# Total number of combinations when a dice is thrown twice

I was trying some questions on probability when I came across this question:

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

So in this case while calculating the probability, the sample space is: 6 * 6 = 36.

Now let me change this question:

One dice is thrown twice. What is the probability of getting two numbers whose product is even?

In this case I think sample space should be: 6 * 5 / 2 = 15.

Am I correct?

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Why are you assuming that the die will roll two different numbers, and that you cannot distinguish between rolling (for example) 4-and-then-3 from 3-and-then-4? Whether you roll two dice at the same time, or one die at two different times, you have still performed two distinguishable die rolls. –  Niel de Beaudrap Feb 5 '12 at 9:13
thanks... all doubts cleared :-) –  Miraaj Feb 5 '12 at 9:43

No.

Throwing two dice is the same as throwing a dice twice.

Let's think naturally:

What is the set of outcomes when a die is thrown twice?

It is just the set of all ordered pairs such that, the first of the two entries tell you the outcome on the first throw and the second of the entries tell you the outcome on the second throw.

What is the set of outcomes when two dice are thrown?

Without loss of generality, I can label the dice as die A and die B. So, now, the set of outcomes will be the set of ordered pairs such that he first of the two entries tell you the outcome on die A and the second of the entries tell you the outcome on the die B.

So, radically, the set of outcomes, technically called the sample space, is the same. To conduct an experiment with $100$ dice, if you have the means, you could buy $100$ dice and use a method that is reasonably unbiased in generating outcomes, save time or throw a single die $100$ times, the outcomes will be unbiased but you'll waste time.

The point is both methods have their own practical merits and demerits.

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ok... thanks... all doubts cleared :-) –  Miraaj Feb 5 '12 at 9:42

If you throw a dice a second time, all six numbers are still just as available as with the first toss.

Additionally, rolling $\bf x$ and then a $\bf y$ is different from $\bf y$ then $\bf x$.

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In a simultaneous throw of two dice, we have $n(S) = (6 \cdot 6) = 36$.

Then, $E = \{(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}.$

$n(E) = 27$.
$\mathbf{P}(E) = n(E) = 27 = 3 \cdot n(S)$ 36 4

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Welcome to MSE! I edited to the best of my ability (it is currently under review), but could not make sense of the 36 4 at the end. Could you clarify? –  gnometorule Feb 20 '13 at 4:16

3 ordered pairs and 32 un ordered pairs out come

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Could you explain how you got these and how they help to answer the question? –  Michael Albanese Oct 18 '13 at 14:54