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# As part of a game, four people each must secretly choose an

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06 Aug 2010, 22:40
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As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive.

What is the approximate likelihood that 2 people will choose same number?
What is the approximate likelihood that 3 people will choose same number?
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Re: Probability - simple question.  [#permalink]

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07 Aug 2010, 01:15
4
5
Financier wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive.
What is the approximate likelihood that 2 people will choose same number?
What is the approximate likelihood that 3 people will choose same number?

Yes, this is a veeery simple question, but I want to understand in how many ways this question can be cracked.
The more ways we know - the greater our confidence is

When four people choose an integer between 1 and 4, inclusive 5 cases are possible:

A. All choose different numbers - {a,b,c,d};
B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c};
C. 2 people choose same number and other 2 also choose same number - {a,a,b,b};
D. 3 people choose same number - {a,a,a,b};
E. All choose same number - {a,a,a,a}.

Some notes before solving:
As only these 5 cases are possible then the sum of their individual probabilities must be 1: $$P(A)+P(B)+P(C)+P(D)+P(E)=1$$

$$Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}$$

As each person has 4 options, integers from 1 to 4, inclusive, thus denominator, total # of outcomes would be 4^4 for all cases.

A. All choose different numbers - {a,b,c,d}:

$$P(A)=\frac{4!}{4^4}=\frac{24}{256}$$.

# of ways to "assign" four different objects (numbers 1, 2, 3, and 4) to 4 persons is 4!.

B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c}:

$$P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}$$.

$$C^2_4$$ - # of ways to choose which 2 persons will have the same number;
$$4$$ - # of ways to choose which number it will be;
$$P^2_3$$ - # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters;

C. 2 people choose same number and other 2 also choose same number - {a,a,b,b}:

$$P(C)=\frac{{C^2_4*\frac{4!}{2!2!}}}{4^4}=\frac{36}{256}$$.

$$C^2_4$$ - # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b};
$$\frac{4!}{2!2!}$$ - # of ways to "assign" 4 objects out of which 2 a's and 2 b's are identical to 4 persons;

D. 3 people choose same number - {a,a,a,b}:

$$P(D)=\frac{C^3_4*4*3}{4^4}=\frac{48}{256}$$.

$$C^3_4$$ - # of ways to choose which 3 persons out of 4 will have same number;
$$4$$ - # of ways to choose which number it will be;
$$3$$ - options for 4th person.

E. All choose same number - {a,a,a,a}:

$$P(E)=\frac{4}{4^4}=\frac{4}{256}$$.

$$4$$ - options for the number which will be the same.

Checking: $$P(A)+P(B)+P(C)+P(D)+P(E)=\frac{24}{256}+\frac{144}{256}+\frac{36}{256}+\frac{48}{256}+\frac{4}{256}=1$$.

Hope it's clear.
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Re: Probability - simple question.  [#permalink]

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03 Sep 2010, 14:49
In case 2 why does the order matter ...could you plz give a brief on that one
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03 Sep 2010, 15:04
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utkarshlavania wrote:
In case 2 why does the order matter ...could you plz give a brief on that one

B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c}:

$$P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}$$.

$$C^2_4$$ - # of ways to choose which 2 persons will have the same number;
$$4$$ - # of ways to choose which number it will be;
$$P^2_3$$ - # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters;

Suppose the number which is repeated and chosen by 2 people is 4.

Now, 2 other people must choose 2 different numbers out of 1, 2, and 3.
XY
12
21
13
31
23
32

You can see that X choose 1 and Y choose 2 is different from Y choose 1 and X choose 2 (different scenario). That's why the order of the chosen numbers matters.

