S2T2

Jim Randell 8:33 am on 2 February 2021 Permalink Reply
Tags: by: Ian Kay ( 5 )   

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  Jim Randell 8:35 am on 2 February 2021 Permalink | Reply

  The fair die has a probability of 1/6 of showing any particular face (scores 1-6).

  The unfair die has a probability of p of showing 6, and (1 − p) / 5 of showing the other faces (scores 1-5).

  If we consider the probability of throwing a total of t (scores 2-12) using both dice:

  There are { n = (t − 1) if t ≤ 6; otherwise n = (13 − t) } different ways of achieving the total.

  And with two fair dice (which have 36 equally like outcomes) the probability of throwing t is n / 36.

  And we want to find when the probability of throwing t with one fair and one unfair dice is twice this value: i.e. when the probability is n / 18.

  If t ≤ 6, then neither die can show a 6, so the probability of throwing t is:

  P = n ((1 − p) / 5) (1/6)

  which we can solve for P = n / 18:

  n ((1 − p) / 5) (1/6) = n / 18
  3(1 − p) = 5
  p = −2/3 (not possible)

  If t > 6, then 6 can show on the unfair dice, so we have two ways to make the total:

  P = [(n − 1) ((1 − p) / 5) (1/6)] + [p (1/6)]

  and we can solve this for P = n / 18:

  (n − 1) ((1 − p) / 5) (1/6) + p / 6 = n / 18
  3(n − 1)(1 − p) + 15p = 5n
  18p − 3np − 3 = 2n
  p = (2n + 3) / (18 − 3n)

  Also for t > 6 we have: n = 13 − t, so:

  p = (2(13 − t) + 3) / (18 − 3(13 − t))
  p = (29 − 2t) / (3t − 21)

  And we are interested in cases where t = 7 .. 12 and 0.5 < p ≤ 1:

  t = 7 ⇒ p = undefined (not possible)
  t = 8 ⇒ p = 13/3 (not possible)
  t = 9 ⇒ p = 11/6 (not possible)
  t = 10 ⇒ p = 1
  t = 11 ⇒ p = 7/12
  t = 12 ⇒ p = 1/3 (not permitted)

  So we find there are 2 possible situations:

  (1) The loaded die always shows 6, and the setter threw 10.
  The probability of throwing 10 with two fair dice is 1/12.
  The probability of throwing 10 with one fair die and the loaded die is 1/6.

  (2) The loaded die has a probability 7/12 of showing 6, and the setter threw 11.
  The probability of throwing 11 with two fair dice is 1/18.
  The probability of throwing 11 with one fair die and the loaded die is 1/9.

  I suspect we are to ignore that case where the loaded die always shows 6, as this would be quite obvious. (Although I would have thought a die that shows 6 more than half the time would be fairly obvious anyway). The setter could have explicitly excluded this case by specifying “an equal non-zero chance of scoring 1 – 5″ for the loaded die.

  Solution: The probability of scoring 6 with loaded die is 7/12.

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  John Crabtree 2:16 am on 4 February 2021 Permalink | Reply

  I was fortunate and took the probability of the throwing 6 with the biased die as n times that of any other number ie n>5 and p = n/(n+5).
  I reached t = 10 + 9/(n+2) when Jim’s two situations are immediately apparent.

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