Solution to Puzzle #78: Probability of #6

This was a tricky question…somewhat similar to the Monty Hall Problem (please see https://alokgoyal1971.com/2013/06/09/puzzle-20-the-monty-hall-problem-very-interesting-must-try/). I got a number of answers and a few of them correct in my opinion. The ones that sent the correct answers include Vishv Vivek Sharma (my nephew in Canada) and Girish Tutakne from Singapore.

I am taking the liberty of copying the answer from Girish:

P(Die is actually a 6 / He reports it as 6) = P(Die is a 6 and he reports it as 6) / P(He reports it as 6).

P(Die is a 6 and he reports it as 6) = 1/6 * 3/4 (3/4 is the probability of his speaking the truth)

Denominator is (5/6×1/4)+(1/6×3/4) = 1/3 (first expression is the probability of it not being a 6 [5/6] and then the person lying [1/6] added to the numerator)

Doing the calculations will yield the answer as 3/8.

Hope you enjoyed the puzzle!

 

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6 Responses to Solution to Puzzle #78: Probability of #6

  1. Girish Tutakne says:

    Hi Alok: Actually the denominator should be (5/6×1/4×1/5)+(1/6×3/4); the additional multiplication by 1/5 to cater for the fact that in the 5/6 cases that it is not 6 and he lies, the lie can be any of the other 5 numbers. Taking the calculation to the end, the actual probability will then work out to 3/4 (and not 3/8), which interestingly is the probability of his speaking the truth!
    Logically, the answer seems accurate because all events are equally likely, and hence, the probability of it being a 6 when he says it is a 6 should be 3/4.
    Btw, I am now in Manila 🙂

  2. Rakesh Garg says:

    Hi Alok – this is a nice trick puzzle. But I agree with Girish in his answer – that it should be 3/4. I view it very simply – It doesn’t matter whether the guy is rolling a dice or a roulette table. His probability of speaking the truth is 3/4. Therefore, if he says the result is a 6, the probability of it actually being a 6 is 3/4. All other analysis is redundant.
    – Rakesh

  3. Shyam baddepudi says:

    To compute the fact that he calls 6 when it is not 6 should be

    is (5/6×1/4) * 1/5 as he could have called any number other than the actual number and assuming he calling 6 is random..

  4. Actually I used the following logic and want to know what is wrong?

    P(6) = P(actually it was 6 and he called 6) + P(actually it was not 6 and he called 6)

    1/6*3/4 + 5/6*1/5*1/4

    3/24 + 1/24

    1/6 which is the original throw of dice !

    • Alok Goyal says:

      Many of you have sent a comment on this answer and the underlying issue of whether or not a probability factor of 1/5 should be added whenever the person guesses 6 when it is not 6. In my opinion, there is an ambiguity here – the puzzle does not specifically say that when the person lies, the lie is “random”. As such, I have not applied that factor, though I completely appreciate the point that everyone has made.

  5. Girish Tutakne says:

    Shyam: what is wrong in your analysis is that you are answering a different question. The question you are answering is what is the probability of a 6. The actual question is what is the probability it is actually a 6 when he says it is a 6.

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