rupee and paise coins probability question 
Initially a bag was known to contain some one rupee ('0' or more) and some fifty paisa ('0' or more) coins. In all bag was known to have 4 coins. Two coins were randomly drawn from the bag and both found to be one rupee coin. If these coins are replaced, what is the probability that next drawn coin is fifty paisa coin ?



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*(Assumption : Initially all number of rupee coins in the bag are equiprobable)


My try : Let $P$ denotes $50 $ paise coin and $R$ represent $1$ rupee coin. So cases would be as follows 


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*$PPPP$

*$PPPR$

*$PPRR$

*$PRRR$

*$RRRR$
But as the question says there were $2$ $R$ coins so case $1.$ and $2$. get rejected.So remaining cases are $3$.
So getting paise coin probabillity can be written as $\frac{1}{3}$.$\frac{2}{4}$ (case 3) + $\frac{1}{3}$.$\frac{1}{4}$ (case 4). i.e $\frac{1}{4}$ but answer is $\frac{1}{8}$
I think my method is wrong. Please help.
 A: The problem is that seeing $RR$ in the first two draws causes us to revise our estimates for the probabilities.
Assume that our prior was that the five states had equal probability (maybe not a great assumption, but I don't see how else to proceed).  Then the total probability that you draw $RR$ initially is $$\frac 15\times \left(\sum_{i=2}^4\frac {\binom i2}{\binom 42}\right) = \frac 13$$
We use Bayes Theorem to revise our probability estimates and we get the new probabilities $$P(PPRR)=.1\quad P(PRRR)=.3\quad P(RRRR)=.6$$
Where do these numbers come from?  Well, each of them just represents that portion of the total probability which is explained by being in the specified state.  Thus, for example, $$P(PPRR)=\frac {\frac 15\times \frac {\binom 22}{\binom 42}}{\frac 13}=\frac 35 \times \frac 1{6}=\frac 1{10}$$
We then can read off the answer $$\frac {1}{10}\times \frac 12+\frac {3}{10}\times \frac 14+\frac 6{10}\times 0=\boxed {\frac 18}$$
A: The 5 cases are equally likely at the beginning, but once you've seen two R coins, they are no longer equally likely (it's more likely you have a bag with lots of R's than only a few).
We have
\begin{align*}
P(\text{case 3}|2R) &= \frac{P(\text{case 3 and } 2R)}{P(2R)}\\
&= \frac{P(2R|\text{case 3})P(\text{case 3})}{P(2R|\text{case 3})P(\text{case 3}) + P(2R|\text{case 4})P(\text{case 4})+P(2R|\text{case 5})P(\text{case 5})}\\
&= \frac{\left(\frac{2}{4}\cdot \frac{1}{3}\right) \cdot \frac{1}{5}}{\left(\frac{2}{4}\cdot \frac{1}{3}\right)\cdot \frac{1}{5}+ \left(\frac{3}{4}\cdot \frac{2}{3}\right) \cdot \frac{1}{5}+1\cdot \frac{1}{5}}\\
&= \frac{1/30}{20/60}\\
&= \frac{1}{10}
\end{align*}
You can find the probability of case 4 given 2R similarly. Then repeat the calculation you did with the probabilities $P(\text{case 3}|2R),P(\text{case 4}|2R)$ instead of $\frac{1}{3}$.
