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The probability that Xera has a four is: \(1:6=\dfrac{1}{6}\)
The probability that Yeta has a four is: \(4:52=\dfrac{1}{13}\) (A standard deck has 4 cards of four)
So the probability that Yeta wins is: \(\dfrac{1}{13}:\left(\dfrac{1}{6}+\dfrac{1}{13}\right)=\dfrac{6}{19}\)

FA Liên Quân Garena 30/12/2017 at 21:49
The probability that Xera has a four is: 1:6=16
The probability that Yeta has a four is: 4:52=113
(A standard deck has 4 cards of four)
So the probability that Yeta wins is: 113:(16+113)=619

FC Alan Walker 16/03/2018 at 13:31
We have: \(y^2=x\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
\(\Leftrightarrow y^2+1=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(\Leftrightarrow y^2+1=\left(x^2+3x+1\right)^2\)
\(\Rightarrow y=0\) (Because \(y^2\) and \(\left(x^2+3x+1\right)^2\) is two consecutive square numbers)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\\x=3\end{matrix}\right.\)
So x=0; x=1; x=2; x=3.