What will be the solution to this probability question?

Question

Both Alexa and Shelly have an infinite bitstring. Alexa’s bitstring is denoted by a1a2a3 . . ., whereas Shelly’s bitstring is denoted by s1s2s3 . . .. Alexa can see her bitstring, but she cannot see Shelly’s bitstring. Similarly, Shelly can see her bitstring, but she cannot see Alexa’s bitstring. The bits in both bitstrings are uniformly random and independent. The ladies play the following game: Alexa chooses a positive integer k and Shelly chooses a positive integer l. The game is a success if sk = 1 and al = 1. In words, the game is a success if Alexa chooses a position in Shelly’s bitstring that contains a 1, and Shelly chooses a position in Alexa’s bitstring that contains a 1

• Assume Alexa chooses k = 4 and Shelly chooses l= 7. Determine the probability that the game is a success.

• Assume Alexa chooses the position, say k, of the leftmost 1 in her bitstring, and Shelly chooses the position, say l  of the leftmost 1 in her bitstring. –

If k is not equal to l, is the game a success? –

Determine the probability that the game is a success.

Answer 1

Alexa knows that, counting from the left, she is is selecting s4 and Shelley knows she is selecting a7.

Since s4 and a7 can each be a 1 or zero, the game is a success only if these are both ones. There is an equal probability of selecting 1 or 0, so there are 4 possible outcomes for s4a7: 00, 01, 10, 11. Therefore the probability of 11 is ¼ or 0.25.

Assume a4 and s7 (for example) are the selected bits, then a4=s7=1 by definition of selection. Since 4 is not equal to 7 then the conditions of selection have been satisfied. But a4 and s4, and a7 and s7 are not related, so the probability of s4=a7=1 is unaffected. If k=l then the game is a success with a probability of 1. If k≠l the game is unaffected. So the probability remains at 0.25.