StudiodeKadent
Well-Known Member
OK, you all know how the Las Vegas Club's "World's most liberal blackjack" pays 2:1 on a suited BJ but even money on an unsuited (thus making the average BJ payout 125%, i.e. only slightly (5%) better than 6:5 BJ)?
I recently was contemplating what a similar system would do to a house edge. Since I am rotten at mathematics, will someone please answer and/or correct my maths?
I was thinking about a BJ game where Unsuited BJ pays 7:5 (i.e. 140%), suited pays 2:1 (i.e. 200%). Since only 1/4 of BJ's are suited, this makes the average BJ payout = (140%+140%+140%+200%)/4 = 155%. Standard BJ payout (3:2) as we all know is 150%.
If this were implemented in a 6-deck BJ game, what would it do to the house edge?
(Off topic, if the above rules were changed to make Unsuited pay 6:5 and Suited pay 2:1, that would be equivalent to all BJ's paying 7:5, at least according to my very amateur calculations).
I recently was contemplating what a similar system would do to a house edge. Since I am rotten at mathematics, will someone please answer and/or correct my maths?
I was thinking about a BJ game where Unsuited BJ pays 7:5 (i.e. 140%), suited pays 2:1 (i.e. 200%). Since only 1/4 of BJ's are suited, this makes the average BJ payout = (140%+140%+140%+200%)/4 = 155%. Standard BJ payout (3:2) as we all know is 150%.
If this were implemented in a 6-deck BJ game, what would it do to the house edge?
(Off topic, if the above rules were changed to make Unsuited pay 6:5 and Suited pay 2:1, that would be equivalent to all BJ's paying 7:5, at least according to my very amateur calculations).