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OK, so take a game of blackjack, but if you win with 5 or more cards in your hand, you win double what you normally would.

How would I go about adapting basic strategy to account for this? My thinking is that adapting to anything when you have <= 3 cards is likely neutral / negligible, so essentially I would simply have a seperate strategy table to use for situations where I have exactly 4 cards in my hand. What sorts of changes would I make?

If you need a specific ruleset: 6 decks Dealer doesnt peek double any 2 cards double after split is OK no surrender no resplitting split aces get 1 card only

pingu2k4
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It should make you more prone to draw when you have four cards in your hand. You would have to go back to the computation of the tables to see how much. The computation behind the tables takes a certain situation, for example you have $14$ total and the dealer shows a $3$. It computes the chance you win if you draw and compares with the chance you win if you stand. If the chance of winning is higher when you stand, the table tells you to stand. With the new rule, you should draw if the chance of winning on a draw is at least half the chance of winning on a stand.

It should make you very slightly more prone to draw with three cards because of the chance of winning double. It may not be significant enough to show up in the tables.

You should ask what change is made to the rules to pay for this. Increasing the payout in certain situations has to be balanced by reducing it elsewhere to keep the house advantage where it needs to be. I have always suspected that the changes to the rules were generally in the house's favor, but have never done the computation to support it.

Ross Millikan
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