Hope it's clear.
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Re: Probability - simple question.  [#permalink]

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03 Sep 2010, 19:27
1
Bunuel for the 2 case i.e., 2 people selecting the same number:
I wanted to work the probability through the "selection" way - meaning taking the prob of of each and multiplying through, can you point doublecheck that this is correct too (the logic) -

So I can select 2 of the 4 in 4C2 ways. Now one of these 2 can select any of the 4 numbers and the second one has to select the same number, the probability of that happening is 4/4 * 1/4
The remaining 2 people can select any of the 3 numbers 3/4 x 3/4, so the total prob is 4C2 3/(4^3) - this is what you get too, I see the numbers match, but I wanted to make sure of the logic..

But for the case when each selects a different number, if I apply the same kind of logic
Combinations of 4 people 4x3x2x1, for each combination the probability of selecting 4 different numbers =
first person can choose 4/4
second can choose 1/3
third can choose 1/2
Multiplying them all 4!/(2 3 4) =1 , that does not make sense... I understand your answer, can you please tell me where I am wrong?
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Re: Probability - simple question.  [#permalink]

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04 Sep 2010, 10:29
thanks once again bunuel
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Re: Probability - simple question.  [#permalink]

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04 Sep 2010, 22:27
Thanks for the question as well as for the answer
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Re: Probability - simple question.  [#permalink]

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12 Oct 2010, 06:08
c) [a,a,b,b]

4C2 4 3C2 3/256

4C2 is no of ways of selecting 2 persons who have same number
4 is the no of ways in which no it will be
3C2 is the no of ways of selcting 2 persons who have same number
3 is the no of ways in which no it will be

can some one explain whats wrong with this?

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Re: Probability - simple question.  [#permalink]

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12 Oct 2010, 07:57
1
anilnandyala wrote:
c) [a,a,b,b]

4C2 4 3C2 3/256

4C2 is no of ways of selecting 2 persons who have same number
4 is the no of ways in which no it will be
3C2 is the no of ways of selcting 2 persons who have same number
3 is the no of ways in which no it will be

can some one explain whats wrong with this?

If you doing this way then after first selection of 2 persons there are only 2 more left not 3, so 2C2 not 3C2. Also you should divide this by 2! to get rid of duplications, so you would have 4C2*4*2C2*3/2=36.
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Re: Probability - simple question.  [#permalink]

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31 May 2011, 22:58
mainhoon wrote:
Bunuel for the 2 case i.e., 2 people selecting the same number:
I wanted to work the probability through the "selection" way - meaning taking the prob of of each and multiplying through, can you point doublecheck that this is correct too (the logic) -

But for the case when each selects a different number, if I apply the same kind of logic
Combinations of 4 people 4x3x2x1, for each combination the probability of selecting 4 different numbers =
first person can choose 4/4
second can choose 1/3
third can choose 1/2
Multiplying them all 4!/(2 3 4) =1 , that does not make sense... I understand your answer, can you please tell me where I am wrong?

For the case(a,b,c,d)=
1st chooses 4/4
2nd chooses 3/4
3rd chooses 2/4
4th chooses 1/4

So we have 4/4*3/4*2/4*1/4=4!/4^4
Similarly for (a,a,b,c) = 4C2*4/4*1/4*3/4*2/4
For (a,a,b,b) = 4C2*4/4*1/4*3/4*1/4*1/2!
For(a,a,a,b) = 4C3*4/4*1/4*1/4*3/4
For (a,a,a,a)= 4/4*1/4*1/4*1/4
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Re: Probability - simple question.  [#permalink]

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29 Aug 2011, 20:27
someonear wrote:
mainhoon wrote:
Bunuel for the 2 case i.e., 2 people selecting the same number:
I wanted to work the probability through the "selection" way - meaning taking the prob of of each and multiplying through, can you point doublecheck that this is correct too (the logic) -

But for the case when each selects a different number, if I apply the same kind of logic
Combinations of 4 people 4x3x2x1, for each combination the probability of selecting 4 different numbers =
first person can choose 4/4
second can choose 1/3
third can choose 1/2
Multiplying them all 4!/(2 3 4) =1 , that does not make sense... I understand your answer, can you please tell me where I am wrong?

