0
$\begingroup$

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$?$

$\endgroup$
1
  • 1
    $\begingroup$ Yes, that is the meaning. $\endgroup$ – true blue anil Jan 11 '16 at 10:12
2
$\begingroup$

$A$ 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....

$0.5(\sqrt5 - 1)$

$\endgroup$
2
  • $\begingroup$ 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 $\endgroup$ – diya Jan 11 '16 at 10:30
  • $\begingroup$ No, $A$ wins only by tossing $T$, $B$ wins only by tossing $H$. $\endgroup$ – true blue anil Jan 11 '16 at 11:51
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $p_A=\frac12$
$\endgroup$
1
  • $\begingroup$ @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. $\endgroup$ – drhab Jan 11 '16 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.