A cricketer is having a lucky day today. On every ball he faces the batsman scores either 1,2,3 or 4 runs each with equal probability of 0.25. Assume that the batsman does not get out and there are infinite overs in game play. What is the probability that the cricketer scores a double century(>=200 runs) with a boundary (that is by hitting 4 runs).
Attempt- Intuitively I thought that once the player reaches 196, 197, 198 or 199 he just needs to score a boundary with probability=0.25 and, P(reaching 196,197,198 or 199)=1, so required probability should be 0.25. But this is not the answer
If no score below 196 can make the double century, then only the scores from 196 to 199 need be considered. Because the number of ways of achieving a number by summing elements in ${1,2,3,4}$ freely chosen is really big after comparatively few elements are summed, the probabilities that each of these numbers is our starting point are almost identical, so we'll act like they're all equally likely.
196, there is only one way to make it: a four. The other three balls take the batsman to one of the next scores.
197, either a three or a four will get there. This gives us two ways by a four and once by a three, so far.
198, a single won't do, the rest can. That's three fours, two threes, and one two as an outcome.
199, it wouldn't matter what he gets. That's now one way by a one, two ways by a two, three by a three, and four ways by a four.
This is a total of ten ways to achieve the double century, so the probability it happens as a boundary is $$P=\frac{4}{10}=0.4$$