For the case(a,b,c,d)=
1st chooses 4/4
2nd chooses 3/4
3rd chooses 2/4
4th chooses 1/4

So we have 4/4*3/4*2/4*1/4=4!/4^4
Similarly for (a,a,b,c) = 4C2*4/4*1/4*3/4*2/4
For (a,a,b,b) = 4C2*4/4*1/4*3/4*1/4*1/2!
For(a,a,a,b) = 4C3*4/4*1/4*1/4*3/4
For (a,a,a,a)= 4/4*1/4*1/4*1/4

where did you get the 4c2 in the AABC case?

i had the following in place of the 4c2 which gives the wrong answer, I can't for the life of me figure out why it's 4c2

then you can re-arrange everything since the order in this case matters, since the order represents different people

[(4/4) * (1/4) * (3/4) * (2/4)] * (4! / 2!)

where 4! allows me to re-arrange everything and 2! removes the duplicates for person 1 and 2.

am i doing something wrong?
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Re: Probability - simple question.  [#permalink]

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30 Aug 2011, 07:20
pinchharmonic wrote:
someonear wrote:
mainhoon wrote:
Bunuel for the 2 case i.e., 2 people selecting the same number:
I wanted to work the probability through the "selection" way - meaning taking the prob of of each and multiplying through, can you point doublecheck that this is correct too (the logic) -

But for the case when each selects a different number, if I apply the same kind of logic
Combinations of 4 people 4x3x2x1, for each combination the probability of selecting 4 different numbers =
first person can choose 4/4
second can choose 1/3
third can choose 1/2
Multiplying them all 4!/(2 3 4) =1 , that does not make sense... I understand your answer, can you please tell me where I am wrong?

For the case(a,b,c,d)=
1st chooses 4/4
2nd chooses 3/4
3rd chooses 2/4
4th chooses 1/4

So we have 4/4*3/4*2/4*1/4=4!/4^4
Similarly for (a,a,b,c) = 4C2*4/4*1/4*3/4*2/4
For (a,a,b,b) = 4C2*4/4*1/4*3/4*1/4*1/2!
For(a,a,a,b) = 4C3*4/4*1/4*1/4*3/4
For (a,a,a,a)= 4/4*1/4*1/4*1/4

where did you get the 4c2 in the AABC case?

i had the following in place of the 4c2 which gives the wrong answer, I can't for the life of me figure out why it's 4c2

then you can re-arrange everything since the order in this case matters, since the order represents different people

[(4/4) * (1/4) * (3/4) * (2/4)] * (4! / 2!)

where 4! allows me to re-arrange everything and 2! removes the duplicates for person 1 and 2.

am i doing something wrong?

Think of AABC has a word with repeating alphabets A. So AABC can be arranged in 4!/2! ways
Now as ordering is important here we have
4!/2! * (4/4*1/4*3/4*2/4)*1/2 as final answer where we can say 4!/2! * 1/2 is nothing but 4C2
Hope this helps
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Re: Probability - simple question.  [#permalink]

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30 Aug 2011, 12:21
Quote:
Think of AABC has a word with repeating alphabets A. So AABC can be arranged in 4!/2! ways
Now as ordering is important here we have
4!/2! * (4/4*1/4*3/4*2/4)*1/2 as final answer where we can say 4!/2! * 1/2 is nothing but 4C2
Hope this helps

thanks, but i'm on point with everythign you said, but where is the 1/2 from? I also re-arranged the 4 letters using (4!/2!), but I can't figure out why you have the 1/2 at the end.
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Re: Probability - simple question.  [#permalink]

