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The problem goes as following : Which of the following has a higher probability of happening :
a) person rolls 4 dice at once, one of them must must fall at 6
b) person rolls 2 dices 24 times, at least one of those 24 times the combo 1-1 must fall?

Any help will be greatly appreciated.

  • 2
    Hint: in both cases it is easier to work from the complement. That is, compute the probability that player A does not get any sixes, and the probability that player B never gets $(1,1)$. – 2017-02-01
  • 1
    Perhaps of some interest: this is the "Chevalier de Mere's problem": http://www.cut-the-knot.org/Probability/ChevalierDeMere.shtml#explain – 2017-02-01

3 Answers 3

1

If you work by complement the solutions are much easier

a) $p_a = 1 - p(\text{ no dice falls on 6 }) = 1 - \left(\frac{5}{6}\right)^4 \approx 0.517$

b) $p_b = 1 - p(\text{ no 1-1 combination }) = 1 - \left(\frac{35}{36}\right)^{24} \approx 0.491$

  • 0
    should it not be 34/36 ? – 2017-02-01
  • 1
    $$ p(\text{ 1-1 combination }) = p(\text{ 1st die falls on 1 }) * p(\text{ 2nd die falls on 1 }) = \frac{1}{6} * \frac{1}{6} = \frac{1}{36} $$ $$ p(\text{ no 1-1 combination }) = \frac{35}{36} $$ – 2017-02-01
0

Case 1-

$p(6) = \frac16$

Then,

$q(Other) = 1 - \frac16 = \frac56$

Now according to bionamial distribution

$P(X=r) = \binom nr × (q)^{n-r} × (p)^r$

Here X = 1 and n = 4.

$P(X=1) = \binom 41 × (\frac 56)^3 × (\frac 16)$

Case 2-

At least 1 pair of (1,1) = 1 - None cases with (1,1)

$= 1 - (\frac{35}{36})^{24}$

-1

Case 1 - Two dices each with a probability of rolling 6 at 1 / 6th. Since you need just one of them to be 6. Total probablility is

$$\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{2}{3}$$ $$66.66\%$$ There is a 66.66% probability that one of your two dices will roll a six.

Case 2 -

2 dices rolled 24 times,
number of combinations / roll is given by $\frac{1}{6^{2}}$ since we have 6 possibilities and each with a combination of six more possibilities on die two. Do this 24 times and total probability becomes.

$$=24 * \frac{1}{6^{2}}$$ $$=\frac{3}{4}$$ $$=75\%$$ There is a 75% probability that one of your 24 rolls will result in a 1,1 roll.

Case 2 has a higher probability.

Edit: If you feel my answer is wrong I'd like to know why?

  • 0
    Your first case doesn't address the OP's question, since that asks about 4 single-die throws not 2. In both cases, your calcualtions are wrong- I suggest you read the other answers to this post to see what the correct answer is. – 2017-02-02
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    I agree on the first one. Now it's answer can be 2 / 3 from my point of view. I feel the question did not ask for **just** one single 6 among all of them. In that case I can say there is a $\frac{4}{6^{4}}$ probability. But I can see one of them must be a 6. I felt that 4 6's, 3 6's, 2 6's, 1 6 or with any combinations will have no effect. Thus they can be treated as disjoint maybe? and thus a 2 /3 answer. – 2017-02-03
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    I did the same with the second problem. If I'd have considered that there is a need to calculate combinations then I'd have done the same as others?! – 2017-02-03
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    So if you rolled six dice, the chance of getting at least one 6 would be $\frac{1}{6}\times 6 = 1$? Does that seem a credible answer? How about rolling twelve dice, the chance of at least one 6 would be $\frac{1}{6}\times 12 = 2$ by your method. Have you studied the other answers to and comments on this question, where the correct reasoning is explained? – 2017-02-03
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    I have looked at other answers and they do not look intuitive to me, so I have to consider all of them as a sequence like, one probability over 6 and for each of those there is one more.. and there are $(\frac{1}{6})^{4}$? Every one is talking about the reverse approach, why doesn't any one talk about the straight forward approach? – 2017-02-03
  • 1
    For the first part, your formula would have been true if when you roll the dices, you cannot fall twice on the same value. as this is not the case you cannot simply add the probabilities You basically made the same error on the second part by considering the 24 rolls at $\frac{1}{36}$ probability made a $\frac{24}{36}$ probability – 2017-02-03
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    okay, I get why I am wrong. Great, @BigPanda why did you use a reverse approach? Why is the straight forward approach hard? Is it like drawing a probability tree and selecting all branches that have atleast a one 6? Our professor has taught us. Can you explain how you ended up getting your approach? – 2017-02-04
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    When you need to have at least one result among others, your global result is not the combination of independent events anymore. A good rule of thumb is to use a reverse approach when you want to know the probability of at least one event among several tries (because when you reverse it the problem becomes determining the probability of the event never happening). – 2017-02-06