# Players $A$ and $B$ alternately toss a biased coin,with $A$ going first.$A$ wins if $A$ tosses a tail before $B$ tosses a head;otherwise $B$ wins.

Players $A$ and $B$ alternately toss a biased coin,with $A$ going first.$A$ wins if $A$ tosses a tail before $B$ tosses a head;otherwise $B$ wins.If the probability of a head is $p$,find the value of $p$ for which the game is fair to both players.

I do not understand what its mean by the game is fair to both players.Does it mean the probability of winning is same for both the players$?$

• Yes, that is the meaning. Commented Jan 11, 2016 at 10:12

$A$ wins on the first toss if $T$, probability $(1-p)$

$B$ wins on the second toss if $HH$, with probability $p^2$

If neither wins in two tosses, we are back to the start.

Thus for both to have equal chances, $(1-p) = p^2$

Proceed....

$0.5(\sqrt5 - 1)$

• I also do not understand the condition of winning of $B$.Does it mean $B$ wins if $B$ tosses a head before $A$ tosses a head?@trueblueanil
– diya
Commented Jan 11, 2016 at 10:30
• No, $A$ wins only by tossing $T$, $B$ wins only by tossing $H$. Commented Jan 11, 2016 at 11:51

Yes, a fair game is a game where all players' expected payoff is the same.

Hints:

• If $p_A$ denotes the probability that $A$ wins the game then:$$p_A=(1-p)+p(1-p)p_A$$(do you understand why?)
• The game is fair if $p_A=\frac12$
• @trueblueanil I don't understand what you are trying to say. Substitution of $p_A=\frac12$ leads eventually to the equality in $p$ that is also mentioned in your answer, and that can be solved. Commented Jan 11, 2016 at 12:35