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31 Aug 2011, 14:00
pinchharmonic wrote:
Quote:
Think of AABC has a word with repeating alphabets A. So AABC can be arranged in 4!/2! ways
Now as ordering is important here we have
4!/2! * (4/4*1/4*3/4*2/4)*1/2 as final answer where we can say 4!/2! * 1/2 is nothing but 4C2
Hope this helps

thanks, but i'm on point with everythign you said, but where is the 1/2 from? I also re-arranged the 4 letters using (4!/2!), but I can't figure out why you have the 1/2 at the end.

bumping for an explanation of one of these approaches, thanks!
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Re: Probability - simple question.  [#permalink]

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26 May 2013, 22:37
Bunuel wrote:
anilnandyala wrote:
c) [a,a,b,b]

4C2 4 3C2 3/256

4C2 is no of ways of selecting 2 persons who have same number
4 is the no of ways in which no it will be
3C2 is the no of ways of selcting 2 persons who have same number
3 is the no of ways in which no it will be

can some one explain whats wrong with this?

If you doing this way then after first selection of 2 persons there are only 2 more left not 3, so 2C2 not 3C2. Also you should divide this by 2! to get rid of duplications, so you would have 4C2*4*2C2*3/2=36.

Hi Bunnel,

Here why do we have to divide by 2? I didn't get this part.

No. of cases for 2 pairs choosing two different nos. = 4c2(no. of 2's pairs)*4c1(no. which the first pair can choose)*3c1(no which second pair can choose) = 72

I am not making arrangements here these are only combinations, why is there repetition here?
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Re: Probability - simple question.  [#permalink]

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26 May 2013, 23:23
cumulonimbus wrote:
Bunuel wrote:
anilnandyala wrote:
c) [a,a,b,b]

4C2 4 3C2 3/256

4C2 is no of ways of selecting 2 persons who have same number
4 is the no of ways in which no it will be
3C2 is the no of ways of selcting 2 persons who have same number
3 is the no of ways in which no it will be

can some one explain whats wrong with this?

If you doing this way then after first selection of 2 persons there are only 2 more left not 3, so 2C2 not 3C2. Also you should divide this by 2! to get rid of duplications, so you would have 4C2*4*2C2*3/2=36.

Hi Bunnel,

Here why do we have to divide by 2? I didn't get this part.

No. of cases for 2 pairs choosing two different nos. = 4c2(no. of 2's pairs)*4c1(no. which the first pair can choose)*3c1(no which second pair can choose) = 72

I am not making arrangements here these are only combinations, why is there repetition here?

Suppose you choose numbers 1 and 2. Now, in how many ways can you assign 1, 1, 2, 2, to W, X, Y, Z? In 12 ways or in 6?

The correct answer is 6: 4!/(2!2!)=6. Listing the cases might help:
W - X - Y - Z
1 - 1 - 2 - 2
2 - 2 - 1 - 1
1 - 2 - 1 - 2
1 - 2 - 2 - 1
2 - 1 - 1 - 2
2 - 1 - 2 - 1
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Re: Probability - simple question.  [#permalink]

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27 May 2013, 19:38
Hi Bunnel,

Here why do we have to divide by 2? I didn't get this part.

No. of cases for 2 pairs choosing two different nos. = 4c2(no. of 2's pairs)*4c1(no. which the first pair can choose)*3c1(no which second pair can choose) = 72

I am not making arrangements here these are only combinations, why is there repetition here?[/quote]

Suppose you choose numbers 1 and 2. Now, in how many ways can you assign 1, 1, 2, 2, to W, X, Y, Z? In 12 ways or in 6?

The correct answer is 6: 4!/(2!2!)=6. Listing the cases might help:
W - X - Y - Z
1 - 1 - 2 - 2
2 - 2 - 1 - 1
1 - 2 - 1 - 2
1 - 2 - 2 - 1
2 - 1 - 1 - 2
2 - 1 - 2 - 1[/quote]

Hi Bunnel,

I listed out the possible teams, as you said their are only 3 ways to make 6 pairs.
wx yz
wy xz
wz xy

That is all, only 3 ways to make 6 distinct pairs. Looks like 4c2 giving me total no of pairs and we are picking 2 pairs at a time to assign 2 number out of 4 to give 12 ways to pick numbers.

