A bag contains equal numbers of red, blue, green and yellow balls. Four balls are drawn at random, with replacement after each ball is drawn. One way to classify the possible outcomes is:

  1. 4 balls of the same colour.
  2. 3 balls of one colour and 1 of another.
  3. 2 balls of one colour and 2 of another.
  4. 2 balls of one colour, 1 of another, and 1 of yet another.
  5. 4 balls, all of different colours.

What is the best way to find the probability of outcome 4?

I reasoned as follows. Altogether there are $4^4=256$ possibilities. The 2 balls of one colour could be of any of the 4 colours. Considering the draws in order, those two balls could be drawn in $\binom{4}{2}=6$ ways. The first of the other balls could be any of the 3 remaining colours, and the second could be any of the 2 remaining colours. The probability is therefore:

$$P(\text{outcome 4}) = \frac{4 \times 6 \times 3 \times 2}{256} = \frac{144}{256} = 0.5625$$

I wonder whether this is correct as it seems (to me at least) slightly counter-intuitive that the probability of just one of the 5 outcomes should be more than $0.5$.

Prompted by Ned's comment, my calculations of the number of possibilities falling into each of the patterns are respectively 4; 48; 36; 144; 24, which do sum to 256. But it still seems surprising that one pattern has so many more possibilities than any of the others.

  • $\begingroup$ It's not that counter intuitive considering the first outcome has only $4\over 256$ chance. $\endgroup$ – cr001 Feb 14 at 16:02
  • 1
    $\begingroup$ To verify it, compute how many of the 256 fall into each of the five patterns. $\endgroup$ – Ned Feb 14 at 19:59

Here's some intuition, maybe, using conditional probability. After three draws, the only possibility that rules out the case (2,1,1) is three of the same color, which only happens 1/16 of the time, so 15/16 of the time, you have a chance of hitting (2,1,1) on the last draw.

With (2,1) on the first three, you have 50% chance of (2,1,1) and with (1,1,1) you have 75% chance of (2,1,1). So 15/16 of the time, you approach the last ball with at least 50% chance of getting (2,1,1).

  • $\begingroup$ So writing subscript $_4$ for events when drawing 4 balls and subscript $_3$ for events when drawing 3 balls, we have: $P(2,1,1) _4 = P[(2,1,1) _4 | (2,1) _3] + P[(2,1,1) _4 | (1,1,1) _3] = (0.5)(9/16) + (0.75)(6/16) = 9/32 + 9/32 = 0.5625$. That makes it more intuitive. $\endgroup$ – Adam Bailey Feb 15 at 12:03

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