0
$\begingroup$

Probability Puzzle - secret four-digit binary message

I too came across the same question as asked here in that app itself. But what I thought is significantly different from the answer accepted.

I believe that since unscrambled message and scrambled message have a one-one relation - any pair of u,s will have a probability of occurrence of 1/16. So i think answer should be 1/16. Why should Eva have any role to play here? She sees the scrambled message and the q asks probability of unscrambled being 1111 which is only possible if pattern is ___X which has a probability of 1/16. Any ideas on this? Thanks!

$\endgroup$
1
  • $\begingroup$ You have found the probability that Eve sees 1111 given that the secret message is 1110 (you are correct; this is 1/16). What the question is asking for is the opposite: the probability that the secret message is 1110 given that Eve has seen 1111. $\endgroup$
    – jacob
    Jul 27, 2022 at 19:55

1 Answer 1

0
$\begingroup$

The pads are uniformly random, but there is no reason that the unscrambled messages are uniformly random. The implicit assumption in the question is that Eve's estimates are valid, i.e., we can assume that the unscrambled message is generated so that it has chance $0.6$ to be $1111$. The calculation then shows that knowing the scrambled message seen by Eve has no effect on the distribution of the unscrambled message (The two messages are independent due to the independence and uniformity of the pad). This yields the claimed answer of $0.6$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .