GATE 2024 DA Q.36: solution and answer

Official answer: CSource: Official answer key, GATE 2024 DA (IISc Bengaluru)

1. Each throw is even (2, 4 or 6) with probability $p=\frac{1}{2}$ and odd with probability $\frac{1}{2}$, independently of the other throws. We wait for two even throws in a row.

2. Define two states. $E_0$ is the expected number of further throws when the last throw was not even (or at the start). $E_1$ is the expected number of further throws when the last throw was even.

3. From state 0 one throw is made. With probability $\frac{1}{2}$ it is even and we move to state 1. With probability $\frac{1}{2}$ it is odd and we stay in state 0: $$E_0=1+\frac{1}{2}{E}_1+\frac{1}{2}{E}_0.$$

4. From state 1 one throw is made. With probability $\frac{1}{2}$ it is even and we stop. With probability $\frac{1}{2}$ it is odd and we fall back to state 0: $$E_1=1+\frac{1}{2}\cdot 0+\frac{1}{2}{E}_0.$$

5. Substitute $E_1$ into the first equation: $$E_0=1+\frac{1}{2}\left(1+\frac{1}{2}{E}_0\right)+\frac{1}{2}{E}_0=\frac{3}{2}+\frac{3}{4}{E}_0.$$

6. So $\frac{1}{4}{E}_0=\frac{3}{2}$, which gives $E_0=6$ (and $E_1=4$). The die starts in state 0, so the expected number of throws is $6$.

7. Check with the standard result for $r$ consecutive successes with success probability $p$: $$E=\frac{1-p^r}{p^r(1-p)}=\frac{1-\frac{1}{4}}{\frac{1}{4}\cdot \frac{1}{2}}=\frac{3/4}{1/8}=6.$$ This agrees, so the answer is option (C).