# Find alpha for this dice rolling game. There is the point A(10) on the number line. Let's following the die rolling game rules like below with its pips are 1 to 6 as commonly we've known.

Find $\alpha$ for this dice rolling game.
There is the point A(10) on the number line. Let's following the die rolling game rules like below with its pips are 1 to 6 as commonly we've known.
- The 1st rule) A point moves as much as +2[positive direction with magnitude 2] for the even pips like 2,4 and 6
- The 2nd rule) A point moves as much as −1[negative direction with magnitude 1] for the odd pips like 1,3 and 5.
After doing this trial 100 times, "A" lied on the right direction compared with 90. We get $P\left(Z\ge \alpha \right)$ considering the continuity correction. Find the real number $\alpha$. (Z is the random variable following normal distribution.)
From here, my solution starts. Let the ${X}_{i}$ be the movement of a point for ith trial (Here Each i, $1\le i\le 100$)
$\begin{array}{clcr}{X}_{i}& 2& -1& \\ p\left({X}_{i}\right)& 1/2& 1/2& 1\end{array}$
From the table I got $E\left({X}_{i}\right)=\frac{1}{2}$ and $V\left({X}_{i}\right)=\frac{9}{4}$
So only have to consider is $X\left(={X}_{1}+{X}_{2}+...+{X}_{100}\right)>80$, Plus $E\left(X\right)=50$ and $V\left(X\right)={15}^{2}$
Therefore $P\left(X\ge 80.5\right)=P\left(Z\ge \frac{80.5-50}{15}\right)$ from continuity correction for $P\left(X>80\right)$. (In other words, $\alpha =\frac{61}{30}\left(=\frac{30.5}{15}\right)$ )
The answer sheet claimed the $\alpha =2.1$. I can't find any errors in my solution. Please let me know which points did I wrong.
p.s.) The answer sheet solution. (The reason why the author claimed 2.1).Say Y be the number of the even pips doing the 100 times trials, Then only just find $P\left(10+2Y-\left(100-Y\right)>90\right)$
In other words $P\left(Y\ge 61\right)$. Since the Y~B(100,1/2), $P\left(Y\ge 61\right)=P\left(Y\ge 60.5\right)=P\left(z\ge 2.1\right)$
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Dana Simmons
Step 1
Note that $P\left(Y>60\right)=P\left(X>90\right)$. The difference only comes into play when you add the continuity correction factor. The standard deviation of X is three times as great as the standard deviation of Y (because the distance between -1 and 2 is three times as great as the distance between 0 and 1). Therefore the continuity correction on X is $\frac{1}{3}$ as significant as the continuity correction on Y.
Step 2
So I would say that both the "official" answer ($\alpha =2.1$) and your solution ($\alpha =2.03333...$) are reasonable approaches, with the difference being due to the relative importance of 1 unit of X compared to 1 unit of Y.