Solving “Bearly Balanced”

Since this new gamemode is purely mathematical, I tried to solve it, which was not as easy as I first thought.

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First of all, the strategem is crazy. This is nothing new to anyone who can count, but only after thinking through it I actually realized how crazy it is:

If blue uses the 5 point strategem and wins with 5 or less points, blue cannot lose unless blue starts discarding bears or doesnt play a couple during round 2. Because if blue wins with 5 or less points, the worst way for blue would be to win with 5 points. But then, both decks going into round 2 are equal. So, since red still has to win 2 rounds, and both have EXACTLY the same amount of points in their deck, red is mathematically inable to win both rounds, resulting in a forced win for blue.

So my first realization was, that blue will always play cards in round 1, without surpassing the 5 point threshold. If he would, hed simply waste a winning position. And if he doesnt have the card to keep the small 5 point distance? Hell just pass, since if he goes 6 or more points up, red will just pass and force a draw or a favorable position.

Now onto red. Red can therefore not let blue win by 5 points or less, if he does, the game already is over. Thats why red HAS to push round 1 (If both players play optimally) and win by the smallest advantage he can get. However, even if red won round 1 by 1 point, he now has at least 6 point less in the deck for the rest of the match. Saying this is a unfavorable position, is an understatement.

So, is red doomed? Not quite yet! We still havent looked on one big point in this game: Passing early. If red passes with 8, cards, he can mulligan out his worst card and be back in the game. But ONLY, if blue doesnt play his worst cards in round 1. Thats why blue optimally should have the 1, 2, and 3 bear on his hand in round 1. Only then, blue wont give red an advantage in the form of an early pass. This basically extends to wanting the lowest cards on your hand in round 1. Always. And I'll explain why.

If blue plays a 1 in round 1 and red passes with 10 cards, both can force their worst cards back into the deck, meaning blue will have the 2 and 3 in his deck, and red the 1, 2 and 3. However, since red had to play at least 1 point more in round 2 in order to win, red will have at least 1 point less for round 3. So passing with 10 cards, when blue played the 1 in round 1, will also force blue to win (If both mulligan their lower cards into the deck). This extends to blue playing the 2 as his second card in round 1 and his 3 after that. Only then, red cannot pass early on him whilst gaining an advantage.

But if blue plays a, lets say 5 as his very first card, red gains an advantage by passing with 10 cards. Because then, red will leave out his 1, 2 and 3 in his deck, whilst blue leaves out his 1 and 2. HOWEVER, he already lost 5 points by playing round 1 like that. Red on his side, only lost the '3' which he now also cant play (The 1's and 2's left in both players decks cancel each other out). So red technically won 2 points total by passing with 10 cards on hand. Which is a HUGE improvement compared to the otherwise -6 points (best case).

So in the end (if both players play optimally), red should instantaneously pass, if blue plays a 4 or higher in his very first 3 cards. Otherwise red should try to stay up by 1 point, which is a loosing position, but no forced loss (considering blue can draw bad for round 2 and is somewhat dorced to win by 6 or more points).

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Thats my observation, however Im fairly certain that I left some things out. Feel free to correct me, Im really interested in seeing what I left out!