There are two tetrahedra, on each face of which the numbers 1, 2, 3 and 4 are applied. When ejected, the tetrahedron is equally likely to fall on the surface with any face. Both tetras are thrown simultaneously. How many bits of information will carry the message that one of the tetrahedrons fell to the face with the number 1 while the second tetrahedron fell to the face with the number 2. In the answer, indicate an integer.

I solved this problem like this: log(2)4 + log(2)4 = 4 бита .

Where is the mistake? Why is this decision wrong?

    2 answers 2

    The answer, of course, is different. Your decision and the @teveakea solution is correct, if the condition is understood so that the first tetrahedron gave 1, and the second - 2. And in the condition one tetrahedron gave 1, and the other - 2. Thus, 2 of the 16 experimental results are suitable (1, 2 ) and (2, 1). Well, now read the logarithm :)

    • Please write in detail. - Eugene536
    • One is optional first, the other is optional second. Only 16 outcomes of the experiment, pairs (X, Y), where X and Y are a number from 1 to 4. Your solution uses only a pair (1, 2), while the condition implies any pair (1, 2) or (2, 1 ). That is, the probability of a given set condition is not 1/16, but 2/16. - northerner
    • Exactly, thanks a lot, it turns out log (2) 8 = 3 bits, and that is exactly the answer. - Eugene536

    You can do this: we encode the state of one tetrahedron bits

    • dropped 1 - 0001
    • dropped 2 - 0010
    • dropped 3 - 0100
    • dropped 4 - 1000

    Now we describe event 1 tetrahedral fell 1, and the second 2 = 00010010 - that is, we encode the state of one tetrahedral with 4 digits.

    Accordingly, your event carries information in 8 bits = 1 byte.

    • Generally some kind of crap, the answer is not close - Eugene536