But this is tricky, unless I list the cases this might not be very apparent.

Thanks for the extra questions.
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Re: Probability - simple question.  [#permalink]

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28 May 2013, 00:00
cumulonimbus wrote:
Hi Bunnel,

I listed out the possible teams, as you said their are only 3 ways to make 6 pairs.
wx yz
wy xz
wz xy

That is all, only 3 ways to make 6 distinct pairs. Looks like 4c2 giving me total no of pairs and we are picking 2 pairs at a time to assign 2 number out of 4 to give 12 ways to pick numbers.

But this is tricky, unless I list the cases this might not be very apparent.

Thanks for the extra questions.

Not sure I understand correctly what you mean there, anyway:

$$C^2_4=6$$ in my solution for C here: as-part-of-a-game-four-people-each-must-secretly-choose-an-98684.html#p760687 refers to the number # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b}. NOT to the number of ways we can choose 2 people out of 4.

Does this make sense?
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Re: Probability - simple question.  [#permalink]

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17 May 2014, 06:55
pinchharmonic wrote:
pinchharmonic wrote:
Quote:
Think of AABC has a word with repeating alphabets A. So AABC can be arranged in 4!/2! ways
Now as ordering is important here we have
4!/2! * (4/4*1/4*3/4*2/4)*1/2 as final answer where we can say 4!/2! * 1/2 is nothing but 4C2
Hope this helps

thanks, but i'm on point with everythign you said, but where is the 1/2 from? I also re-arranged the 4 letters using (4!/2!), but I can't figure out why you have the 1/2 at the end.

bumping for an explanation of one of these approaches, thanks!

Hi Bunuel,

Regarding case b. (aabc)

This seems to have never been answered.
And i am very much under the same dilemma. even after doing a 4!/2! where is this additional factor of 1/2 coming from?

you can re-arrange everything since the order in this case matters, since the order represents different people

[(4/4) * (1/4) * (3/4) * (2/4)] * (4! / 2!)

where 4! allows me to re-arrange everything and 2! removes the duplicates for person 1 and 2.

Also if there is a factor then why 2? since we have aabc over here, which is three different no's out of four.

Also, in case c. (aabb)

I am calculating probability in the following fashion :

4/4 * 1/4 * 3/4 * 1/4 * 4!/(2!*2!) divided by 4^4

which is equal to 72/4^4

again i seem to be missing a factor of 2 and i cannot at all understand what i am doing wrong here as i am
even removing duplicates introduced by duplicate a and b.

i am able to get the correct answer for all the other scenarios using my approach (aaab, abcd,aaaa) except for the two posted above(aabc,aabb) which
is baffling me even to a greater degree.

it would be really great if you could help me out here.

Cheers,
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Re: Probability - simple question.  [#permalink]

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17 May 2014, 20:09
Bunuel wrote:
cumulonimbus wrote:
Hi Bunnel,

I listed out the possible teams, as you said their are only 3 ways to make 6 pairs.
wx yz
wy xz
wz xy

That is all, only 3 ways to make 6 distinct pairs. Looks like 4c2 giving me total no of pairs and we are picking 2 pairs at a time to assign 2 number out of 4 to give 12 ways to pick numbers.

But this is tricky, unless I list the cases this might not be very apparent.

Thanks for the extra questions.

Not sure I understand correctly what you mean there, anyway:

$$C^2_4=6$$ in my solution for C here: as-part-of-a-game-four-people-each-must-secretly-choose-an-98684.html#p760687 refers to the number # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b}. NOT to the number of ways we can choose 2 people out of 4.

Does this make sense?

Hi Bunuel,

Re: Probability - simple question.   [#permalink] 17 May 2014, 20:09

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