# Why is the answer to this probability question not $\frac{1}{2}$?

I was trying to solve the following homework probability question which has the following setup:

We have $$2$$ dice: $$A$$ and $$B$$. Die $$A$$ has $$4$$ red faces and $$2$$ white faces, whereas die $$B$$ has $$2$$ red faces and $$4$$ white faces. On each turn, a fair coin is tossed. If the coin lands on heads then die $$A$$ is thrown, but if the coin lands on tails then die $$B$$ is thrown. After this the turn ends, and on the next turn the coin is once again tossed to determine the die (i.e. we throw the coin and the corresponding die on each turn).

From this game I'm asked to answer $$2$$ questions:

1. Show that the probability of obtaining a red face on any $$n$$-th throw is $$\frac{1}{2}$$.
2. If the first $$2$$ consecutive die throws result in red faces, what is the probability that the third throw is also red?

To answer part $$1$$ I used the law of total probability. Denoting obtaining a red face on the $$n$$-th die throw as $$P(R_n)$$ I get

\begin{align*} P(R_n) &= P(R_n \vert A) P(A) + P(R_n \vert B) P( B) \\ & = \left(\frac{4}{6}\right)\left(\frac{1}{2}\right) + \left(\frac{2}{6}\right)\left(\frac{1}{2}\right)\\ & = \frac{1}{2} \end{align*}

But on question $$2$$ is where I started running into trouble. Using the same notation, what I want to calculate is $$P(R_3 \vert R_2 R_1)$$. And recalling that for events $$E_1$$ and $$E_2$$ we can say that $$P(E_2 \vert E_1) = \frac{P(E_2E_1)}{P(E_1)}$$ I get that $$P(R_3 \vert R_2 R_1) = \frac{P(R_3 R_2 R_1)}{P(R_2 R_1)} \tag{1}$$ and from here I obtain 2 different solutions using $$2$$ distinct methods:

Using that on the first part of the question we showed that $$P(R_n) = \frac{1}{2}$$, and noticing that the die throws are independent since what I threw before does not affect how I throw the next coin toss or how I roll the next die, from equation $$(1)$$ I get \begin{align*} P(R_3 \vert R_2 R_1) &= \frac{P(R_3 R_2 R_1)}{P(R_2 R_1)}\\ &= \frac{P(R_3) P(R_2) P(R_1)}{P(R_2) P(R_1)}\\ & = P(R_3) = P(R_n) = \frac{1}{2} \end{align*}

Using the law of total probability on $$(1)$$ I get \begin{align*} P(R_3 \vert R_2 R_1) & = \frac{P(R_1 R_2 R_3 \vert A)P(A)+ P(R_1 R_2 R_3 \vert B)P(B)}{P(R_1 R_2 \vert A ) P(A)+ P(R_1 R_2 \vert B) P(B)}\\ &= \frac{\left[\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\right]\frac{1}{2}+ \left[\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\right]\frac{1}{2}}{\left[\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\right]\frac{1}{2}+\left[\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\right]\frac{1}{2}}\\ & = \frac{3}{5} \end{align*}

To me both of the previous answers seem to be following coherent logic, but since I didn't get the same answer I knew one of them was wrong. I decided to write a program to simulate the game and I found out that the correct solution was $$P(R_3 \vert R_2 R_1) = \frac{3}{5}$$. But even though I verified this answer to be correct I couldn't seem to understand what part of my analysis is wrong on Answer 1.

So my question is, why is $$P(R_3 \vert R_2 R_1) \neq \frac{1}{2}\quad ?$$

• I'm confused about the nature of this experiment. Are you repeatedly tossing a coin then rolling the corresponding die $n$ times (i.e. $n$ coin tosses and $n$ dice rolls), or are you tossing a coin once, then rolling the designated die $n$ times (i.e. $1$ coin toss, and $n$ die rolls)? – Theo Bendit Mar 23 at 5:56
• It's the first option. We have one coin toss and one dice roll per turn. I'll edit the problem to clarify this, thank you! – Robert Lee Mar 23 at 5:59
• In your computation, $A$ determines the successive $3$ roll of the die. – Oolong milk tea Mar 23 at 6:07
• If it's one coin toss and one dice roll per turn, then every (coin toss + dice roll) is independent of all the others, and the probability is $1/2$ every time. If one coin toss determined multiple dice rolls, then those dice rolls would not be mutually independent, and you would get something greater than $1/2$ (because there would be a positive correlation between rolls using the same coin toss). But if you're saying it's one coin toss and one dice roll per turn, then your simulation is wrong. – mjqxxxx Mar 23 at 6:24
• I don’t quite get the logic in answer 2. To get consecutively 2 times red, you have AA, BB, AB, BA 4 paths with their own probabilities, would expect 1/2x1/2x2/3x2/3 for AA path,... and we should have 4 terms in the denominator. Do I miss something? – BStar Mar 23 at 6:38

The flaw in your second answer is that it is not necessarily the same die that is thrown after each coin toss. The way you have written the law of total probability is acceptable in your first answer, but not in the second, because it is only in the second answer that you conditioned on the events $$A$$ and $$B$$, which represent the outcomes of the coin toss. Consequently, the result is incorrect because it corresponds to a model in which the coin is tossed once, and then the corresponding die is rolled three times.

