I suspect that one or two more is possible, but that may require a ridiculously complicated ruleset. As if the following wasn't bad enough already.
First 3 tricks:
Cost 4 cards each.
The second player will always get at least one guess right by always guessing on the cards of the same group as the first card played, 1 and 2 higher respectively for the second and third card played, if possible, and if that fails deduce the value of the fourth card from the above rules.
Next 3 tricks:
Cost 3 cards each.
The remaining 28 cards are implicitly enumerated 1 through 28, this requires that both players remember all the played cards. They are similarly split into 4 virtual groups of 7 cards each.
The first player finds one of the following combinations in hand:
- 3 cards from the same group.
- 2 cards from the same group, one card from the group below, and one card from the group either 1 or 2 above.
- Depending on the combination, the cards are played like so:
- Just like case 1 in the first 4 tricks, only the difference can be no larger than 2.
- Play the first card of the pair, then the second if they are 1 apart, inject the card from the group below if they are 2 apart, or the card from the group above if they are 3 apart.
Last 4 tricks:
Using the last 5 cards.
The remaining cards are enumerated 1 through 10, then the first player plays card 1 if he has it, and this step is repeated with the cards enumerated 1 through 8, 6, 4 and finally 2.
When the first player doesn't have card 1, the other cards form a loop of 9, 7, 5, 3 or 1 card. He chooses the one starting point in this loop where the following is possible.
He plays the first of a series of consecutive cards, the second player may not guess this card.
He plays the last card of the same series. Since all the cards in between are in the first players hand, they cannot be drawn from the pile, so the second player can always identify this card.
If there is more than one card between the series and the following series, each extra card in the gap is marked by playing one of the cards that the second player already know that he has, then the last card of the following series is played, and the process is repeated for each series and gap.
Example last 4 tricks:
After enumeration the cards in hand are: 2, 5, 6, 9, 10. The pile has 1, 3, 8, 4, 7 in that order.
Player 1 has to start with 9 in this case, the pile plays 1. Player 2 guess 1, so this trick is lost.
Player 1 mark the end of his series from 9 to 2 by playing 2, the pile plays 3. For player 1 to have played 3 in this scenario he must have had 10, 2 and 3 on hand, but that is impossible since either 2 or 3 must have come from the pile, so player 2 can conclude that he has to guess 2. He also now know that player 1 has 10 in hand as that is between 9 and 2.
Player 1 plays 10 to mark the double gap between 2 and his next card 5. The pile plays 8. Player 2 of course guess 10, and since a known card was played, rather than a card from the next series, player 2 now know that there is an extra gap after 2, so he will be looking for 5 or above in the next trick.
Player 1 plays 6, to mark the end of the 5 to 6 series, the pile plays 4. Player 2 knows that player 1 hasn't got 4, since he marked that by playing the known 10, so he looks for 5 or above, the closest card for that criteria is 6, so that is his guess. Player 2 now know that player 1 has 5 in hand, as that is must be that start of the series when the gap between the first and the second series is 2.
Player 1 plays 5, the pile plays 7. Player 2 know that Player 1 has 5, so that is his guess.