Doubling a 6:5 Blackjack

[Sir Counts-A-Lot]

Does anyone know if it might be a higher EV move to double down on a BJ if you are playing a 6:5 game? Seems like treating it as an 11 and doubling might yield better results than a 6:5 payout in some cases, ie high count and the dealer showing a 5 or a 6.

I realize the smart move would be to ignore the 6:5 game and find something much better to play...this is a suppose you are stranded on a desert island and all they have is 6:5 blackjack and you want to beat it by any means necessary question.

[shadroch]

The question has been asked and answered several times on these forums, including once by myself. The short answer is no, it isn't.

[callipygian]

Originally Posted by Sir Counts-A-Lot

Does anyone know if it might be a higher EV move to double down on a BJ if you are playing a 6:5 game? Seems like treating it as an 11 and doubling might yield better results than a 6:5 payout in some cases, ie high count and the dealer showing a 5 or a 6.

If you accept a blackjack, you automatically gain 120% of your bet, or an EV of +1.20.

The EV on doubling would be 2*Win-2*Loss; assuming no pushes, it's roughly 4*Win-2.

If you want greater EV on doubling, 4*Win-2>1.20, or Win > 0.8. You have to win more than 80% of your doubles in order for this move to be profitable.

If you don't recognize how absurdly high that is, note that winning a hard 20 vs. dealer 6 only occurs about 78% of the time.

[k_c]

Originally Posted by Sir Counts-A-Lot

Does anyone know if it might be a higher EV move to double down on a BJ if you are playing a 6:5 game? Seems like treating it as an 11 and doubling might yield better results than a 6:5 payout in some cases, ie high count and the dealer showing a 5 or a 6.

I realize the smart move would be to ignore the 6:5 game and find something much better to play...this is a suppose you are stranded on a desert island and all they have is 6:5 blackjack and you want to beat it by any means necessary question.

It's not close. Count would need to be very high to double.

[Cardcounter]

Originally Posted by callipygian

If you accept a blackjack, you automatically gain 120% of your bet, or an EV of +1.20.

The EV on doubling would be 2*Win-2*Loss; assuming no pushes, it's roughly 4*Win-2.

If you want greater EV on doubling, 4*Win-2>1.20, or Win > 0.8. You have to win more than 80% of your doubles in order for this move to be profitable.

If you don't recognize how absurdly high that is, note that winning a hard 20 vs. dealer 6 only occurs about 78% of the time.

A push on doubling on blackjack would be a lose of 1.2 units because you would of won 1.2 units had you taken the blackjack. A loss would be a loss of 3.2 units the 1.2 units you would have won minus the 2 units you actually lose. A win would only be a win of .8 units because 1.2 is guranteed to you and you can only win 2 units by doubling down. Just because you might win the hand most of the time when you double doesn't mean it is the right play. It is like betting on 2 12 number slots on roulette you will probably win one of them with a 24/37 chance of winning one unit or 64% but the times you lose two units will wipe you out in the end.