Permalink Submitted by Anonymous on January 11, 2012

As per above:
"When should Player A announce a double, supposing that she owns the doubling cube? If P(t) < 1-a and she announces a double, the opponent will accept the double on the grounds that his expected payoff is greater than -S. This is so because his probability of winning, which is 1-P(t), exceeds a. As the expected payoff for Player B is minus the expected payoff for Player A, this is undesirable for Player A. Hence, she must wait until P(t) >= 1-a, in which case Player B will refuse the double and she wins the stake. This argument shows that necessarily b=1-a = 4/5"

I think this is flawed reasoning, specifically the statement that "as the expected payoff for Player B is minus the expected payoff for Player A, this is undesirable for Player A". This misses the distinction that while Player B has to compare the expected payoff from accepting to -S, the payout from not accepting, Player A has to compare the expected payoff from doubling to P(t) x S, the expected payoff from not doubling.

This means that in any case where the expected payoff for B from accepting the double is between -S and -(P(t) x S), it is both correct for A to double and for B to accept the double.

For instance, if P(t) > a such that E(B|accept double) = -0.9S, then A should double as he increases his payoff from <0.8S to 0.9S, but B should accept the -0.9S expected payout because its better than rejecting and forfeiting S.

## I think there is a flaw in the "when should A double" logic

As per above:

"When should Player A announce a double, supposing that she owns the doubling cube? If P(t) < 1-a and she announces a double, the opponent will accept the double on the grounds that his expected payoff is greater than -S. This is so because his probability of winning, which is 1-P(t), exceeds a. As the expected payoff for Player B is minus the expected payoff for Player A, this is undesirable for Player A. Hence, she must wait until P(t) >= 1-a, in which case Player B will refuse the double and she wins the stake. This argument shows that necessarily b=1-a = 4/5"

I think this is flawed reasoning, specifically the statement that "as the expected payoff for Player B is minus the expected payoff for Player A, this is undesirable for Player A". This misses the distinction that while Player B has to compare the expected payoff from accepting to -S, the payout from not accepting, Player A has to compare the expected payoff from doubling to P(t) x S, the expected payoff from not doubling.

This means that in any case where the expected payoff for B from accepting the double is between -S and -(P(t) x S), it is both correct for A to double and for B to accept the double.

For instance, if P(t) > a such that E(B|accept double) = -0.9S, then A should double as he increases his payoff from <0.8S to 0.9S, but B should accept the -0.9S expected payout because its better than rejecting and forfeiting S.