Chance of having score of 63 Assume a batsman has an equal chance of getting a score of 1,2,3,4 and 6 and that he has 100% chance of eventually having a score greater than 63. What is the probability of the batsman having exactly a score of 63. For example if the batsman is on a total of 62, then there is only 0.1667 chance of him getting 63. What is the chance assuming his starting total is 0?
 A: As @JMoravitz says, this is a Markov chain problem. But a $69\times 69$ matrix is unwieldy.
A good approximation is obtained by saying that, with every score $1,2,3,4,6$ equally likely, on average you will get $5$ scores in every total of $16\; (=1+2+3+4+6)$. E.g. if your first five scores were $1,2,3,4,6$ then you have gotten totals of $1,3,6,10,16$. Thus,
\begin{eqnarray*}
P(63) &\approx & \dfrac{5}{1+2+3+4+6} = \dfrac{5}{16} = 0.3125
\end{eqnarray*}
The reason, I think this is a good approximation is that the further away from $0$ your target score is, the more even become the probabilities. That is, $P(1), P(2), P(3)$ might be quite varied, but $P(61), P(62), P(63)$ become almost the same and they would approach that limiting value above.
It's not hard to use that method in a more realistic model where scores $1,2,3,4,6$ are not equally likely. Let's say, we estimate their individual probabilities as:
\begin{eqnarray*}
P(1) &=& 0.4 \\
P(2) &=& 0.2 \\
P(3) &=& 0.1 \\
P(4) &=& 0.25 \\
P(6) &=& 0.05
\end{eqnarray*}
Then our estimate would be:
\begin{eqnarray*}
P(63) &\approx & \dfrac{1}{ 1\times 0.4 + 2\times 0.2 + 3\times 0.1 + 4\times 0.25 + 6\times 0.05} = \dfrac{1}{2.4} \approx 0.4167
\end{eqnarray*}
The denominator there is the expected score on any one shot.
A: Mick A's approximation is borne out by simulation.   Checking if 63 is reached at some point  in the cumulative sums of a sample from $\{ 1, 2, 3, 4, 6 \}$, for one million possible samples, is a one-liner in R:
table(replicate(10^6, 63 %in% cumsum(sample(c(1,2,3,4,6), 63, TRUE))))

and this returns for me
 FALSE   TRUE 
687399 312601

Note that this depends on 63 being large compared to the possible scores.  For example the probability of eventually reaching 1 is 0.2, the probability of eventually reaching 2 is 0.24 (your sequence has to start with 2 or with 1, 1), and so on.  But apparently by the time you get to 63 this washes out.
