# 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 $P(Y_B = 225) = 11\%.$

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.

• To be fair, the question did not ask to use normal approximations for part (c). The central limit theorem (at least, this version) is fairly lousy at calculating quantities of the form $P(X=x)$ anyway. Aug 12 '17 at 20:47
• @Jason: Inconsistent with title, but you're right. How then would you do part (c)? Aug 12 '17 at 22:03
• Directly. As callculus notes in their answer, this is just the probability that player B gets exactly $25$ heads and $25$ tails. You could also use a local central limit theorem in a pinch, but this seems beyond the level of the OP. Aug 12 '17 at 22:09

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 ?

• No. Something wrong there. Do you know what $\Phi$ is? Aug 12 '17 at 18:41
• Ah forgot about the Φ. 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%. Φ is the PDF of a normal distribution right? Aug 12 '17 at 18:48
• @DanielRobotics At b) I have the same intermediate result. You can use some more decimal places. $1-\Phi(-3.735582)=1-0.00009=99.991\%\approx 100\%$ Aug 12 '17 at 20:02
• Ah alright, so the negative number will not affect the Φ-value? Aug 12 '17 at 20:10
• @DanielRobotics If the z-value is negative then $\Phi(z)<0.5$. But if the z -value is very negative, $z<-3.5$, then $\Phi(z)\approx 0$. This is the case in b). Aug 12 '17 at 20:17