SOLUTIONS FOR ALL SCENARIOS

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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out of curiosity guys, sometimes there are questions that we will have to look for the opposite and then subtract that answer from 1. but how come it doesn't work with this problem? for example:

what is the probability that they will choose the same number instead of different numbers:

4/4 * 1/4 * 1/4 * 1/4 = 1/64

so 1 - 1/64 = 63/64 which is not even close to 9%. how come this doesn't work? cause i always confuse between when to use the opposite and when not. anybody?

Possible combinations of each choosing a diffrent number (favorable events):
4*3*2*1=24

Total possible combinations (total events): 4*4*4*4=256

Favorable events/total events = 24/256 approx 10%

Is this a correct approach?

very quickly: total area of probability equals 4*4*4*4=4^4
at the numerator: 4! (if the first choose one number, there are 3 possible alternatives remaining...)
so, 4!/4^4=0.093....

p for all of them select 1 = 1/4^4

p for all of them select 1 or 2 or 3 or 4 = 4*1/4^4= 1/4^3

As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

a) 9%
b) 12%
c) 16%
d) 20%
e) 25%

Soln:
Chances that all four choose different numbers is
Assuming first person chooses any of 4 numbers, second can choose the 1 from the left over 3, third can choose one number from the left over two and the last person chooses the last number. Hence
4 * 3 * 2 * 1 = 24 ways

Total possible chances is 4 * 4 * 4 * 4 = 256

Thus probability is = (24/256) * 100 = 9%

Hi Bunuel,

For the case C, I am going wrong somewhere. Could you point out where please?

# of ways to chose 2 people from the group = 4C2
# of ways for this group to select one number out of 4 numbers = 4
# of ways to select 2 people out of remaining 2 = 2C2
# of ways for this group to select a number from the remaining 3 = 3

Hence total number of ways = 4C2*4*2C2*3 = 72.

Thanks, makes a lot more sense looking at it this way.
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Way cool. Wish I could bookmark just this response. Many times, I bookmark a topic, but forget that it was actually a constituent post way way down that really prompted me to do so...

Bunuel,

I am teribbly sorry to revive this old post but I am still trying to find out to out compute options (B) and (C) that you have laid out.

For (B) this was my thought process- - {a,a,b,c}:

(4C1)(4C2)(3C1)(2C1)(2C1)(1C1)=288 ( You got 144, which is half my answer, why do I need to divide by 2?)

(# of ways to pick the paired number)(# of ways to place the pair within the 4 slots) (# of ways to pick the first non-pair number)(# of ways to place first non-pair number in the remaining two slots)(# of ways to pick the second non-pair number)(# of ways to place the second non-pair in the last slot)

For (C) this was my thought process - {a,a,b,b}:

(4C1)(4C2)(3C1)(2C2)=72

(# of ways to pick the first pair)(# of ways to place first pair number in the 4 slots)(# of ways to pick the second pair )(# of ways to place second pair in the remaining 2 slots)


What am I doing wrong here? Help Bunuel!
Thank you!