Paul have no edges between them (sure!) and each one has no edge facing Ringo (and vice-versa) . Let´s see from (say) Paul´s perception:

> If John chooses like Ringo (John chooses 2), Paul may win (choosing 1) or may let them go to the next round (choosing 2) :: there is an edge of Paul over Ringo.
> If John does not choose like Ringo (John chooses 1), Paul may be beaten by Ringo (Paul choosing 1) or he may be beaten by John (Paul choosing 2) :: there is an edge of Ringo over Paul.

In short: Paul has no edge over Ringo (and vice-versa), and the same reasoning applies to conclude that John also does not have an edge over Ringo (and vice-versa).


Conclusion: all of them have the same probability of winning the game, hence 1/3 is the correct answer.


Now let´s see a much clearer solution, but in the end I will use the sum of a convergent geometric sequence with infinite number of terms... that´s out of GMAT´s scope...

\({\rm{Total}}\,\,\left( {R,J,P} \right) = \left( {2,J,P} \right)\,\,:\,\,\,1 \cdot 2 \cdot 2 = 4\,\,{\rm{equiprobable}}\,\,{\rm{possibilities}}\,\,{\rm{per}}\,\,{\rm{round}}\,\,\,\left( * \right)\)

\({\left( {R,J,P} \right)_N}\,\,\,{\rm{indicates}}\,\,{\rm{choices}}\,\,{\rm{in}}\,\,{\rm{round}}\,\,N\)

\({\rm{Favorable}}\,\,:\,\,\,\left\{ \matrix{
\,{\left( {R,J,P} \right)_1} = \left( {2,1,1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{1 \over 4} \hfill \cr
\,{\left( {R,J,P} \right)_1} = \left( {2,2,2} \right)\,\,\,{\rm{and}}\,\,\,\,{\left( {R,J,P} \right)_2} = \left( {2,1,1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{1 \over 4} \cdot {1 \over 4} \hfill \cr
\,{\left( {R,J,P} \right)_1} = \left( {2,2,2} \right)\,\,\,{\rm{and}}\,\,\,\,{\left( {R,J,P} \right)_2} = \left( {2,2,2} \right)\,\,\,{\rm{and}}\,\,\,{\left( {R,J,P} \right)_3} = \left( {2,1,1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{1 \over 4} \cdot {1 \over 4} \cdot {1 \over 4} \hfill \cr
\, \ldots \hfill \cr} \right.\)

\(? = {1 \over 4} + {\left( {{1 \over 4}} \right)^2} + {\left( {{1 \over 4}} \right)^3} + \ldots = {{{1 \over 4}} \over {1 - {1 \over 4}}} = {1 \over 3}\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

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New post 17 Jan 2019, 19:23
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Hi All,

We're told that three children - John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number DIFFERENT from those of the other two children, he is declared the winner. If all of the children choose the SAME number, the process repeats until one child is declared the winner. We're told that Ringo ALWAYS chooses 2 and the other children select numbers randomly. We're asked for the probability that Ringo is declared the winner. This question can be approached in a number of different ways. Since the number of possible outcomes is so small, you can actually just list them all out to determine the answer to this question:

Ringo ALWAYS chooses 2...
IF>..
John = 1 and Paul = 1, then Ringo WINS
John = 1 and Paul = 2, then Ringo loses
John = 2 and Paul = 1, then Ringo loses
John = 2 and Paul = 2, then the game is REPLAYED.

We're told that the game is played (and replayed) until there is a winner, so there really aren't 4 outcomes - there are only 3 outcomes. Since each winning outcome is equally likely, Ringo will win 1/3 of them.

Final Answer:

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Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

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New post 21 Jan 2019, 18:58
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SajjadAhmad wrote:
Three children, John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number different from those of the other two children, he is declared the winner. If all of the children choose the same number, the process repeats until one child is declared the winner. If Ringo always chooses 2 and the other children select numbers randomly, what is the probability that Ringo is declared the winner?

A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3


Since Ringo always chooses the number 2, there are only 4 possible outcomes for any round of the game. We list Ringo’s choice first, followed by John’s and then Paul’s:

(2, 1, 1) Ringo wins

(2, 1, 2) Ringo loses (John wins)

(2, 2, 1) Ringo loses (Paul wins)

(2, 2, 2) No one wins, so another round is played. When this happens, the same 4 outcomes occur for the next round and for subsequent rounds.

We see that in any given round, each boy has a 1/3 chance of winning. It doesn’t matter if Ringo always chooses 2 or just randomly chooses a number, his probability of winning is 1/3.

Answer: C
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child   [#permalink] 21 Jan 2019, 18:58
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