A local casino near me deals Spanish 21 from a six-deck show. They have a peculiar player-friendly rule where if you're dealt two identical face cards (say, two Queens of Spades), you win automatically--even if the dealer has a blackjack. The Wizard of Odds says this reduces the house edge by 0.15%.
What I can't figure out is the second part of the following calculation: there are 288 cards in a Spanish 21 six-deck shoe (not 312, because all the 10s are removed). So you have a 1 in 4 chance of your first card being a face card, and when that happens, you have a 5/287 (0.0174) chance of it being matched, for an automatic win.
However, I'm stumped as to how to figure out how much this improves your result. Obviously, a hard 20 is already a good hand; just as obviously, you don't always win with it. So how much improvement is there in EV when a hard 20 is "promoted" to an automatic winner? That number would have to be multiplied by 0.0174 to get the Wizard's 0.15 number, and though I could do the math in reverse, I'm more interested in the process you would take to get there. How much better than a hard 20 is an automatic win?