Originally Posted by
Dog Hand
PeterLee,
Ahh... I see now why you're asking this.
I ASSUME that, on an unequal split, the rider (big) bet goes on the "first" hand, so if the dealer has a BJ, that first-hand rider bet is automatically lost.
If you could somehow get the rider bet on an unequal split placed on the "second" hand, then it would probably be worthwhile to alter the play of the hands to avoid busting, since if neither busts and the dealer gets a BJ, the rider bet is NOT lost: this would be especially true vs. a dealer's Ace, since the BJ possibility is typically >30% (unless the count is really negative, that is). However, I suspect this option is not readily available for precisely this reason.
Barring the above paragraph, with a SINGLE unequal split, you should do what I said originally: play the hands however your strategy suggests. Thus, if the strategy says to Stand on 15 vs. X on hand #2 if hand #1 didn't bust, then do so. Note, though, that you would still play the Big Bet hand #1 according to normal B.S. (since playing conservatively would only provide protection for the small hand #2 bet), so you would (probably) hit if that one is 15 vs. X.
However, if you have a round with MULTIPLE Unequal Splits, you might want to alter your Hand #1 play. For example, you're dealt 8-8 vs. A, you PU, on the first hand you get another 8, and you PU again. Now (counting from 1st base) hands #1 and #2 have rider bets, and hand #3 does not (by the way, I ASSUME this is the order you'd get: I've never played such a game myself where the order of the hands matters). Here, it might be worthwhile to alter the play of hand #1, since if it doesn't bust, it provides some "protection" for the big bet on hand #2. As above, if hand #1 doesn't bust, you'd follow the conservative strategy on hand #2. Assuming hand #2 also doesn't bust, though, the only advantage to using a conservative strategy on hand #3 is the chance to save its small bet in the event of a BJ: hand #3 in this case is not protecting any big bet from a dealer BJ.
This BB+1 problem is much more intricate than the ones I have studied: in my previous work, the results of one hand of a split had no effect on the other hand(s). Thus, I have couched my replies in conditional terms, since I do not have the results. I would recommend that you work out the rules yourself: you seemed to be on the right track in your above post.
Hope this helps!
Dog Hand
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