May 1st, 2015 at 7:41:07 AM
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I'm trying to understand the expected returns in the tables listed at Blackjack Payout Guide. The guide is specifically for the early payout variant but it should apply generally to the expected return playing with basic strategy. I threw together a little simulation and I'm able to reproduce everything in the 'Stand' table, except for situations where the dealer has a ten or ace. In those cases I get significantly lower expected returns for the player.

One scenario where this is especially clear is when the player stands on 21 and the dealer has an ace. I would expect that 4/13 of the time the dealer will draw a ten next and the outcome will be a push. Even if the player wins the other 9/13 of the time he would only have a maximum return of 9/13=0.69, but the table lists the expected return as 0.9. My first thought is that maybe they're taking into account the 3 to 2 payout for blackjacks but there are significant discrepancies for all of the situations where the dealer has a ten or ace.

Does anybody have any ideas about what they might be doing differently to get those numbers? Any help on this would be greatly appreciated.

One scenario where this is especially clear is when the player stands on 21 and the dealer has an ace. I would expect that 4/13 of the time the dealer will draw a ten next and the outcome will be a push. Even if the player wins the other 9/13 of the time he would only have a maximum return of 9/13=0.69, but the table lists the expected return as 0.9. My first thought is that maybe they're taking into account the 3 to 2 payout for blackjacks but there are significant discrepancies for all of the situations where the dealer has a ten or ace.

Does anybody have any ideas about what they might be doing differently to get those numbers? Any help on this would be greatly appreciated.

May 1st, 2015 at 8:21:15 AM
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Seems like a math error on their end. It looks like they're using probabilities from American Rules.

The .9 EV is correct for that hand if we know the dealer doesn't have a 10 under their ace (since they checked for BJ and didn't have it with American rules)

Therefore, the probabilities of the dealer drawing to a given hand assuming you have a 21 are:

The values here are slightly off since I'm assuming an infinite deck. I'm also assuming the dealer stands on all 17s.

Note that the odds of the Dealer drawing to 21 are quite low since they now need to do it with a combination of 2+ cards without exceeding a 16 before hitting their 21.

Edit: I may have also made a math mistake so it'd be good if someone checked my work. Take the above with a grain of salt.

The .9 EV is correct for that hand if we know the dealer doesn't have a 10 under their ace (since they checked for BJ and didn't have it with American rules)

Therefore, the probabilities of the dealer drawing to a given hand assuming you have a 21 are:

The values here are slightly off since I'm assuming an infinite deck. I'm also assuming the dealer stands on all 17s.

Note that the odds of the Dealer drawing to 21 are quite low since they now need to do it with a combination of 2+ cards without exceeding a 16 before hitting their 21.

Edit: I may have also made a math mistake so it'd be good if someone checked my work. Take the above with a grain of salt.

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