Free Ace--Math Question

At the casino I work at; we run an annual "free blackjack ace" promotion. A player presents his coupon and sets it down on the table; in effect; his first card is an automatic ace. The maximum bet on this coupon is $25.

What is the player advantage gained (if any) on this; and how much would a player potentially be giving up by only betting $5 (the minimum).

Free aces are worth about half of face value. So if the player bets $25, it is worth $12.50. If he bets $5, it is worth $2.50.

Quote: Wizard

Free aces are worth about half of face value. So if the player bets $25, it is worth $12.50. If he bets $5, it is worth $2.50.

It has to be taken into consideration that free Ace coupons attract deuces and threes.

That makes me wonder--are even hands like soft 13 and soft 14 winners, on the average, or do you have to have, say, soft 18 or better before A-x is a winner against a random dealer upcard?

Yes, you have to have soft 18 or better to have a postive EV against a random dealer upcard.

Using the data from https://wizardofodds.com/blackjack/appendix5.html for A,7 for an S17 game, soft 18 would be worth an average of 0.073 assuming the dealer does not have blackjack. However, before the dealer peeks, soft 18 would be worth an average of 0.031.

Assuming the dealer does not have blackjack, soft 13 is actually a winner with an average of 0.015. But before the dealer peeks, soft 13 is a loser at -0.028.

The free ace is kind of a come-on that isn't as good as it seems, unless you get a blackjack. I've used them a couple times, with big bets ($50-$100 -- that's big for me) out. If you don't get the blackjack, and the dealer has a weak upcard, you're faced with the choice of having to double a large amount of money. I wouldn't double a A-3 against a 4, or similar soft doubles, for instance. I think Grosjean talks about this in one of his books. Of course, if you get a A-4 or A-5 against a dealer high card, you're a net loser, because those are crummy hands to begin with.