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If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?

I think the permutations formula is needed i.e. $n!/(n-r)!$ because order matters but I'm not sure if n is 3 or 6 and what would r be?

Any help would be much appreciated!

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5 Answers 5

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Let's calculate the probability, then convert that to odds.

On a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $\dfrac{1}{2}$.

The probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $\dfrac 1 2$ of obtaining the desired result. So, we have:

$$P(E,E,O) = \dfrac 1 2 \cdot \dfrac 1 2 \cdot \dfrac 1 2 = \dfrac 1 8$$

Now, the probability of that not happening is $$1-\dfrac 1 8 = \dfrac 7 8$$

So, the odds are 7:1 against the desired outcome.

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I think you might be over complicating things.

It has to be even on the first two, the probability of this is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.$

The probability of odd on the third roll is also $\frac{1}{2}$ so your final probability is $\frac{1}{8}.$

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  • $\begingroup$ This is simpler. $\endgroup$
    – Snowcrash
    Sep 21, 2018 at 14:44
  • $\begingroup$ ... this is using odds... not probability $\endgroup$
    – Jason Kim
    Sep 22, 2018 at 3:36
  • $\begingroup$ @jason Kim what do you think the distinction is..? $\endgroup$
    – MRobinson
    Sep 22, 2018 at 7:59
  • $\begingroup$ Odds for is (prob)/(1-prob) which demonstrates the difference... $\endgroup$
    – Jason Kim
    Sep 24, 2018 at 0:58
  • $\begingroup$ @JasonKim if the probability is $\frac{1}{8}$ then the odds are $1$ in $8$. If you want to write the odds like a betting shop then you'll have $1:7$. I struggle to see how you believe I have used odds and not probability. $\endgroup$
    – MRobinson
    Sep 24, 2018 at 7:32
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Given a fair die, the probability of any defined sequence of three evens or odds is the same, namely, $1/2 \times 1/2 \times 1/2 = 1/8$. So the odds are $7$ to $1$ against.

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Assuming the die is just marked even and odd rather than with numbers, there are eight orderings. They range from all odd to all even: OOO, OOE, OEO, EOO, OEE, EOE, EEO, EEE. In your formula:

$$\frac{n!}{(n-r)!}$$

You are missing that you don't care about the orders of the duplicates. So you need

$$\frac{n!}{(n-r)!r!}$$

The $n$ is the total number of rolls, the $r$ is the number of either evens or odds (it's symmetric). But to get the total number of orderings, you need to add these:

$$\sum\limits_{r=0}^n\frac{n!}{(n-r)!r!}$$

Now substitute 3 for $n$.

$$\sum\limits_{r=0}^3\frac{3!}{(3-r)!r!}$$

Unrolling that, we get

$$\frac{3!}{3!0!} + \frac{3!}{2!1!} + \frac{3!}{1!2!} + \frac{3!}{0!3!}$$

or

$$1 + 3 + 3 + 1$$

So we have one all odds, three with one even, three with two evens, and one all even. That's eight total.

Another way of thinking of this is that there is only one ordering of all odd numbers or all even numbers while there are three places where the lone odd or even number can be.

Of those eight, how many fit your parameters? Exactly one, OOE. So one in eight or $\frac{1}{8}$.

As others have already noted, you could get that much more easily by simply figuring that you have a one in two chance of getting the result you need for each roll. There's three rolls, so $(\frac{1}{2})^3 = \frac{1}{8}$.

If you want to treat 1, 3, and 5 as different values and 2, 4, and 6 as different values, you can. But it is much easier to think of them as just odd or even. Because you don't want to try write this out for $6^3 = 216$ orderings. And in the end, you will get the same basic result. You will have twenty-seven OOE orderings, which is again one eighth of the total. This is because there are three possible values for each, 1, 3, and 5 for the two odds and 2, 4, and 6 for the even. And $\frac{27}{216} = \frac{1}{8}$.

Permutations leads you down a harder path. It's easier to think just in terms of probability or even ordering.

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All of the above or below :-} answers are correct that is 50:50 for each throw so your "formula" is

  n (2 -1):1 against where n is the number of throws

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    $\begingroup$ Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ Sep 21, 2018 at 22:08

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