First, let us do the calculation the proper way. We want $$\Pr[R_3 \mid R_1, R_2] = \frac{\Pr[R_1, R_2, R_3]}{\Pr[R_1, R_2]}$$ as you wrote above. Now we must condition on all possible outcomes of the coin tosses, of which there are eight: \begin{align} \Pr[R_1, R_2, R_3] &= \Pr[R_1, R_2, R_3 \mid A_1, A_2, A_3]\Pr[A_1, A_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid A_1, A_2, B_3]\Pr[A_1, A_2, B_3] \\ &+ \Pr[R_1, R_2, R_3 \mid A_1, B_2, A_3]\Pr[A_1, B_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid A_1, B_2, B_3]\Pr[A_1, B_2, B_3] \\ &+\Pr[R_1, R_2, R_3 \mid B_1, A_2, A_3]\Pr[B_1, A_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid B_1, A_2, B_3]\Pr[B_1, A_2, B_3] \\ &+\Pr[R_1, R_2, R_3 \mid B_1, B_2, A_3]\Pr[B_1, B_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid B_1, B_2, B_3]\Pr[B_1, B_2, B_3] \\ \end{align} and since each of the $$2^3 = 8$$ triplets of ordered coin tosses has equal probability of $$1/8$$ of occurring, $$\Pr[R_1, R_2, R_3] = \tfrac{1}{8}\left((\tfrac{2}{3})^3 + 3(\tfrac{2}{3})^2(\tfrac{1}{3}) + 3(\tfrac{2}{3})(\tfrac{1}{3})^2 + (\tfrac{1}{3})^3\right) = \tfrac{1}{8}.$$ A similar (but simpler) calculation for the denominator yields $$1/4$$, and the result follows.

Of course, none of this is necessary; it is only shown here to illustrate how the calculation would be done if it were to be done along the lines of your second answer.

• I think I'm starting to understand where my lack of understanding is. I believe I may not understand what $P(R_3 \vert R_2 R_1)$ is to begin with. Let's say I list all my throws as red or white accordingly. If I then count how many chains of 3 consecutive reds there are on my list, and I also count the number of times 2 reds are succeded by a white, is $$P(R_3 \vert R_2 R_1) \sim \frac{\text{#RRR}}{\text{#RRW} + \text{#RRR}}$$ in the sense that $\sim$ means the RHS approaches $P(R_3 \vert R_2 R_1)$ as I take more throws? Or does $P(R_3 \vert R_2 R_1)$ mean something else? – Robert Lee Mar 23 at 7:41
• I ran out of space in my previous comment, but #RRR stands for the number of chains of 3 consecutive reds on my list of throws, and #RRW is the number of chains where the first 2 places are red and the third consecutive throw is white, again on my list of throws. Just to clarify what I meant in the last comment. – Robert Lee Mar 23 at 7:46
• @RobertLee It is unnecessary and mathematically inappropriate to think of the problem in terms of long-run averages of more than three rolls, because the entire question can be answered precisely under a strict interpretation of the model, in which there are only ever three rolls, each of which is preceded by a coin toss. – heropup Mar 23 at 7:57
• I agree that it's not necessary to think of this mathematically. I only ask this because of my previously described attempt to verify the theoretical result with a simulation, which led me to an incorrect result. I'm trying to understand if what I programmed is in fact not trying to calculate the probability $P(R_3 \vert R_2 R_1)$ that I want, or if I'm still not understanding what the probability in itself means. – Robert Lee Mar 23 at 8:07
• @RobertLee The frequentist approach via simulation would be as follows. [1] Locate all occurrences of two consecutive red dice rolls. [2] For each such occurrence, look at the outcome of the die roll immediately after the pair of reds. [3] Tally the number of outcomes where the third die is red. Divide the number obtained in step 3 by the total number of red pairs observed in step 1. – heropup Mar 23 at 8:18

With the second approach:

To have first two die throws resulting in red, there are 2x2 paths: AA,AB,BA, BB $$\to$$

$$P(R1\cap R2)=(\frac{1}{2})^2(\frac{2}{3})^2$$+2$$(\frac{1}{2})^2\frac{2}{3}\frac{1}{3}$$+$$(\frac{1}{2})^2(\frac{1}{3})^2$$ =$$\frac{1}{4}$$

To have three die throws resulting in red, there are 2x2x2 paths:AAA,BBB,AAB,ABA,BAA,BBA,BAB,ABB $$\to$$

$$P(R1\cap R2\cap R3)= (\frac{1}{2})^3(\frac{2}{3})^3$$+3$$(\frac{1}{2})^3(\frac{2}{3})^2\frac{1}{3}$$+3$$(\frac{1}{2})^3(\frac{1}{3})^2\frac{2}{3}$$+ $$(\frac{1}{2})^3(\frac{1}{3})^3$$ =$$\frac{1}{8}$$

$$\frac{P(R1\cap R2\cap R3)}{P(R1\cap R2)}$$ =$$\frac{1}{2}$$

In this way, you get consistent result compared to first approacch.