What is the approximate likelihood that 2 people will choose same number?

Is it 6/256, if yes then I can explain.

duongthang wrote:38. 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?

For this we have two cases to consider.

1. 2 ppl will select same and 2 ppl will select different digits [A,A,B,C]

Probability = [ (No. of ways to choose which 2 persons will have the same number) * (No. of ways to choose which number it will be) * (No. of ways to choose 2 different numbers out of 3 left ) * (No. of ways to choose which person it will be out of the left 2)] / [Total possibilities]

So, No. of ways to choose which 2 persons will have the same number = C(4,2) = 6
No. of ways to choose which number it will be = C(4,1) = 4
No. of ways to choose 2 different numbers out of 3 left = C(3,2) = 3
No. of ways to choose which person it will be = C(2,1) = 2
Total possibilities = 4^4 = 256

Thus, P1 = (6*4*3*2)/256 = 144/256

2. 2 ppl will select same and 2 ppl will select other two same digits [A,A,B,B]

Probability = [ (No. of ways to choose which 2 persons will have the same number) * (No. of ways to choose which number it will be) / [Total possibilities]

No. of ways to choose which 2 persons will have the same number = C(4,2) = 6
No. of ways to choose which number it will be = C(4,2) = 6

Thus, P2 = 6*6/256 = 36/256.

Required Answer = 144/256 + 36/256 = 180/256

duongthang wrote:What is the approximate likelihood that 3 people will choose same number?

Probability = [ (No. of ways to choose which 3 persons out of 4 will have same number) * (No. of ways the number will be selecetd) * (No. of ways last person will be selected)] / [Total possibilities]

No. of ways to choose which 3 persons out of 4 will have same number = C(4,3) = 4
No. of ways the number will be selecetd = C(4,1) = 4
No. of ways last person will be selected = 3

Answer = 3*4*4/256 = 48/256

Probability of 2 people choosing same number = No of ways only 2 people can choose a number/ Total no of possible ways.

Since there are 4 people I gotta fill 4 slots.

Let the first person choose anything from 1 to 4.Then no of ways he can choose =4
The second person chooses the same number as the first person.Therefore he as only 1 possible way =1
No of ways that this 3rd person cannot choose the same number as first 2 persons. =3 (because 3 no.s are left now)
No of ways the 4th person can choose a number,which is different from the one chosen by 3rd and first 2 people=2

Logic :

The first person can choose 1 or 2 or 3 or 4 [4 choices available]
if no 4 is chosen then the second person can only choose no 4. [ only 1 choice available]
The third person could choose [1 or2 or 3].lets assume he chooses 3.[3 choices available]
The fourth person can only choose 1 or 2.[only 2 choices available]

Therefore no of ways only 2 persons can choose = 4*1*3*2=24.

Total no of possible ways = 4 *4*4*4 [anybody can choose any number,no restriction on whether duplication is allowable]

Probability = 24/256= 3/32

Final probability = 4!/2!2! *3/32= 6*3/32= 9/16

Reasoning : If you have 5 boys and 4 girls.The probability of choosing 1 boy and 1 girl = BG or GB = 5/9* 4/9 + 4/9 *5/9 = 40/81.

This is equivalent to : BG = 5/9 *4/9 *2!/1!1!=40/81 [ Treat B and G as letters in a words and divide the factorial of total of words by repetion of each word]

Similarly for 3 people choosing same no : 4*1*1*3 = 12

Probability = 12/256 = 3/64

Final probability = 4!/3!1! *3/64= 3/16

2 persons = 9/16 ; 3 persons = 3/16

What are the answers?

HI! Please explain Guys,

clear my confusion ,

In total 256 ways,
There will be 24 ways in which everybody have a different number, 4 * 3 * 2 * 1 = 24
There will be 48 ways in which exactly 3 people have the same number,

Number of ways to choose 3 people from 4 is 4 C 3 = 4
They have 4 number to choose from,
Now the last person has 3 different numbers to choose from

so 4 * 4 * 3 = 48

Everybody will have the same color 4 ways,

in total 24 + 48 + 4 = 76 ways,

so 256 - 76 = 180 ways in at least 2 people will have same color,

Now MY problem is you have calculated

shovan85 wrote: 2. 2 ppl will select same and 2 ppl will select other two same digits [A,A,B,B]

Probability = [ (No. of ways to choose which 2 persons will have the same number) * (No. of ways to choose which number it will be) / [Total possibilities]

No. of ways to choose which 2 persons will have the same number = C(4,2) = 6
No. of ways to choose which number it will be = C(4,2) = 6

Thus, P2 = 6*6/256 = 36/256.

That 144 is same as yours,
Number of ways to choose 2 people from 4 is 6
They have 4 Numbers to choose from so 4
Third person choose any number from 3 numbers so 3
Last person will have 2 numbers to choose from so 2

6 * 4 * 3 * 2 = 144 in total

Number of ways to choose 2 people from 4 is 4 C 2 = 6
They can choose any number from 4 Numbers so 4 ways,
Rest 2 people can choose only 1 number from 3 remaining numbers so 3 C 1 = 3

72 ways,

Now i have calculated these people Twice , But I am not able to understand where I have counted them twice,
Please Explain What i am doing wrong, What i should not do , so i don't do this mistake again....

goyalsau wrote: Number of ways to choose 2 people from 4 is 4 C 2 = 6
They can choose any number from 4 Numbers so 4 ways,
Rest 2 people can choose only 1 number from 3 remaining numbers so 3 C 1 = 3

72 ways,

Now i have calculated these people Twice , But I am not able to understand where I have counted them twice,
Please Explain What i am doing wrong, What i should not do , so i don't do this mistake again....

See the structure here

AABB: Means only 2 numbers are to be selected here. Say out of four ppl 2 ppl selected some SAME number (say both selected 1) and remaining 2 ppl selected some other SAME number (say both selected 2).

Thus,
Number of ways to choose 2 people from 4 is 4 C 2 = 6
Number of ways to choose 2 numbers from 4 is 4 C 2 = 6

Thus 6*6 = 36.

You have not counted the people twice but the digits you are miscalculating. Here it cannot be choose any number from 4 Numbers so 4 ways as simultaneously 2 numbers are to be selected.

shovan85 wrote:

goyalsau wrote: Number of ways to choose 2 people from 4 is 4 C 2 = 6
They can choose any number from 4 Numbers so 4 ways,
Rest 2 people can choose only 1 number from 3 remaining numbers so 3 C 1 = 3

72 ways,

Now i have calculated these people Twice , But I am not able to understand where I have counted them twice,
Please Explain What i am doing wrong, What i should not do , so i don't do this mistake again....

See the structure here

AABB: Means only 2 numbers are to be selected here. Say out of four ppl 2 ppl selected some SAME number (say both selected 1) and remaining 2 ppl selected some other SAME number (say both selected 2).

Thus,
Number of ways to choose 2 people from 4 is 4 C 2 = 6
Number of ways to choose 2 numbers from 4 is 4 C 2 = 6

Thus 6*6 = 36.

You have not counted the people twice but the digits you are miscalculating. Here it cannot be choose any number from 4 Numbers so 4 ways as simultaneously 2 numbers are to be selected.

Thanks for the above explanation, But i Just don't understand Why the First 2 people will not have 4 choices to choose from, Can we do this problem that way,

God Help me,,,,,,,,,,,,,,,,,,,,,,