Two players put a dollar in a pot. They decide to throw a pair of dice alternatively. The first one who throws a total of $ 5$ on both dice wins the pot. How much should the player who starts add to the pot to make this a fair game?
So my interpretation of this problem is that I first throw a dice and then the other person throws a dice afterwards. E.g. if I throw a 1, and if the other person throws a 4 wins the pot. So we are trying to find the expected payout?
On
First, find the probability $p$ that a $5$ is thrown and define $q = 1 - p$. The probability that the first player wins is $C=p+pq^2+pq^4+\ldots$; calculate this value. Now set $C=(1-C)(1+x)$ and solve for the extra contribution $x$.
On
Building on your interpretation, let us define the rules as follows: A and B alternately throw a single die. If the sum of the last two throws is 5, the one who just threw wins. Note that as stated, an initial 5 could win, or a series of throws of a different length than 2. The analysis would change, but follow the same route. The below requires precisely two throws to add to 5.
Let p be the probability that the first player wins. Let q be the probability that the next player to throw wins given that he has received a chance of winning, that is that the last throw is less than 5. Then if you receive a throw of 5 or 6 your chance of winning is p. So q=1/6 (that you win on this throw) + (1-p)/3 (that you throw 5 or 6 and then win) +(1-q)/2 (that you throw <5, don't win this throw, but finally win. p=(1-p)/3 (that you throw 5 or 6 and win)+2(1-q)/3 (that you throw <5 and win).
$q=\frac{1}{6}+\frac{1-p}{3}+\frac{1-q}{2}$
$p=\frac{1-p}{3}+\frac{2(1-q)}{3}$
If I have the algebra right, p=15/32 and q=9/16, so the first player should withdraw $2/17.
Player 1 throws on odd throws and player 2 throws on even throws. This is a geometric distribution. So his distribution is $\left(\frac{8}{9} \right)^{2k} \cdot \frac{1}{9}$ and player 2's distribution is $\left(\frac{8}{9} \right)^{2k-1} \cdot \frac{1}{9}$. So player 1 must give $1/8$.