Question: If a game of backgammon were to be played between opponents, where one used standard dice and the other a pair of Sicherman dice, who if either would have the probabilistic advantage?
Assume that all other factors and rules were/remained the same.
References to literature, especially which explore this problem or an analogue, are also acceptable.
For those not familiar with either or both of these topics:
Background: Backgammon is a historic game in which two players try to move their pieces past each other. Noteworthy rules include
rolling two six-sided dice each turn, moving individual pieces by moves of length equal to one or both of the numbers shown. In particular, where a double is rolled, the number of moves is also doubled - rolling e.g. $22$ gives that player up to four movements of length 2
in a turn, a player must move pieces by the greatest possible total length. That is, turns which don't use all possible movement are not permissible; where no movement is possible, no movement occurs.
the ability to hit single opponent pieces, returning them to the bar, from which they must begin the movement process again. This is the primary method of gaining some advantage over the opponent.
the inability to land on, or hit, opponent pieces which have formed a point (wherein at least two pieces are stacked in the same spot). This is a significant part of strategy.
Background: Sicherman dice are a pair of non-identically-numbered six-sided dice $$(1,2,2,3,3,4)+(1,3,4,5,6,8)$$ with the property that the distribution of the sums of rolling them together, is the same as that of two standard six-sided dice, $$(1,2,3,4,5,6)+(1,2,3,4,5,6)$$ This can be shown by (among several methods) creating a table of sums, and counting frequencies, for each pair of dice.