# Probability role of die of two events A and B

In a roll of a die, let a "six" be the face with six spots. Consider the two events below.

Event A: more than 16,000 sixes in 100,000 rolls Event B: more than 160,000 sixes in 1,000,000 rolls.

Pick the right option? P(A)=P(B) or P(A) < P(B) or P(A) > P(B).

"Law of averages"

As you keep rolling, in the long run you get about 1/6 with six Spots.

Event A: P(A more than) = 100k rolls 1/6%(+-) 0.67%=k=0.159967....0.173367 Chance 0.833333

Event B: P(B more than)= 1,000k rolls 1/6% (+-) 6.7%=k=1.59967...1.73367 Chance 0.3230112

P(A)=0.83333 > P(B)=0.3220112

• Why have you said $P[A] = \frac{16000}{100000}$? By that logic, what would the probability of there being more than 1 six out of 100,000 rolls be? Commented May 3, 2013 at 23:02
• "Law of averages" As you keep rolling, in the long run you get about 1/6 with six Spots. Event A: P(A more than) = 100k rolls 1/6%(+-) 0.67%=k=0.159967....0.173367 Chance 0.833333 Event B: P(B more than)= 1,000k rolls 1/6% (+-) 6.7%=k=1.59967...1.73367 Chance 0.3230112 P(A)=0.83333 > P(B)=0.3220112 Commented May 4, 2013 at 0:08

In $n$ independent rolls, the total number of sixes is an integer random variable that follows binomial distribution $\operatorname{Bin}(n,p)$, where $p = \frac{1}{6}$ is the probability of rolling 6.

The question now asks you to compute $\Pr(A) = \Pr(X>16,000)$, where $X \sim \operatorname{Bin}\left(10^5,\frac{1}{6}\right)$ and $\Pr(B) = \Pr(Y > 160,000)$ where $Y \sim \operatorname{Bin}\left(10^6,\frac{1}{6}\right)$.

We could use Mathematica to estimate these:

In[17]:= Probability[X > 16000,
X \[Distributed] BinomialDistribution[10^5, 1/6]] <
Probability[Y > 160000,
Y \[Distributed] BinomialDistribution[10^6, 1/6]]

Out[17]= True


Intuitively these are clear. Mean and variances of $X$ and $Y$ are $$\begin{split} \mathbb{E}(X) = 10^5 \times \frac{1}{6} \approx 16666.7 \quad \mathbb{Var}(X) = \sqrt{ 10^5 \frac{1}{6} \frac{5}{6} } \approx 117.85 \\ \mathbb{E}(Y) = 10^6 \times \frac{1}{6} \approx 166666.7 \quad \mathbb{Var}(X) = \sqrt{ 10^6 \frac{1}{6} \frac{5}{6} } \approx 372.678 \end{split}$$ 16,000 is 5.65 standard deviations away from mean of $X$, while 160,000 is 17.8 standard deviations away from the mean of $Y$.

• Explain Please in Conclusion that P(A)>P(B) Commented May 4, 2013 at 0:19
• @statistics-student13: We have $\Pr(B)\gt \Pr(A)$. Both are very close to $1$, but $\Pr(B)$ is closer. Commented May 4, 2013 at 0:26

Hint: how many 6's do you expect out of 10,000 and 100,000 rolls. You can use a Gaussian approximation so you know you are more likely to have a certain relative error with fewer throws.

On expectation, in both case you should get $1/6>16\%$ of the rolls giving a 6. So the question is to compare the probabilities of deviating from the expectation by a given fraction $\delta$ of the draws ($\delta=1/6-0.16$). Do you think it is the same? Does the fact that you do more trials affect this (eg, by "smoothing out" the random fluctuations)?

• Therefore P(A)>P(B) by Law of Averages could You Check? Commented May 4, 2013 at 0:13
• It is the other way. Commented May 4, 2013 at 0:19
• Thanks Andre Nicolas Can You Explain P(A)<P(B)? Commented May 4, 2013 at 0:23
• $\bar{A}$ is the event of having less than 16,000 sixes in 100,000 rolls -- that is to deviate quite a lot from the expected number. Same thing for $\bar{B}$, with 160,000 sixes and 1,000,000 rolls. As explained, the more realizations of the random variable you get, the more "concentrated around the expected value" the fraction of sixes get. So $\mathbb{P}\bar{A} > \mathbb{P}\bar{B}$, and thus $\mathbb{P}{A} < \mathbb{P}{B}$. Commented May 4, 2013 at 0:30