Probability of successfully deciding which of the two six-face dice (Fair and Rigged) was rolled?

Question

Question statement: You are handed a dice which is either fair (all outcomes with probability $\frac{1}{6}$) or produces only the outcomes 2, 4, 6, each with probability $\frac{1}{3}$. If you can roll the dice once, what is the probability for you to successfully decide which of the two it is?

My approaches so far:

1. Probability of picking Fair Dice and Rigged Dice is $\frac{1}{2}$ and $\frac{1}{2}$ respectively.

2. If you picked up Fair Dice, probability of getting {1, 3, 5} is $\frac{1}{2}\cdot \frac{1}{2}$ = $\frac{1}{4}$ and I since this is the only case in which I will certainly know what dice is decisively, I thought about it as my final answer but I still could not convince myself of that fully as I feel it is disregarding probability of unfair dice. Similarly, probability of getting {2, 4, 6} is $\frac{1}{2}\cdot \frac{1}{2}$ = $\frac{1}{4}$.

3. If you picked up Rigged Dice, probability of getting {1, 3, 5} is Non existent.Similarly, probability of getting {2, 4, 6} is $\frac{1}{2}\cdot 1$ = $\frac{1}{2}$

4. I wrote a Python code to simulate it a 1000 times and got probability as 0.25...., but that was more of to get an idea or estimation and I am really unsure about that too because it says in the question what is probability rolling the dice once.

I would appreciate the help very much, I tried searching it around the web but I could not found much accessible content regarding Rigged Dice.

Answer 1

We have to start with a couple of assumptions, without which I don't see how we can make progress.

1. Each die has the same probability $(1/2)$ of being handed to us.
2. We will decide the die is fair if an odd number shows up. And that it is loaded if an even number shows up.

This is the optimal strategy.

Let

• $X$ be the event of the die being the loaded/rigged one.
• $Y$ be the event of an even number showing up.

Let's calculate $P(X|Y)$, the probability of the die being a loaded one if an even number shows up.

Clearly, $P(Y|X) = 1$. The loaded die throws up even numbers only.

Also, $P(Y|X^c) = 1/2$

Now, using the law of total probability, we have

$$\begin{align*} P(Y) &= P(X) \cdot P(Y|X) + P(X^c) \cdot P(Y|X^c) \\[0.3cm] &= \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2} \\[0.3cm] &= \frac{3}{4} \end{align*}$$

And using the Bayes' theorem,

$$\begin{align*} P(X|Y) &= \frac{P(X) \cdot P(Y|X)}{P(X)} \\[0.3cm] &= \frac{\frac{1}{2} \cdot 1}{\frac{3}{4}} \\[0.3cm] &= \frac{2}{3} \end{align*}$$

Alright, our probability of making the right decision is

$$\begin{align*} P &= \frac{3}{4} \cdot \frac{2}{3} + \frac{1}{3} \cdot 1 \\[0.3cm] &= \frac{3}{4} \end{align*}$$

The first term is our probability of being right when an even number shows up and the second term is that of being right when an odd number shows up.

Answer 2

First, if you quoted the problem correctly and you have the option of not using the text that posed the problem in this way, don’t use it. A text on probability that makes the very basic mistake of asking you for an a posteriori probability without specifying the a priori probabilties is likely to confuse you in other places, too.

Second, simplify the problem. The odd and even numbers are just distractions; the problem is equivalent to one where you have a fair coin and a coin that shows heads on both sides.

So let’s assume, since it wasn’t specified, that the coin isn’t handed to you by a game show host pursuing a certain strategy, but by someone who uniformly randomly picked it out of a box containing a fair coin and a coin with heads on both sides.

Then you have $4$ equiprobable elementary events, corresponding to all possible combinations of the two coins and the two sides. If your strategy is to guess that the coin is fair exactly if it shows tails (which is the optimal strategy), then you’ll guess correctly in $3$ of those $4$ equiprobable cases. Thus, your probability to guess correctly is $\frac34$.