First to the sequence HT between two players

Question

Two players, A and B, alternatively toss a fair coin (A tosses first and then B). The sequence of heads and tails is recorded. If there is a head followed by a tail (HT subsequence), the game ends and the person who tosses the tail wins. What is the probability that A wins the game?

The solutions states that $P(A) = 1/2*P(A|H) + 1/2*P(A|T)$ and takes $P(A|T) = 1-P(A)$ which makes sense. However, I can't understand why it takes $P(A|H) = 1/2*(1-P(A|H))$. I know the part $(1-P(A|H))$ has to do with the case when B gets a H, but I don't understand the logic behind this term.

Thank You

Thank You

Answer 1

PLEASE REVIEW:

Can someone please help me spot the error in following approach. Let $A$ be the event when player A wins and $B$ be the event when player B wins. And $H_A$ represents Heads drawn by A and $H_B$ represents Heads drawn by B (and likewise Tails).

So, $P(A) = \frac 12P(A|H_A) + \frac 12P(A|T_A)$

$\implies P(A) = \frac 12P(A|H_A) + \frac 12(1 - P(A))$ [$\because$ by symmetry, $P(A|T_A) = P(B) = 1 - P(A)$]

$\implies 3P(A) = 1 + P(A|H_A)$ ... (Eq. 1)

$\implies 3P(A) = 1 + (\frac12 P(A|H_AH_B) + \frac12 P(A|H_AT_B))$ [$\because$ B's turn to flip]

$\implies 3P(A) = 1 + (\frac12 P(A|H_AH_B) + 0)$ [$\because$ B has won in event of $H_AT_B$ ]

$\implies 3P(A) = 1 + \frac12 (\frac12 P(A|H_AH_BT_A) + \frac12 P(A|H_AH_BH_A))$ [$\because$ A's turn to flip]

$\implies 3P(A) = 1 + \frac12 (\frac12 *1 + \frac12 P(A|H_AH_BH_A))$ [$\because$ A has won in event of $H_AH_BT_A$ ]

$\implies 3P(A) = 1 + \frac14 (1 + P(A|H_A))$ [$\because P(A|H_AH_BH_A) = P(A|H_A)$]

Using eq.1, $3P(A) = \frac 54 + \frac14 (3P(A) - 1)$

$\implies \frac{9}{4}P(A) = 1$

$\implies P(A) = \frac49$

Answer 2

The game begins with player A, who either tosses a head or a tail.   By the total probability theorem:$$\mathsf P(A)=\tfrac 1 2 \mathsf P(A\mid H)+\tfrac 1 2 \mathsf P(A\mid T)$$

If A first tosses a tail, then the game repeats with player B starting first.   By symmetry then: $$\mathsf P(A\mid T) = 1 - \mathsf P(A)$$

If A first tosses a head, then the next player to toss a tail wins. If B does not do so on the next toss, then at that point B has the same probability of winning as A did before, thus: $\mathsf P(A\mid HH)=1-\mathsf P(A\mid H)$, and hence:

$$\mathsf P(A\mid H) = \tfrac 12 (1-\mathsf P(A\mid H))$$

Answer 3

Start with a simpler problem: let $\phi$ be the probability that $A$ is the first to throw a Tails. That is easy to compute...we look at one toss. $A$ wins if it comes up T and, should it come up H, the game restarts with $A$ as the second player. Thus $$\phi=\frac 12\,1+\frac 12\,(1-\phi)\;\;\Rightarrow\;\;\phi=\frac 23$$ Now back to your game. Let $p$ be the answer you seek. Again, we consider one toss. If it is T then the game restarts with A as the second player (so A's odds of winning along this path are $1-p$). If the first toss is H then the game is won by whomever throws the first T; A is tossing second so from our earlier result A's probability of winning along this path is $\frac 13$. Thus $$p=\frac 12\,(1-p)+\frac 12\,\frac 13\;\;\Rightarrow\;\;p=\frac 49$$

To understand the solution you sketched, note that B can either throw a T or an H. If it is a T, then B wins (so A has probability of $0$ of winning via this route). If it is an H then A is now in the same position B was in a moment ago.

Answer 4

Alternatively, this can be modelled with a Markov Chain.

Let $a_{i}$ denote the probability of reaching state $(H_B,T_A)$ (i.e. B tosses H and then A tosses T, i.e. A wins the game), given initialization in state $i$. Then using the law of total expectation, the following relations arise:

• $a_{T_A} = \frac{1}{2} a_{Start} + \frac{1}{2} a_{H_B}$
• $a_{H_B} = \frac{1}{2} + \frac{1}{2} a_{H_A}$
• $a_{Start} = \frac{1}{2} a_{T_A} + \frac{1}{2} a_{H_A}$
• $a_{H_A} = \frac{1}{2} a_{H_B}$

Solving yields $a_{Start} = \frac{4}{9}$.