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The question was: "If rolling two 6-sided dice what is the probability of getting at least one six? One third? Why or why not?"

In order to answer that I wrote a simulation in Python and the results are a pretty convincing NO.

The results:

Ser. 0 D1=6 16671 ; D2=6 16622 ; D1 AND D2 = 6 2721 ; D1 OR D2 = 6  30572 / 100000
Ser. 1 D1=6 16629 ; D2=6 16782 ; D1 AND D2 = 6 2739 ; D1 OR D2 = 6  30672 / 100000
Ser. 2 D1=6 16618 ; D2=6 16651 ; D1 AND D2 = 6 2754 ; D1 OR D2 = 6  30515 / 100000
Ser. 3 D1=6 16596 ; D2=6 16957 ; D1 AND D2 = 6 2792 ; D1 OR D2 = 6  30761 / 100000
Ser. 4 D1=6 16575 ; D2=6 16665 ; D1 AND D2 = 6 2822 ; D1 OR D2 = 6  30418 / 100000
Ser. 5 D1=6 16630 ; D2=6 16634 ; D1 AND D2 = 6 2759 ; D1 OR D2 = 6  30505 / 100000
Ser. 6 D1=6 16585 ; D2=6 16706 ; D1 AND D2 = 6 2809 ; D1 OR D2 = 6  30482 / 100000
Ser. 7 D1=6 16707 ; D2=6 16762 ; D1 AND D2 = 6 2811 ; D1 OR D2 = 6  30658 / 100000
Ser. 8 D1=6 16568 ; D2=6 16618 ; D1 AND D2 = 6 2694 ; D1 OR D2 = 6  30492 / 100000
Ser. 9 D1=6 16705 ; D2=6 16695 ; D1 AND D2 = 6 2826 ; D1 OR D2 = 6  30574 / 100000

So every result is I expected: 16% for either die to be six 2.7% for both dice to be six

except for at least one being six.

What bothers me is WHY? And how would I derive the result from formal probability without doing the simulation?

rioZg
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3 Answers3

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The probability of getting no $6$ is easy to see; $\frac56$ for both dice. Thus, the chance you do get a $6$ is simple $1-\frac56\frac56=\frac{11}{36}$.

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probability of having at least $1$ six is equal to one minus the probability that no sixes occur.

$$1- \left( \frac56\right)^2=\frac{36-25}{36}=\frac{11}{36} < \frac{12}{36}$$

Alternatively, let $A_i$ be the event that $6$ appears in the $i$-th dice.

\begin{align} P(A_1 \cup A_2)&=P(A_1)+P(A_2)-P(A_1 \cap A_2) \\ &= \frac13 - P(A_1)P(A_2) \end{align}

Siong Thye Goh
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  • Thanks, it was exactly the formal logic that I wanted. In the meantime I worked it out from the numbers above.

    In the output above it is: D1 + D2 - (D1 AND D2) = (D1 OR D2)

     – rioZg Apr 12 '18 at 11:13
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Instead of a python script randomly evaluating throws of two dice, and getting a somewhat certain answer, how about you use a python script (or pen and paper) to go through all the 36 equally likely results when throwing two dice, and count how many of them contain at least one 6, and get an absolutely exact answer?

The reason this happens is that getting a double 6 doesn't count twice when counting the number of throws which contains a 6. So while die 1 gets a 6 one sixth of the time, and die 2 gets a 6 one sixth of the time, they don't add up to one third of the time in total, since there is some overlap. If you do count the double 6's twice (or, alternatively, don't count any die throw with a pair), then you will see 1/3 pop back up.

Arthur
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  • +1 I like this. Sample spaces give the intuition in my opinion. – Karl Apr 12 '18 at 09:11
  • It is exactly what I did. It is an excerpt of the sample space that is relevant for the question. – rioZg Apr 12 '18 at 09:13
  • @rioZg It doesn't seem to be what you've done. You seem to have simulated dice rolls many many times. This is not what Arthur means by looking at the sample space. The sample space only has $36$ outcomes in it, whereas you seem to have simulated a million trials. – Theoretical Economist Apr 12 '18 at 09:24
  • I understand what he means. I simulated million (or 100000) trials multiple times of the whole sample space. So it is actually a list with radnomly thrown dice, for example 2, 5 and so on million times. Out of that I counted only the ones that satisfy what is of interest in this question. – rioZg Apr 12 '18 at 11:11