Probability Calculation for Die Rolling Sequences

Let $p$ be the probability of rolling a red face (R) and $q$ be the probability of rolling a green face (G) on a six-faced cubical die. Since the die has both red and green faces, we know that $0 < p < 1$ and $0 < q < 1$, with $p + q = 1$. The experiment stops when 4 red faces appear.

Probability of Sequence A

Sequence A is GRRRR. This sequence requires 5 throws:

• Throw 1: G (Probability $q$)
• Throw 2: R (Probability $p$)
• Throw 3: R (Probability $p$)
• Throw 4: R (Probability $p$)
• Throw 5: R (Probability $p$) - This is the 4th Red face, stopping the experiment.

The probability of Sequence A, $P(A)$, is calculated as:

$ P(A) = q \times p \times p \times p \times p = q \times p^4 $

Probability of Sequence B

Sequence B is GRGRRR. This sequence requires 6 throws:

• Throw 1: G (Probability $q$)
• Throw 2: R (Probability $p$)
• Throw 3: G (Probability $q$)
• Throw 4: R (Probability $p$)
• Throw 5: R (Probability $p$)
• Throw 6: R (Probability $p$) - This is the 4th Red face, stopping the experiment.

The probability of Sequence B, $P(B)$, is calculated as:

$ P(B) = q \times p \times q \times p \times p \times p = q^2 \times p^4 $

Comparing Probabilities P(A) and P(B)

We need to compare $P(A) = q \times p^4$ and $P(B) = q^2 \times p^4$.

Since $p \ne 0$, we can divide both probabilities by $p^4$. The comparison reduces to comparing $q$ and $q^2$.

Given that $0 < q < 1$ (because not all faces are red, meaning at least one face is green, and not all faces are green, meaning at least one face is red), it is always true that $q > q^2$.

Therefore, $q \times p^4 > q^2 \times p^4$, which means $P(A) > P(B)$.

Conclusion

Sequence A is more probable than Sequence B.