# A dice and Law of Large Numbers

The dice A has 2 red faces and 4 green, and the dice B conversely: 4 red and 2 green. We toss a symetric coin; if come up heads, we choose the dice A, otherwise - the dice B. Next we execute a series of throws by the chosen dice. Let: $X_n= 0$, if a red face in a n-th throw and $X_{n}=1$, otherwise. Find the limit of $\frac{X_1+X_2+\dots+X_n}{n}$ in a.s. sense.

Because $(X_n)$ are iid and $\text{E}X_1=\frac{1}{2}$, we can apply SLLN. Hence:

$$\frac{X_1+X_2+\dots+X_n}{n} \to \frac{1}{2}$$ almost sure.

I'm not sure if $X_n$ are independent. Are they?

• No, $(X_n)$ is not independent, for example $$P(X_1=X_2=1)=\frac{1}{2}\cdot\left(\frac{1}{3}\right)^2+\frac{1}{2}\cdot\left(\frac{2}{3}\right)^2=\frac{5}{18}\ne\frac{1}{2}\cdot\frac{1}{2}=P(X_1=1)\cdot P(X_2=1).$$ And as a matter of fact, the conclusion of the SLLN does not hold (as mathematics and common sense both indicate). – Did Jun 14 '15 at 19:47
• 12 minutes... Hmmm... – Did Jun 14 '15 at 19:49
• @Did Perhaps I had misunderstood the question. It seems with your calculation above that you think that a single same coinflip determines the result of the choice of die for the first and second die, yet I was thinking that the coin was flipped each time the die was to be rolled to redetermine which die to use. If the interpretation is that a single coinflip determines the choice of all dice (which upon rereading the question seems to be the case), then my answer as given was incorrect. I am unable to delete an accepted answer, so instead I shall update the answer. – JMoravitz Jun 15 '15 at 4:30
• @JMoravitz Yes I "think" that, just because it is written in the question. – Did Jun 15 '15 at 5:34
• @Did and I thank you again for pointing out my shortcomings. I hope that in the future you may continue to help me (and others) avoid such mistakes in the future. – JMoravitz Jun 15 '15 at 5:36

## 1 Answer

Try to calculate what $Pr(X_i = 1)$ is. You would get $Pr(X_i=1) = Pr(Coin=heads)\cdot Pr(dice A = green) + Pr(Coin=tails)\cdot Pr(dice B = green) = \frac{1}{2}\cdot \frac{4}{6} + \frac{1}{2}\cdot \frac{2}{6} = \frac{1}{2}$.

Similarly $Pr(X_i=0)$ will also be $\frac{1}{2}$.

Now, for purpose of argument, calculate $Pr(X_i=1 | X_j=1)$ (with $i\neq j$). Note that the way the problem is worded, a single coinflip will determine the choice of which die to use for all throws of the dice, and that we do not reflip the coin each time. (I had not noticed this when I initially posted my answer.). As such, as explained by another user in a comment to the question, we calculate $Pr(X_i=1 | X_j=1) = \frac{Pr(X_i=X_j=1)}{Pr(X_j=1)} = \frac{\frac{1}{2}(\frac{1}{3})^2 + \frac{1}{2}(\frac{2}{3})^2}{\frac{1}{2}} = \frac{5}{9}\neq \frac{1}{2} = Pr(X_i=1)$

Noting that $Pr(A|B)=Pr(A)\Leftrightarrow A$ and $B$ are independent events, the above shows that $X_i=1$ and $X_j=1$ are not independent events. You will get then that the SLLN does not apply in this case.

Re limits, given that the coinflip was heads, the limit is $\frac{1}{3}$, and, given that the coinflip was tails, the limit is $\frac{2}{3}$.

• Actually I did compute Pr(Xi=1|Xj=1) (with i≠j) and I did not find that it equals the same thing as before. – Did Jun 14 '15 at 19:53
• @Did you are absolutely correct, and I apologize about not being aware of that sooner. I had misread the problem. By all means, please feel free to cast downvotes when you notice mistakes in posts to draw the attention the OP and answerer to such mistakes being made. By all rights, I should have received several downvotes for this. – JMoravitz Jun 15 '15 at 4:44
• Three points. 1. You should be aware now that upvotes and downvotes on the site are erratic. In particular, wrong answers written in a seemingly correct mathematical style often attract upvotes. 2. I did draw the attention of the OP and the answerer to the mistake in the answer, by two comments. 3. When one reproduces quasi verbatim an argument due to somebody else, one usually mentions the fact. – Did Jun 15 '15 at 5:29