Waiting for a Six

In general, ${\mathbb E}[ X | Y=y ] = \frac15 \sum_{k=1}^{y-1} k(4/5)^{k-1} + (4/5)^{y-1}(y+6).$ The first part is the contribution from the cases where we roll a six before a five, and the second part is the contribution from the cases where we roll a five first; in the latter case, there is no conditioning on the rolls after the $y$ th roll, so the expected total number of rolls if we haven't hit a six before a five is $y+6.$

When $y=1,$ we get $7$ as expected, so $\alpha = 7.$ On the other hand, $\begin{aligned} {\mathbb E}[ X | Y = 2 ] &= 33/5 > 6 \\ {\mathbb E}[ X | Y = 3 ] &= 157/25 > 6 \\ {\mathbb E}[ X | Y = 4 ] &= 753/125 > 6 \\ {\mathbb E}[ X | Y = 5 ] &= 3637/625 < 6 \end{aligned}$ so $\beta = 5,$ and the answer is $\fbox{35}.$

Edit: as Mark Hennings points out, there is an easy formula for ${\mathbb E}[ X | Y=y ],$ which I can derive from my formula as follows: since $\sum\limits_{k=0}^{y-1} kx^{k-1}$ is the derivative of $\sum\limits_{k=0}^{y-1} x^k = \frac{x^y-1}{x-1},$ we get $\sum\limits_{k=0}^{y-1} kx^{k-1} = \frac{(y-1)x^y - yx^{y-1} + 1}{(x-1)^2}.$ Plugging in $x=4/5$ gives $25((y-1)(4/5)^y - y(4/5)^{y-1} + 1).$ Plugging that into the original formula gives $\begin{aligned} {\mathbb E}[ X | Y=y ] &= 5((y-1)(4/5)^y - y(4/5)^{y-1} + 1) + (y+6)(4/5)^{y-1} \\ &= (4y-4)(4/5)^{y-1} - 5y(4/5)^{y-1} + 5 + (y+6)(4/5)^{y-1} \\ &= 2(4/5)^{y-1} + 5. \end{aligned}$

I wonder if it's easy to compute this in some other way?

Then $\beta$ is the first value of $y$ such that $(4/5)^{y-1} \le 1/2,$ which is easy to compute: since $(.8)^3 = .512 > .5$ but $.8^4 = .4096 < .5,$ $\beta = 5.$