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Conditioned die

Posted by Chris on January 11, 2011 – 8:23 am

I have a perfectly standard die. i.e. If I roll it I will get a 1, 2, 3, 4, 5 or 6 with equal probability.

Q1. If, out of my sight, someone rolls that die and announces that it isn’t showing a 5, what is the probability that it is showing a 6?

You and your colleague have agreed in advance, that should a 5 be rolled, then your colleague will say nothing, and will roll the die again. He will keep on rolling the die until it isn’t showing 5. Your colleague doesn’t lie or cheat or any other such trickery.

Q 2. As before but now the colleague will announce when the die isn’t showing 4 or 5 (and he’ll only do that when the die isn’t showing a 4 or a 5 . What is the probability that it is showing 6?

Q3. Same idea, but not showing 3, 4 or 5. What is the probability of it showing 6?

Q4. Same idea, but not showing 2, 3, 4 or 5. What is the probability of it showing 6?

and finally,
Q5. Same idea, but not showing 1,2,3,4 or 5. What is the probability of it showing 6?


This post is under “Logic, MathsChallenge” and has 5 respond so far.
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  1. 1. DP Said:

    1: 1 in 5 (1, 2, 3, 4 and 6 are the only possibilities)
    2: 1 in 4 (1, 2, 3 and 6 are the only possibilities)
    3: 1 in 3 (1, 2 and 6 are the only possibilities)
    4: 1 in 2 (1 and 6 are the only possibilities)
    5: it has to be a 6

  2. 2. Chris Said:

    Thanks DP. I expect that you might have a good idea where this one is going :)

  3. 3. Chris Said:

    Obviousl I posted this problem in order to lay another line of attack for the Girls, girls, girls problem.

    First the conventional analysis.

    A1. Using P’(N) to represent the conditional probability that an N will be rolled given that a 5 wasn’t rolled we must have P’(1) + P’(2) + P’(3) + P’(4) + P’(6) = 1, and P(1) = P’2) = P’(3) = P’(4) = P’(6). Solving gives P’(n) = 1/5 for n = 1,2,3,4,6.

    Similar arguments for all the other cases.

    Now for a list version. For a normal die the possible throws are 1,2,3,4,5,6. There are 6 equallly likely outcomes, so the probability of rolling any specified value (and a 6 in particular is 1/6).

    A2. We now know that we didn’t throw a 5. So that means the list is now 1,2,3,4,6. That’s 5 equally likely outcomes, so probabilty of throwing 6 is now 1/5. etc.

    Note that this is exactly the same logic that I’ve used in the Girls, girls, girls problem. The 1, 2, 3, 4, 5 and 6 correspond to BBB,BBG,…,GGG. Or if you like, the families corrrspond to the face of a hypothetical 8 sided die. (In fact a spinning top randomiser would be an achievable choice.)

    But sticking with the (non-constructable) die, I now associate (in no particular, but fixed order) a mapping from one to the other. The die has faces BBB,BBG,BGB,BGG,GBB,GBG,GGB,GGG. Corresponding to saying roll the die and get “not a 5″, I’d say roll the die and get “not BBB”. Then you’d get the 1/7 for GGG.

  4. 4. Chris Said:

    Oooh and now for a formal conditional probability version for Q1.

    P(A|B) = P(A&B) / P(B) NB I’m using & to represent logical intersection (AND). Many websites use a sort of upside down U do that. A|B is read “A given B”

    A is rolling 6. B is the condition that we didn’t roll a 5 = we rolled 1+2+3+4+6 (“+” here => logical Union (or OR))

    P(not 5) = 1 – P(5) = 1 – 1/6 = 5/6.

    So P(6 | not 5) = P(6 & (1+2+3+4+6)) / P(1+2+3+4+6)
    = P(6) / P(1+2+3+4+6) = (1/6)/(5/6) = 1/5

  5. 5. Chris Said:

    …just in case you think I’m picking a simple sample case

    P(A|B) = P(A&B) / P(B) NB I’m using & to represent logical intersection (AND). Many websites use a sort of upside down U do that. A|B is read “A given B”

    A is rolling 6. B is the condition that we didn’t roll a 4 or 5 = we rolled 1+2+3+6 (“+” here => logical Union (or OR))

    P(not (4 or 5)) = 1 – P(4) – P(5) = 1 – 2/6 = 4/6
    So P(6 | not 4 or 5) = P(6 & (1+2+3+6)) / P(1+2+3+6)
    = P(6) / P(1+2+3+6) = (1/6)/(4/6) = 1/4

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