Ruin with an edge

Idea. Conditioning on the first round gives a linear recurrence. Its increments form a geometric sequence, so the answer is a ratio of geometric sums.

Step 1: recurrence. With $p=0.6$, $q=0.4$ and $h_i$ the probability of reaching $5$ from $i$, $$h_i=p{h}_{i+1}+q{h}_{i-1},h_0=0,h_5=1.$$

Step 2: increments. Since $p+q=1$, write $h_i=p{h}_i+q{h}_i$ and rearrange: $$p(h_{i+1}-h_i)=q(h_i-h_{i-1}).$$ So the increments $d_i=h_{i+1}-h_i$ satisfy $d_i=r{d}_{i-1}$ with $r=q/p=\frac{2}{3}$, that is $d_i=r^i{d}_0$.

Step 3: boundary values. Summing the increments, $$h_i=d_0(1+r+\cdots +r^{i-1})=d_0\frac{1-r^i}{1-r}.$$ The condition $h_5=1$ fixes $d_0$, and $$h_i=\frac{1-r^i}{1-r^5}.$$

Step 4: numbers. With $r=\frac{2}{3}$ and $i=2$: $$h_2=\frac{1-\frac{4}{9}}{1-\frac{32}{243}}=\frac{5/9}{211/243}=\frac{135}{211}\approx \mathrm{0.640.}$$

Comment. A fair game would give $\frac{2}{5}=0.4$. A $60\%$ edge per round raises it to about $0.64$; with a larger target the edge matters even more.