The type of deck makes no difference...

The probability of drawing $oros$ is $\frac{1}{4}$ , and not drawing $oros$ is of course $\frac{3}{4}$ .

Antonio goes first. He may win in the 1st turn, or in the 5th provided the rest of the players lose in their respective turns, or in the 9th... $p_A=\dfrac{1}{4}+\left( \dfrac{3}{4} \right)^4\times \dfrac{1}{4}+\left( \dfrac{3}{4} \right)^8\times \dfrac{1}{4} +\ldots=\dfrac{1}{4} \times \dfrac{1}{1-(\frac{3}{4})^4}=\dfrac{64}{175}$

For Benito to win, Antonio must lose his respective turns. Benito's turns are the 2nd, the 6th, the 10th... : $p_B=\dfrac{3}{4}\times\dfrac{1}{4}+\left( \dfrac{3}{4} \right)^5\times \dfrac{1}{4}+\left( \dfrac{3}{4} \right)^9\times \dfrac{1}{4} +\ldots=\dfrac{3}{4}\times\dfrac{1}{4} \times \dfrac{1}{1-(\frac{3}{4})^4}=\dfrac{48}{175}\quad \textcolor{#3D99F6}{ \text{Note that }p_B=\frac{3}{4}p_A}$

The pattern repeats for Carlos, $p_C=(\frac{3}{4})^2 p_A=\dfrac{36}{175}$ and Diego, $p_D=(\frac{3}{4})^3p_A=\dfrac{\textbf{27}}{175}$ .

Therefore $a=64,b=48,c=36$ and $\boxed{d=27}$ .

$\text{Table}\left[\frac{1}{4} \left(\frac{3}{4}\right)^p \sum _{i=0}^{\infty } \left(\frac{3}{4}\right)^{4 i},\{p,0,3\}\right] \Longrightarrow \left\{\frac{64}{175},\frac{48}{175},\frac{36}{175},\frac{27}{175}\right\}$

The probability a person will be able to draw in this round of four draws: $\text{Table}\left[\left(\frac{3}{4}\right)^p,\{p,0,3\}\right] \Longrightarrow \left\{1,\frac{3}{4},\frac{9}{16},\frac{27}{64}\right\}$ .

The probability of winning if you get to draw in this round of draws: $\frac14$ .

The probability of winning in this round of draws: $\frac14 \left\{1,\frac{3}{4},\frac{9}{16},\frac{27}{64}\right\} \Longrightarrow \left\{\frac{1}{4},\frac{3}{16},\frac{9}{64},\frac{27}{256}\right\}$ .

Lastly, we need the infinite sum of the factors of surviving an infinite series of rounds: $\sum _{i=0}^{\infty } \left(\frac{3}{4}\right)^{4 i} =\sum _{i=0}^{\infty } \left(\frac{81}{256}\right)^i \Longrightarrow \frac{256}{175}$ .

Finally, $\frac{256}{175} \left\{\frac{1}{4},\frac{3}{16},\frac{9}{64},\frac{27}{256}\right\} \Longrightarrow \left\{\frac{64}{175},\frac{48}{175},\frac{36}{175},\frac{27}{175}\right\}$

The key observation is that $p_B=\frac{3}{4}p_A$ , $p_C=\frac{3}{4}p_B$ and $p_D=\frac{3}{4}p_C$ .

Using this and the fact that $p_A+p_B+p_C+p_D=1$ , we have

$p_A \left( 1+\frac{3}{4} + \left(\frac{3}{4} \right)^2 + \left(\frac{3}{4} \right)^3 \right) = 1$

Solving gives $p_A = \frac{64}{175}$ ; we then get $p_B = \frac{48}{175}$ , $p_C = \frac{36}{175}$ and $p_D = \frac{27}{175}$ . The only odd numerator is $\boxed{27}$ .