Here, Fishy, Fishy! 2
A player rolls a standard six-sided die with five main colors (red, purple, blue, yellow, green) and a sixth "wildcard" color (white). The goal is to catch one fish of each of the five main colors. Rolling one of the five main colors allows you to catch the fish of that color, if you have not already caught it. Rolling a white allows you to catch an uncaught fish of whichever color you pick.
Assuming he plays optimally, what is the expected number of rolls before the player catches one fish of each of the five colors?
Tip: For this problem, drawing a tree diagram helps in visualizing the problem.
The answer is 8.7.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Solution using Linearity of expectation
Suppose we have caught x different fish already. The number of favourable outcomes is 6 − x and total outcomes is 6. So probability of catching a new fish is 6 6 − x And the corresponding expected number of moves is 6 − x 6 .
Thus the answer is 6 − 0 6 + 6 − 1 6 + 6 − 2 6 + 6 − 3 6 + 6 − 4 6 = 1 + 1 . 2 + 1 . 5 + 2 + 3 = 8 . 7