Bayes theorem with information cascade Urn 1: 3 blue balls, 1 red ball
Urn 2: 1 blue ball, 3 red balls
It is equally likely to choose either of the two types of urns. In sequence, each player is asked to take a ball at random, and based on the color of the ball, guess if that urn is majority blue or majority red. After a player draws a ball, the ball is replaced. The subsequent player sees the previous player's guess but not their ball. In this experiment, what is the probability that a wrong cascade will occur after the first two people guess?
A cascade can happen if the first two people drew blue balls, so the third person will guess that this is Urn 1 regardless of whether he/she drew a red or blue ball. (vice-versa for red balls as well). The third person guesses after his/her draw.
A wrong cascade is when the third person incorrectly guesses the urn based on a cascade. 
The answer is $0.06$ but I keep getting $0.2$, my thoughts:
$Answer = P(U_1|2R) + P(U_2|2B)= \frac{P(2R|U_1) * P(U_1)}{P(2R)} + \frac{P(2B|U_2) * P(U_2)}{P(2B)}$
$ = 2* \frac{\frac{1}{16} * \frac{1}{2}}{\frac{5}{16}} = .2 $
 A: I wanted to write a bit more on this question. 
First the answer. Our model of the situation is that an urn is selected from urns 1 or 2 with equal likelihood. We are told that if the same color of ball is drawn twice at the start of our game we will experience a cascade failure. (assuming this color is the minority color in our urn) We are asked to find the probability that such a thing will occur.
\begin{equation}
P(F)=P(F|1)P(1)+P(F|2)P(2)
\end{equation}
Where $F$ corresponds to the event of a Cascade Failure and $1/2$ correspond to the chosen urn. We can now calculate this:
\begin{eqnarray}
P(1)&=&P(2)=1/2\\
P(F|1)&=&P(2R|1) = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}\\
P(F|2)&=&P(2B|2) = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}\\
P(F)&=&\frac{1}{16}\cdot\frac{1}{2}+\frac{1}{16}\cdot\frac{1}{2}=\frac{1}{16}
\end{eqnarray}
I did not want this to be the end of the story since the premise of a cascade was kind of curious to me. Say we return to the drawing situation again. Obviously player one's best guess is the color they drew. Let's take a look at how that influences the other players:
\begin{eqnarray}
P(1|B) = \frac{P(B|1)P(1)}{P(B)} &=& 3/4\\
P(1|1,B) = \frac{P(1,B|1)P(1)}{P(1,B)} &=& \frac{9}{10}\\
P(1|1,R) = \frac{P(1,R|1)P(1)}{P(1,R)} &=& \frac{1}{2}\\
\end{eqnarray}
Now we are a bit stuck since the second player can either call 1 or 2 and have an equal chance of being correct from their perspective. From the description of the problem it sounds like such issues are resolved where the player will guess the opposite of the previous player as any other condition would weaken the likelihood of a cascade failure. Thus consider player 3:
\begin{eqnarray}
P(1|1,1,B) = \frac{P(1,1,B|1)P(1)}{P(1,1,B)} &=& \frac{27}{28}\\
P(1|1,1,R) = \frac{P(1,1,R|1)P(1)}{P(1,1,R)}=\frac{9/64}{(9/64)+(3/64)} &=&\frac{3}{4}\\
P(1|1,1,1,R) = \frac{P(1,1,1,R|1)P(1)}{P(1,1,1,R)}=\frac{1}{2}
\end{eqnarray}
This shows that the flipping strategy would also be terminal and not necessarily cascade further.
This assumes that players know all prior calls. We could consider further effects like players only know about the most recent call, but know their place in line. Players only know the most recent call and don't know their place in line. It seems like these would only put more pressure on the participants to trust their own draw though and that again put more pressure on stopping a cascade.
