Bayesian Updating from Two Perspectives Suppose there is a game of luck with chance of winning $p_w = .01$. You can attempt to cheat with success probability $p_c = .005$.  If you successfully cheat, your win probability is $p_{w|c} = .3$.  Players always try to cheat, and if you try to cheat but fail, win probability is $p_w = .01$.  
With a single player, if you observe a win, the posterior probability that the player successfully cheated is:
$$ p_{\{c|w\}} = \frac{.005\times.3}{.005\times.3 + .995\times.01} = .1310044$$  
Now let's introduce a second player.  Suppose all we can observe now is whether anyone won.  That is, we know whether zero people won or at least one person won.  What is the probability that at least one person cheated conditional on at least one win?
My figuring is as follows. The probability that neither player successfully cheats is $.995^2 = .990025$.  In turn, the probability that at least one of them cheats is $p_{\{c>0\}} = 1 - .990025 = .009975$.
Next is the probability that at least one player wins conditional on at least one successful cheater. In two scenarios exactly one succeeds, and in one scenario both succeed.  We have to weight those scenarios by their chances of happening.
\begin{equation*}
\begin{split}
w_1 = \frac{.005\times.995} { 2\times.005\times.995 + .005^2} = .4987469\\
\\
w_2 = \frac{.005^2} { 2\times.005\times.995 + .005^2} = .002506266
\end{split}
\end{equation*}  
The first weights the scenario where one has cheated and the second weights the scenario where both managed to cheat.
Now we have $$p_{\{w>0 | c>0\}}= 2w_1(1 - .7\times.99) + w_2 (1 - .7^2) = .3075088 $$ 
Finally, we have the marginal probability of at least one player winning (i.e., across all scenarios):
$$p_{\{w>0\}}=.995^2(1 - .99^2) + 2(.995)(.005)(1 - .7\times.99) + .005^2(1 - .7^2) = .0227689$$
Putting it all together 
$$p_{\{c>0|w>0\}} = \frac{p_{\{w>0|c>0\}}\times p_{\{c>0\} }}{p_{\{w>0\}}} = \frac{.3075088 \times .009975}{.0227689} = 0.1347189$$
My questions: is this correct?  If not, where did I go wrong?  If yes, is there a better way? Is there a general solution for this problem with an arbitrary number of players $n$?
 A: I agree with your answers!  I did it a brute force method, calculating the probabilities for each of the 16 possible end states, and then via the following:
The probability of someone winning and someone cheating are 2 ways times a winning cheater + 2 ways times a losing cheater and a winning non-cheater minus one way of two cheaters both winning (counted twice in the first part)
$$P[w>0\cap c>0]=2*P[cheat]*P[win|cheat]+2*P[cheat]*P[lose|cheat]*P[nocheat]*P[win|nocheat]-P[cheat]^2*P[win|cheat]^2$$
$$P[w>0\cap c>0]=2*0.005*0.3+2*0.005*0.7*0.995*0.001-0.005^2*0.3^2$$
$$P[w>0\cap c>0]\approx0.0030674$$
$$P[w>0]=P[w>0\cap c>0]+P[w>0\cap c=0]$$
$$P[w>0]=P[w>0\cap c>0]+P[nocheat]^2*(1-P[lose|nocheat]^2)$$
$$P[w>0]=P[w>0\cap c>0]+0.995^2*(1-0.99^2)$$
$$P[w>0]\approx0.00227689$$
$$P[w>0|c>0]=\frac{P[w>0\cap c>0]}{P[w>0]}$$
$$P[w>0|c>0]=\frac{2*0.005*0.3+2*0.005*0.7*0.995*0.001-0.005^2*0.3^2}{2*0.005*0.3+2*0.005*0.7*0.995*0.001-0.005^2*0.3^2+0.995^2*(1-0.99^2)}$$
$$P[w>0|c>0]\approx0.1347189$$
If the number were larger there would be additional binomial terms in the P[w>0 and c>0] term and higher powers in the + term for P[w>0]
