Question

Two players, A and B alternatively toss a fair coin (A tosses the coin first, then B tosses the coin, then A, then B and so on). The sequence of heads and tails is recorded. If there is a head followed by a tail (HT sub-sequence), the game ends and the person who tosses the tail wins. What is the expected length of this game? Let T number of coin tosses until the game ends. Find E(T). Round answer to 3 decimals

Answer 1

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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