Hello, I've studied just enough with the Verite Blackjack software that I feel comfortable getting my first hours in a real casino in.
I found my local casino pays 3:2 when getting a two card 21 on split tens. I'd appreciate if someone would be able to tell me how that effects EV and indices. Rules are 6D, S17, DA2 no hit/resplit aces.
Quoting Stanford Wong's "Professional Blackjack", 6th edition:Originally Posted by Schaster
At the Nevada Club in Reno in 1987, if you split 10-10 and caught an ace, you received 3:2 payment the same as if that 10-ace were a natural. This is not generous enough to cause you to deviate from basic strategy. That is, basic strategy still says do not split 10-10. Thus for a basic strategy player, this rule has no value.
Not to belabor the point, but the difference in expectations of splitting tens versus standing on your hard 20 is so large that the extra 3:2 payoff in the unlikely event that you catch an ace on at least one of those tens will not come close to making this decision preferable to standing. I suspect that the minimum proportion of aces left in the shoe that would make the expectation of splitting tens larger than that of standing on your hard 20 is so huge that any hypothetical side count that identifies them in the un-played shoe would be put to better use tracking them through the shuffle after the cut card comes out.
According to some quick calculations I made using Hi-Lo, the indices for TTv6/5/4 are lower than the traditional ones.
Traditional:
TTv6=+5
TTv5=+5
TTv4=+7
If split tens BJ are paid 3:2 then the indices become:
TTv6=+2
TTv5=+2
TTv4=+4
Hope this helps.
Sincerely,
Cac
Luck is what happens when preparation meets opportunity.
Assume n decks and no knowledge of the dealer's upcard or other cards dealt to other players. Then
P(at least 1 ace dealt on top of a split ten) = 1 - P(no aces dealt on top of a split ten) = 1 - (52n-2-4n)/(52n-2)*(52n-3-4n)/(52n-3)
For n = 1, the desired probability is 0.1551.
For n = 2, the desired probability is 0.1514.
For n = 3, the desired probability is 0.1502.
For n = 4, the desired probability is 0.1497.
For n = 5, the desired probability is 0.1493.
For n = 6, the desired probability is 0.1491.
For n = 7, the desired probability is 0.1489.
For n = 8, the desired probability is 0.1488.
As n approaches infinity, the desired probability approaches 0.1479.
Thus, the likelihood of getting at least one ace in a shoe game actually decreases slowly as the number of decks increases.
The only reason I could fathom using a sim here would be to identify an index of a side-count of aces that would make such a decision to split tens preferable to standing on hard 20 in terms of expected value. I strongly suspect that such a count would be entirely unnecessary for this purpose because the number of aces would be so high in that situation that the AP would be better off using this information to track these unused aces through the next shuffle; furthermore, splitting tens is a suspicious play to the pit in any case, so that any additional value would be offset by the subsequent threat to your longevity in that casino.
21forme,
Actually, the opposite is true: fewer decks gives a higher probability of receiving an Ace. Given the player has 2 tens and with no other information, for SD, 4 Aces are left in the remaining 50 cards: that's 8%. For 8D, 32 Aces are left in the remaining 414 cards: that's 7.73%.
Hope this helps!
Dog Hand
Many years ago, I played at a casino that had this rule along with others that were like being in heaven.
Not only did they consider split hand BJs as BJs, but also, if you tied with the dealer, they paid you half.
Unfortunately, good things don't last long.
Anyway, you could ask the OP where that casino is located
Sincerely,
Cac
Luck is what happens when preparation meets opportunity.
Posting Permissions
Reply With Quote
Bookmarks