[Darryl_P]

I just calculated B's EV under this strategy and it turns out to be 0.4924 -- determined as follows:

If B's hand is less than .5, his EV will be .375 (avg. of .25 and .5, straightforward to show why)

If B's hand is some number b >=.5 his EV will be given by:

b^2 + (1-b)*((1-b)/b)*((1-b)/2)

If A's first card is less than b, he draws and will miss with a probability of b -- that's the first way B can win and that's the first term in the above.

Or if A's first card is more than b (probability of 1-b), then B draws with probability (1-b)/b and wins with probability (1-b)/2 -- that's the second term in the above.

Now that we know B's winning probability as a function of his hand b, we just need to integrate the function from 0.5 to 1 (which gives the area under the graph) and multiply by 2 to scale it up since our interval has length 0.5 and not 1.

Evaluating that integral produces 0.609814 if I didn't make any arithmetical mistakes. Seems reasonable, though, since it basically says B's expectation will be 0.609814 if he's dealt a card higher than 0.5. If bluffing were not allowed it would be 0.625 and so being slightly less seems reasonable.

The final answer of 0.4924 is just the average of 0.6098 and 0.375