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

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mittalg

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18 May 2014, 03:28

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kpali

pinchharmonic

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.

Please please help me understand.

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,
Kriti

Kriti,

For case B) aabc, I advise first you find out the total number of cases. If you see the problem, out of 4 people 2 have to select the same number which is 4C2.

Now these 2 people can select out of 4 available numbers. After they have selected we are left with 3 numbers which is available choice for the next person and after this only 2 numbers are left, which is the choice to the last person. So total number of cases become:

4C2 * 4* 3* 2 = 144

Total number of possible cases are 4^4 = 256.

Therefore, probability is 144/256.

In case c) aabb, we are choosing 2 people out of 4 such that the get the same number. This can be done is 4C2 ways. Now the remaining two are automatically chosen. But there will be repetition as explained below:

Let the first set of people chooses 1 as number and the other set chooses 2 , so we have 1122. Now lets take a case in which 1st set chooses 2 and the second set chooses 1, we have 2211. But the are one and the same thing. Hence, we have to exclude these repetitions by dividing it by 2!

Thus, we can choose groups in 4C2 * 2C2 /2!

First group have 4 choices and 2nd one has 3.

So number of ways become: 4C2*2C2/2! * 4 * 3 = 36

Total number of ways are 4^4 = 256

Hence the probability is 36/256.

Hope it makes things clear for you.

kpali

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18 May 2014, 05:02

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mittalg

kpali

pinchharmonic

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.

Please please help me understand.

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,
Kriti

mittalg

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18 May 2014, 06:02

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kpali

mittalg

kpali

Hi mittalg,

thanks for the post. and i DO understand the solution posted by Bunuel, i am just trying to understand why i am not getting the same answer using my approach thats all.
let me know if you could help me with that.

Cheers.[/quote]

Hi Kriti

I understood your problem now:

You are arranging aabc in 4!/2! ways which is absolutely fine. Lets look at few of the patterns you will get:

aabc
aacb

abca
acba

acab
abac

bcaa
cbaa

baca
caba

baac
caab

Now if you see, we are not bothered about who chooses the number first, i.e. we are not bothered that if b chooses it before c or it is otherwise. Hence, we have to take something like aabc and aacb as the same case. Thus, we divide this by 2. Hope this make sense.
The emphasis is to choose 2 people out of the four and not arranging all the 4.

Likewise, you can understand part C as well.

Cheers
Gaurav

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I'm struggling with where the additional 1/2 factor comes in as well. Unfortunately I cannot grasp Bunnel's approach with the methodologies I've learned to date, hopefully someone can explain within my line of thinking so I can understand the Bunnel approach.

AABC
1) Number of permutations (similar to Mississippi problem) 4!/2!
2) AA - can be 1 of 4 combinations AA,BB,CC,DD
3) B - can be 1 of 3 remaining numbers
4) C - must be 1 of 2 remaining numbers

Saw somewhere to always remember order ALWAYS matters in probability

So I get 4!/2! * 4*3*2 ways out of 4^4 total permutations

Can someone please help explain the additional 1/2 comes into play? I don't see why you would reduce by 2! since we want the total number of permutations in the numerator?

Thanks,

Matt

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Bumping for explanation as well!

pinchharmonic

Quote:

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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Bunuel

Financier

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.

Hi Bunuel,
Can you please clarify C. How come you didn't take into account the options for the numbers whereas you considered it in A, B, D and E?

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31 Oct 2017, 14:13

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kpali

pinchharmonic

Quote:

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.

Please please help me understand.

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,
Kriti

Just my 2 cents regarding the aabc case:
Basically there are 2 choices. 1) which 2 people are going to have the same number and 2) which number are these people going to take and which other 2 numbers the other 2 people are going to take.

When you do: (4/4) * (1/4) * (3/4) * (2/4)] * (4! / 2!),
(4/4) * (1/4) * (3/4) * (2/4) part takes care of the second part, the "which number are these people going to take and which other 2 numbers the other 2 people are going to take part" ie, if the first guy and second guy choose one number(the same number), the other 2 guys can choose the 2 different numbers in 3C2=6 ways. Since the first 2 guys have four options for choosing the "same number" , total ways =6*4=24.

Now you need to take care of the "which 2 people are going to have the same number" part.
So we have 2 people with same number and 2 people with different number. Sort of like 2 groups. The order in the 2 groups doesn't matter. when you are multiplying with (4! / 2!).. you are only taking care of the lack of order in one group. so you need to divide by an additional 2! to take care of lack of order in other group also. That is where the additional 2! in the denominator is coming from.

If the question was how many ways can A &B have same number and C&D have different, then the answer would have been (4/4) * (1/4) * (3/4) * (2/4)]. This already takes care of the different number options. But here, you have to also choose which 2 among a,b,c,d get to have the same number. So to select which 2 people get the same number, (4! / 2!) is not enough coz then you are considering the order in one of the groups. That's why you get a larger answer.

I dunno if that makes sense. But I hope it helps

Hopefully some expert will explain more clearly.

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pinchharmonic

Quote:

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.

Can someone pls explain where is the 1/2 coming from? struggling on the same!!!!!!!!!!

Ksmew

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27 Jul 2020, 00:42

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mittalg

kpali

mittalg

kpali

Hi Kriti

I understood your problem now:

You are arranging aabc in 4!/2! ways which is absolutely fine. Lets look at few of the patterns you will get:

aabc
aacb

abca
acba

acab
abac

bcaa
cbaa

baca
caba

baac
caab

Now if you see, we are not bothered about who chooses the number first, i.e. we are not bothered that if b chooses it before c or it is otherwise. Hence, we have to take something like aabc and aacb as the same case. Thus, we divide this by 2. Hope this make sense.
The emphasis is to choose 2 people out of the four and not arranging all the 4.

Likewise, you can understand part C as well.

Cheers
Gaurav[/quote]

Thank you for this explanation!!!! this is amazing.. i was struggling so much with 1/2, same as Kriti. this simple explanation solved it all.