The E(T) is 4 when A and B, two players, alternately toss a fair coin (A tosses the coin first, then B tosses the coin, then A, then B and so on).
Given that,
A and B, two players, alternately toss a fair coin (A tosses the coin first, then B tosses the coin, then A, then B and so on). The heads and tails are recorded in order. The game is over and the winner is the person who throws the tail if there is a head followed by a tail (HT sub-sequence).
We have to find what is the anticipated game length. Let the game continue for T number of coin tosses. Find E (T).
We know that,
Here,
E(T )=first time sequence HT occurs
E(T) = P(first H)× P(expected number till 1st T |first H)+ P(first T)× P(expected number till 1st HT |first T)
E(T) =0.5×(1+2)+0.5×(1+E(T))
0.5×E(T)=0.5×4
E(T) =4
Therefore, The E(T) is 4 when A and B, two players, alternately toss a fair coin (A tosses the coin first, then B tosses the coin, then A, then B and so on).
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