A coin is tossed if a dice is rolled

Question

I was given this question yesterday.

A dice is rolled. If the number is even, a coin is tossed. If it is odd, the dice is rolled exactly once again and results are recorded. Find the probability that an odd number occurs atleast once on the dice.

I feel the answer is $\frac12$. My teacher (who is an IITian) says that it is $\frac34$.

His approach

He says that we first write the sample space.

$$S={2h,2t,4h,4t,6h,6t,11,12,13,14,15,16,31,32,33,34,35,36,51,52,53,54,55,56}$$

which implies that the number of elements in sample space, $n(S)=24$

Now we define the event $E$ and give a set of favourable outcomes.

$$E={11,12,13,14,15,16,31,32,33,34,35,36,51,52,53,54,55,56}$$

which gives us $n(E)=18$

Now we use the formula $P(E)=\frac{n(E)}{n(S)}=\frac{18}{24}=\frac34$

My approach

I said that, since all outcomes do not have equal probability, we cannot use the formula. I gave another approach.

$$P(odd\ followed\ by\ odd)=\frac12\times\frac12=\frac14$$ $$P(odd\ followed\ by\ even)=\frac12\times\frac12=\frac14$$ $$P(even\ followed\ by\ either\ odd\ or\ even)=0$$

Hence $$P(atleast\ one\ odd)=P(odd\ followed\ by\ odd)+P(odd\ followed\ by\ even)=\frac14+\frac14=\frac12$$

And.....

I was quite sure of my answer, but so was he. He explained that the formula still works. I took up the issue twice, with no conclusion.

Request

Please let me know what the right answer is and why. Also give detailed (but simple) explanation as to why this is true. And if you have any degrees in math, please include that in your answer to make your answer credible.

Answer 1

You are correct. The second roll of the dice does not make a difference in the probability because an odd number will already be rolled, so it is $\frac{1}{2}$. Perhaps you could ask your teacher to consider the possibility that the dice can output numbers up to $1,000,000$ (or some other arbitrarily large even number). Then according to him the chance of rolling an odd number should be very high, when clearly half the rolls will be even and result in no odd numbers.

Answer 2

My answer should be a comment, but I cannot post comments yet.

I think there is a problem in the formulation: with this formulation, you are right, but in fact the coin is useless, and it's the same for the second roll.

But if you replace the sentence by:

A dice is rolled. If the number is even, a coin is tossed. If it is tail, the dice is rolled exactly once again and results are recorded. Find the probability that an odd number occurs atleast once on the dice.

Then the whole exercise makes more sense, and the answer is not the obvious 1/2.

I think the misunderstanding come from here.

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EDIT:

Let's try to change this comment into an answer by searching why your teacher's approach is incorrect.

• The sample space, according to the question, is correct.
• The set of favourable outcomes is correct too.

Then the problem should come from the formula: $$P(E)=\frac{n(E)}{n(S)}$$

The formula exists, but cannot be used with any sample space S: this formula works only if every outcomes of the sample space have the same probability of happening. The sample space must be equiprobable.

In his approach, the sample space your teacher gave is not equiprobable: for instance, $2h$ has a bigger probability to happen than $13$, since it's more likely to get $h$ in $\{h,t\}$ than $3$ in $\{1,2,3,4,5,6\}$ (this statement can be proved if it is not considered obvious).

Since the sample space is not equiprobable, you cannot use that formula, and that is where your teacher went wrong.