Free Bet Blackjack (optional double)

The Wizard of Odds website says to ALWAYS take advantage of a double or split opportunity.

While it makes sense to gamble for more money when it's free, and certainly pays off on the normal double and split opportunities according to the basic strategy, it seems counterintuitive to double a 9 against a dealer 10, GIVEN that you now only receive one card.

If it didn't make sense to double and accept only one card under normal rules, can winning double only a small amount of the time really overcome the "only one card more" handicap? It seems not because otherwise basic strategy would tell us to do it with our own money?

Accepting free money on known doubles and splits is a no-brainer. I worry always doubling a 9,10,11 gives away too much when you can only draw one more card. On top of which, this is also count dependent!

If the Wizard's math says doubling for free on a hard 9 is better than not doubling for free on a hard 9.... double for free on the hard 9.

The reason you always take all free doubles is you win $2 if you win but only lose $1 if you lose. Typically, especially against a picture, you will win $2 anytime the dealer busts and also on other rare occasions when you get a winning 20 (A), 19 (T) etc. It is a better play to double, very roughly, when your odds of winning exceed 33% (as you're getting paid 2/1).

In your title you mention real money doubles -there are some rare cases where you do this (e.g. soft totals vs 5/6) or some weird cases after a free split (e.g. 9A vs 6) (see https://wizardofodds.com/games/free-bet-blackjack/ ) .

Quote: charliepatrick

In your title you mention real money doubles -there are some rare cases where you do this (e.g. soft totals vs 5/6) or some weird cases after a free split (e.g. 9A vs 6) (see https://wizardofodds.com/games/free-bet-blackjack/ ) .

I'm having some trouble wrapping my head around that one, unless it's an edge case that has to do with the count? At a glance, I'd think that if you wouldn't real-double a normal 9A, you wouldn't real-double a free 9A either.

Quote: Venthus

I'm having some trouble wrapping my head around that one, unless it's an edge case that has to do with the count? At a glance, I'd think that if you wouldn't real-double a normal 9A, you wouldn't real-double a free 9A either.

When you're dealing with a free bet, a push is the same as a loss. I think that's why.

Quote: Deucekies

When you're dealing with a free bet, a push is the same as a loss. I think that's why.

That covers why you hit free H17 against (most) high cards, but real doubling a free 9Av6 looks like a case where you're lowering your odds of a win in favor of a bigger win (which makes sense), a different situation.

Quote: Venthus

That covers why you hit free H17 against (most) high cards, but real doubling a free 9Av6 looks like a case where you're lowering your odds of a win in favor of a bigger win (which makes sense), a different situation.

Doubling A9 is optional, but NOT free.

Free doubles are only for HARD totals of 9, 10, 11.

Free splits are only AA thru 99, and I’m not sure about 55.

You can get a free split on 55, but the Wizard recommends you take the free double instead.

Quote: ace7seven

You can get a free split on 55, but the Wizard recommends you take the free double instead.

Unless you're playing the Pot of Gold sidebet at any kind of a "normal" bet ratio.

Quote: DJTeddyBear

Doubling A9 is optional, but NOT free.

Yes, but the curious bit is why it's recommended to pay to double a free 9A, but not a paid 9A (or A9).