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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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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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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:

Cheers,
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 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,
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, 01:45
1
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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Re: Three children, John, Paul, and Ringo, are playing a game. Each child &nbs [#permalink] 18 Sep 2018, 03:22
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