So, if I'm following you correctly:
Player two's response depends on player one's response, and player three's response depends on the responses of players one and two?
If this is the case, you would write for example ''P( P_2=Y | P_1=Y)" to mean the probability that player two says yes given that player 1 said yes. So, with your examples $P( P_2=Y | P_1=N)=.4$?
To find the probability, for example, $P(YNY)$ (that is, the probability that player one says yes and player two says no and player three says yes), you cannot multiply the probabilities that player one says yes, player 2 says yes, and player three says yes. That can be done only when you have independence.
However, you can take the product $ P(YNY)= P(P_1=Y) \cdot P(P_2 = N | P_1=Y) \cdot P(P_3=Y | P_1=Y\ \text{and}\ P_2=N). $
This is called the multiplication rule for probabilities.
Your example probabilities do not make perfect sense to me. You might want to start with:
Player one always says yes with probability $a$ and no with probability $1-a$ (for some $a$. )
If player one says yes, then the probability that player two says yes is ...
If player one says no, then the probability that player two says yes is ...
And you would have four cases for the probability that player three says yes.
Example:
Assume
$P_1$ says yes with probability $.2$.
$P_2$ says yes with probability $.1$, if $P_1$ said yes.
$P_2$ says yes with probability $.3$, if $P_1$ said no.
$P_3$ says yes with probability $.4$, if $P_1$ said yes and $P_2$ said yes.
$P_3$ says yes with probability $.5$, if $P_1$ said yes and $P_2$ said no.
$P_3$ says yes with probability $.6$, if $P_1$ said no and $P_2$ said no.
$P_3$ says yes with probability $.7$, if $P_1$ said no and $P_2$ said yes.
Then $ \begin{alignat*}{2} abc \qquad &P(abc)= P(P_1=a)P(P_2=b|P_1=a) P(P_3=b|P_1=a\ \text{and}\ P_2=b)&\ \\ YYY \qquad & P(P_1=Y)P(P_2=Y|P_1=Y) P(P_3=Y|P_1=Y\ \text{and}\ P_2=Y)& = (.2)(.1)(.4) \\ YYN \qquad &P(P_1=Y)P(P_2=Y|P_1=Y) P(P_3=N|P_1=Y\ \text{and}\ P_2=Y)& =( .2)(.1 )( .6)\\ YNY \qquad &P(P_1=Y)P(P_2=N|P_1=Y) P(P_3=Y|P_1=Y\ \text{and}\ P_2=N)& = ( .2)(.9 )(.5 )\\ YNN \qquad &P(P_1=Y)P(P_2=N|P_1=Y) P(P_3=N|P_1=Y\ \text{and}\ P_2=N)& = ( .2)(.9 )( .5)\\ NNN \qquad & P(P_1=N)P(P_2=N|P_1=N) P(P_3=N|P_1=N\ \text{and}\ P_2=N)& = (.8 )( .7)(.6 )\\ NNY \qquad & P(P_1=N)P(P_2=N|P_1=N) P(P_3=Y|P_1=N\ \text{and}\ P_2=N)& = ( .8)( .7)( .4)\\ NYY \qquad & P(P_1=N)P(P_2=Y|P_1=N) P(P_3=Y|P_1=N\ \text{and}\ P_2=Y)& = ( .8)( .3)(.7 )\\ NYN \qquad & P(P_1=N)P(P_2=Y|P_1=N) P(P_3=N|P_1=N\ \text{and}\ P_2=Y)& = (.8 )(.3 )( .3)\\ \end{alignat*} $
The sum is (adding in pairs starting from the top) $ (.2)(.1)+(.2)(.9)+(.8)(.7)+(.8)(.3)=(.2)+(.8)=1. $