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This conditional probability question has me confused. A worked solution would be much appreciated to help me understand questions of this form. The question is:

A die is rolled until a 3 or 6 appears, otherwise it is rolled four times. Given that a 3 or 6 did not appear in either of the first two rolls, find the probability for that the die was tossed four times.

I don't understand how the die being rolled four times if a 3 or 6 doesnt appear fits into the solution. I have a few more questions like this I can practice on, but would like to properly understand this one before proceeding.

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    $\begingroup$ Hint: Given that a 3 or 6 did not appear in either of the first two rolls, find the probability that the die was NOT tossed four times. $\endgroup$
    – Did
    Aug 11 '16 at 12:26
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The dice is rolled until a 3 or 6 appears, or until four throws have been made, whichever comes earlier. (The fourth throw can be 3 or 6.) If we have already rolled the dice twice, here are the possibilities:

  • 3 or 6 gets rolled on the third roll. This has probability 1/3.
  • The third roll doesn't produce a 3 or 6. This means we go to a fourth and final roll, and the probability of that is 2/3.

Since these are probabilities conditional on the dice being rolled twice without a 3 or 6 showing, 2/3 is the answer.

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BAYES THEOREM

P(A|B) = P(A and B) / P(B)

A = dice tossed 4 times (no 3 or 6 in first 3 rolls)

B = no 3 or 6 in first 2 rolls

now if A is true, then B had to be true, so P(A and B) = P(A) - that's a hard part actually, but the principle is common in Bayes' theorem, conditional probability

P(A and B) = P(A) = (2/3)^3 = 8 / 27

P(B) = (2/3) ^ 2 = 4 / 9

P(A|B) = (8 / 27) / (4 / 9) = 2 / 3

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You can rephrase the question as:

Find the probability that neither 3 nor 6 appeared in the 3rd roll

The answer to that is obviously $1-\frac26=\frac23$.


If you prefer to take the formal route:

  • Let A denote the event in which $4$ rolls where made
  • Let B denote the event in which neither 3 nor 6 appeared during the first $2$ rolls
  • Then $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{\left(1-\frac26\right)^3}{\left(1-\frac26\right)^2}=1-\frac26=\frac23$

Please note that although $A\cap B=A$ in this specific example, it is not necessarily the case.

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