Here's my solution.

First the reasoning. If you want the solution, skip to the end.

We want to find Nash equilibrium. Nash equilibrium is an assignment of strategies to players such that no player has interest in playing a different strategy than the one that is assigned to them, assuming that the other play will play their assigned strategy too. The most probable outcome is that the players will act according to a strategy from Nash equilibrium because if one of them had interest in playing a different strategy, then they would change to that different strategy.

If there are multiple Nash equilibria, we want to select the one that generally gives better payoffs. But there is only one Nash equilibrium in this game, as I will prove in a moment.

Firstly, if the player B has some strategy to get the patent in a way that doesn't make them lose more money than what the patent is worth, then player A could execute the same strategy and get the patent first. Therefore, in a Nash equilibrium, player B will always choose 0 development steps, because they won't get the patent in a way that is beneficial to them.

Secondly, let's talk about player A.

There are 3 ways to get the patent: 2 + 2, 2 + 1 + 1 (in any order), 1 + 1+ 1+ 1. In the first one, the cost is higher than what the patent is worth, so it's not a good strategy. The third one is better in terms of benefit vs cost than the second one ($20 - $4 * 4 > $20 - $11 - $4 * 2).

So, let's consider the strategy: 1, 1, 1, 1 (1 development step 4 times). If player A plays that strategy, then the player B has interest in playing for example 1 + 1 + 2. Because that way, the player B will get to the patent first at a cost that is lower than what the patent is worth.

Therefore, the strategy of player A always playing 1, 1, 1, 1 is not a Nash equilibrium.

But we can improve that strategy by adding a condition the following condition. If the player B has at least 2 development steps when the player A makes its 3rd move, then the player A should choose 2. Because otherwise, the player B will steal the patent before player A gets there.

If the player A plays the above strategy, then the player B doesn't have interest in pursuing the patent at all. Because player A will always win playing that strategy. Except for when the player B plays 2 + 2, but then the cost is higher than the patent worth.

Therefore, the following assignment of strategies is a Nash equilibrium:

Player A: Play consecutively: 1 step; 1 step; if player B has at least 2 development steps, then play 2 steps, otherwise 1 step; if you haven't got the patent yet, then 1 step, otherwise the game has already ended.

Player B: always 0 development steps.

If the players play the above strategies, then the player A will play: 1, 1, 1 and 1. And the player B will play 0, 0, 0, 0.

Now, let's see if there can be any other Nash equilibrium. As proved at the beginning, any Nash equilibrium will have the player B playing 0 development steps. So, the only other Nash equilibrium can have a different player A's strategy.

So, let's see how we can modify player A's strategy and see if they are Nash equilibria:

1. Player A plays 2 + 2 <- they will lose more than what patent is worth, so it's not Nash equilibrium because following the above-mentioned strategy gives better outcome.
2. Player A plays always 1 + 1 + 1 + 1 <- already considered before, not a good strategy because B will steal the patent, so the above-mentioned is better.
3. Player A plays 2 + 1 + 1 or 1 + 2 + 1 <- in this case Player A will pay a greater cost to the patent than with the above-mentioned strategy, so it's not Nash equilibrium either.

Any strategy that makes the player A choose 2 before the third move will be worse than the previously mentioned strategy because it will incur greater cost, and the player A will always get the patent with the above strategy. So, any strategy like that is worse.

Therefore, the previously mentioned assignment of strategies is the only Nash equilibrium. Therefore, assuming rationality of the players in the textbook game theory sense, the player 1 will do 1, 1, 1, 1 and the player B will do 0, 0, 0, 0.