Optimal Proportional Play

Question

You are at a craps table next to a gambler looking for some action. Both the table and the die are fair. She offers you the following "even money" proposition: starting with the next roll (where the sum of the die's faces are tallied) bet whether two consecutive 7s or a 12 is rolled first. You can pick either side and the size of the bet.


What side of the proposition would you take and how would you wager?

Answer

Assuming I'm the 12 bettor:


If a 12 is rolled then I win; if anything but a 7 or 12 is rolled, I'm in the same position as before; if a 7 is rolled and then anything but a 7 or 12 comes up, again I'm in the same position I was in before; if a 7 and then a 12 is rolled, then I win. Mathmatically:


$P(win|bet\ 12)=p=\frac{1}{36}+\frac{29}{36}*p+\frac{1}{6}*\frac{29}{36}*p+\frac{1}{6}*\frac{1}{36}$
$p=\frac{7}{13}$


For completeness, assume I'm the 7 bettor:


$P(win|bet\ 7)=q=\frac{1}{36}+\frac{29}{36}*q+\frac{1}{6}*\frac{29}{36}*q$
$q=1-p=\frac{6}{13}$


Clearly we would want to bet on the 12 at even money, yielding the following PDF:


$PDF(x)=\frac{7}{13}*\delta(x-1)+\frac{6}{13}*\delta(x+1)$


where 


$E[X]=\int x*PDF(x)\ \mathrm{d}x$
$E[X]=\frac{7}{13}-\frac{6}{13}=\frac{1}{13}$


Assuming a logarithmic utility, $U(x)=\ln(x)$, we can choose a fraction of our bankroll (assumed to be 100%) that would maximize our expected utility:


$E[U(x,f)]=\frac{7}{13}*\ln(1+f)+\frac{6}{13}*\ln(1-f)$
$\frac{\partial\ E[U(x,f)]}{\partial\ f}=\frac{7/13}{1+f}-\frac{6/13}{1-f}$
$f^{*}=\frac{1}{13}$


So, wager a 13th of your wealth on the 12 at each coup.