Question

A bag contains 64 balls of eight different colours, with eight of each colour. What is the expected number of balls you would have to pick (without looking) to select three balls of the same color?

Answer 1

Answer:

Step-by-step explanation:

Let N be the number of draws without replacement needed to draw three balls of the same colour. The probability of not drawing three balls of the same colour in n draws is

P(N>n)=(64n)−1∑p=08(8p)(82)p(8−pn−2p)(81)n−2p,

where (64n) is the total number of ways to choose n out of 64 balls, p (for “pair”) is the number of colours of which two balls are drawn, (8p) is the number of ways to select those colours, (82)p is the number of ways to choose the pairs of balls with those colours, n−2p is the number of colours of which a single ball is drawn, (8−pn−2p) is the number of ways to select those colours, and (81)n−2p is the number of ways to choose the single balls with those colours. Thus, the expected number of draws needed to draw three balls of the same colour is

E[N]===≈∑n=0∞P(N>n)∑n=0∞(64n)−1∑p=08(8p)(82)p(8−pn−2p)(81)n−2p9615203985231050472341059.1532

(Wolfram|Alpha computation).

Here's Java code for a simulation that confirms the result.

Note that the expected number of draws required to draw three balls of the same colour is quite close to the maximal number 9 of draws required to draw two balls of the same colour.