Normal approximations: What is the probability that player A will have at least 10 points more than player B? Two players $A$ and $B$ play the following game. Player $A$ uses a fair 8-sided die with numbers of $1, ...,8$ and rolls it to earn points. For each roll player $A$ collects a numbers of points corresponding to the number shown on the die. Player $B$ uses a fair coin and flips it to earn points. If the outcome is "heads", then player B collects $1$ point, if "tails" player $B$ collect $8$ points. This game is as follows: Player $A$ rolls the die $n=50$ times and player $B$ flips the coin $n=50$ times, after which player $A$ has collected $Y_A$ points and player B has collected $Y_B$ points. After $n=50$ trials: 
a) What is the probability that player $A$ will have at least $10$ points more than player $B$? Use normal approximations. 
b) What is the probability that player $A$ and player $B$ together will have more than $340$ points? Use normal approximations.
c) What is the probability that player $B$ will have exactly $225$ points? 

I have calculated the means/expected values to: mean($Y_A$) = 4,5 and mean($Y_B$) = 4,5, the variances I have calculated to: Var($Y_A$)=5,25 and Var($Y_B$)=12,25. 
a) 
Following @callculus answer: 
I have in a table found Φ(.32) to be .6255. So 1 - 0.6255 = 0.3745, will that say that the probability for player A having at least 10 more points than player B is 37%
b) 
Following the method from a): 
$$P(Y_A + Y_B > 340) = 1- P(Y_A + Y_B >= 340) = 1 - Φ((340+0.5-450)/\sqrt{875}) \\= 1 - Φ(-3.7) = 1-0.00009 = 99.991\% \approx 100\%$$
There is almost 100% $P(Y_A+Y_B > 340).$ 
c)
Using the binomial PMF with $50$ trials and $25$ successes. 
$$(50!)/((25!)*(50-25)! * 0.5^{25} * (1-0.5)^{50-25} = 0.112275173.$$ 
So $P(Y_B = 225) = 11\%.$ 
 A: To start I give you some hints: 
a) Firstly the inequlity $Y_A > Y_B + 10$ can be transformed to $Y_A-Y_B>10$. Then using the converse probability:
$P(Y_A-Y_B>10)=1-P(Y_a-Y_B\leq 9)$
The expected values of the random variables are  $E(Y_A)=E(Y_B)=50\cdot 4.5$. 
The variances of the random variables are  $Var(Y_A)=50\cdot 5.25, Var(Y_B)=50\cdot 12.25$
Let $Y_D=Y_A-Y_B$. Then $E(Y_D)=50\cdot 4.5-50\cdot 4.5=0$. 
And $Var(Y_D)=Var(Y_A)+Var(Y_B)=50\cdot 17.5=875$
With the help of the central limit theorem we get
$$P(Y_D\geq 10)=1-P(Y_D\leq 9)\approx 1-\Phi\left(\frac{9+0.5-0}{\sqrt{875}}\right)$$
$+0.5$ is the continuity correction factor.
b) $E(Y_A+Y_B)=E(Y_A)+E(Y_B), Var(Y_A+Y_B)=Var(Y_A)+Var(Y_B)$
c) Here you have to evaluate at what combinations of coin-flips you get exactly 225 points. If I´m right the only combination you get 225 points if you flip 25 times head and 25 times tail. Can you calculate the probability by using the binomial distribution ?
A: Comment (not really an answer): I simulated this mainly because I was skeptical about the normal approximations. I found that $Y_A, Y_A -  Y_B,$
and $Y_A +  Y_B$ are nearly normal, but $Y_B$ is not. So I'd avoid a normally-approximated answer to (c).
In the process, I got approximate
results consistent with $E(Y_A), E(Y_B), Var(Y_A), Var(Y_B),$ and $P(Y_D \ge 10)$
in the Answer by @callculus (+1), which is an excellent guide to finishing
the problem. Other approximate numerical answers in the simulation may be of
interest.
m = 10^6
a = replicate(m,  sum(sample(1:8, 50, repl=T)))
b = replicate(m,  sum(sample(c(1,8), 50, repl=T)))
mean(a);  var(a)
## 224.9918  # aprx E(Y_A) = 225
## 262.9595  # aprx Var(Y_A) = 262.5
mean(b);  var(b)
## 225.0187  # aprx E(Y_B) = 225
## 612.9044  # aprx Var(Y_B) = 612.5
mean(a-b >= 10);  mean(a + b > 340);  mean(b==225)
## 0.37403   # norm aprx P(Y_A - Y_B >= 10) = 0.374
## 0.999915
## 0.112175

Finally, here are plots of simulated distributions of the relevant random
variables, along with the 'best fitting' normal curves.

