# Choosing Random numbers Game

I am trying to solve a question in two different ways.

Que: Consider 4 persons p1,p2,p3,p4, and five numbers n1,n2,n3,n4,n5. In how many ways, at-least two of them choose the same number?

Method 1: The easy way is to figure out the complement,

= 1 - P(none of them choose the same number) = 1 - (5*4*3*2/5*5*5*5) = 1 - (24/125) = 101/125

Method 2, the usual way:

total probability = P(two persons choosing same no) + P(three persons choosing same no) + P(four persons choosing same no)

P(two persons choosing same no) = (4C2 * 5 * 4 * 3)/5^4 = 5*72/5^4

P(three persons choosing same no) = (4C3 * 5 * 4)/5^4 = 5*16/5^4

P(four persons choosing same no) = (4C4 * 5)/5^4 = 5/5^4

total P = 5*89/5^4 = 89/125

I think my solution 2 has some issues, Could someone please point out where i lost the track here?

• You have left off the pattern $AABB$ , $\frac 12\times \binom 42 \times 5\times 4=12\times 5$ – lulu Feb 4 '17 at 20:35
• Method 2 has a lot of double couning. Eg. What if 2 people chose one number and 2 or three choose another? – fleablood Feb 4 '17 at 20:41

$=\dfrac{\dfrac{4C2\times 5 \times 2C2 \times 4}{2!}}{5^4}=\dfrac{12}{125}$