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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