Alternating Coin And Dice Tossing

Let $p$ be the probability that he ends with the coin. He can succeed on the first flip (1/2) or he can fail but also fail on the coin (1/2 x 2/3 = 1/3) in which case we're just back to the beginning! Thus, $p = \frac{1}{2} + \frac{1}{3}p,$ so $p=\frac{3}{4}.$

$\text{probability of him getting head on coin in first throw}=\frac{1}{2}$ $\text{probability of him getting head on coin in second throw}=\left(\frac{1}{2}\times \frac{4}{6}\right)\times \frac{1}{2}$

$\begin{aligned} \text{probability of him getting head on coin in third throw } & = & \left( \frac { 1 }{ 2 } \times \frac { 4 }{ 6 } \right) \times \left( \frac { 1 }{ 2 } \times \frac { 4 }{ 6 } \right) \times \frac { 1 }{ 2 } \\ & = & \left(\frac{1}{2}\times \frac{4}{6}\right)^{2}\times \frac{1}{2} \end{aligned}$ $\vdots$

$\text{probability of him getting head on coin in } k^{th}\text{ throw}=\left(\frac{1}{2}\times \frac{4}{6}\right)^{k-1}\times \frac{1}{2}$

$\therefore P=\frac { 1 }{ 2 } \times \left\{ \left( \frac { 1 }{ 2 } \times \frac { 4 }{ 6 } \right) ^{ 0 }+\left( \frac { 1 }{ 2 } \times \frac { 4 }{ 6 } \right) ^{ 1 }+\left( \frac { 1 }{ 2 } \times \frac { 4 }{ 6 } \right) ^{ 2 }+\dots +\left( \frac { 1 }{ 2 } \times \frac { 4 }{ 6 } \right) ^{ \infty } \right\} =\boxed{0.75}$

The odds that he ends by tossing the coin in the whole game is the same as the odds of doing the same during the first round.

Odds of ending on coins on first round $=(0.5):(0.5)(1/3)=3:1$

The odds that he ends by tossing the coin $=3:1$

Probability that he ends by tossing the coin $=3/(3+1)=3/4=0.75$