Bayesian updating game We have two urns $A$ and $B$. $A$ urn contains $2$ white and $1$ black balls while urn $B$ contains $2$ black and $1$ white balls. Urns are called states of nature and can each happen with the probability of $0.5$.
An urn is selected randomly and $n$ players need to sequentially guess which urn has been selected. Before guessing, they have to pick one of the two choices that should facilitate their guess:
1)  To draw a ball from an urn (get a signal), look at its color and make a guess about the true state of nature. For example, if urn $A$ was randomly selected and a player drew a white ball, an inference that urn $A$ is the true state of nature can be made with the probability of $0.66$.
2)  To look at the history of guesses made by other players. For example, player number $2$ can see the guess of player number $1$. Player number $3$ can see the guesses by players number $1$ and $2$ and so on (note that only guesses which urn is being used can be seen, not the signals). 
Edit: the players are rational, therefore inferences about the signals received can be made from the observed guesses. Additionally, when indifferent between the social information and the private signal, player opts for the latter one.
How can we prove that players $1$, $2$, and $3$ get at least as good chance in guessing the true state of nature by receiving a private signal rather than observing the history of choices, while the rest of the players starting from player $4$ are better off seeing the social information? Note that the player $1$ cannot see the history, therefore the choice is between guessing randomly and drawing a ball.
The Bayesian updating should be used in this case but I fail to go beyond the analysis for player $1$ who, obviously, chooses between 0.5 for a random guess and $0.(66)$ for a draw.
 A: So we've clarified in the comments that the signals can be inferred from the guesses, and if a player is indifferent between the private signal and the social information, they choose the private signal.
It's clear that player $1$ is better off with the private signal and that player $2$ is indifferent. What's interesting and to me quite surprising is that player $3$ is also indifferent; that is, you don't get better information about which urn was selected if you draw two balls than if you draw one ball.
Let $C$ be the event that the player guesses correctly, and let $J$ and $N$ denote the maJority and miNority colours, respectively.
Then after drawing one ball, we have
\begin{eqnarray*}
\def\p#1{\mathsf P\left(#1\right)}
\p C
&=&
\p{C\mid J}\p J+\p{C\mid N}\p N
\\
&=&
1\cdot\frac23+0\cdot\frac13
\\
&=&
\frac23\;,
\end{eqnarray*}
and after drawing two balls we still have
\begin{eqnarray*}
\p C
&=&
\p{C\mid JJ}\p{JJ}+2\p{C\mid NJ}\p{NJ}+\p{C\mid NN}\p{NN}
\\
&=&
1\cdot\frac49+2\cdot\frac12\cdot\frac29+0\cdot\frac19
\\
&=&
\frac23\;,
\end{eqnarray*}
whereas after drawing three balls we have
\begin{eqnarray*}
\p C
&=&
\p{C\mid JJJ}\p{JJJ}+3\p{C\mid NJJ}\p{NJJ}+3\p{C\mid NNJ}\p{NNJ}+\p{C\mid NNN}\p{NNN}
\\
&=&
1\cdot\frac8{27}+3\cdot1\cdot\frac4{27}+3\cdot0\cdot\frac2{27}+0\cdot\frac1{27}
\\
&=&
\frac{20}{27}\;.
\end{eqnarray*}
As I said, I'm quite surprised that drawing another ball can fail to improve the chances; I'd be even more surprised if it could reduce the chances – but even if it did, later players could always simply ignore all but the first three signals to get at least $\frac{20}{27}$, so all players starting with player $4$ will prefer the social information over the private signal.
