Doubling a 6:5 Blackjack

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

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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.
 
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