policy_grad_2-Pixelcopter-PLE-v0

Reinforce Agent playing Pixelcopter-PLE-v0

This is a trained model of a Reinforce agent playing Pixelcopter-PLE-v0 . To learn to use this model and train yours check Unit 4 of the Deep Reinforcement Learning Course: https://huggingface.co/deep-rl-course/unit4/introduction

Some math about 'Pixelcopter' training.

The game is to fly in a passage and avoid blocks. Let we have trained our agent so that the probability to crash at block is p (low enogh, I hope). The probability that the copter crashes exactly at n-th block is product of probabilities it doesn't crash at previous (n-1) blocks and probability it crashes at current block: $$P = {p \cdot (1-p)^{n-1}}$$ The mathematical expectation of number of the block it crashes at is: $$<n> = \sum_{n=1}^\infty{n \cdot p \cdot (1-p)^{n-1}} = \frac{1}{p}$$ The std is: $$std(n) = \sqrt{<n^2>-<n>^2}$$ $$<n^2> = \sum_{n=1}^\infty{n^2 \cdot p \cdot (1-p)^{n-1}} = \frac{2-p}{p^2}$$ $$std(n) = \sqrt{\frac{2-p}{p^2}-\left( \frac{1}{p} \right) ^2} = \frac{\sqrt{1-p}}{p}$$ So difference is: $$<n> - std(n) = \frac{1 - \sqrt{1-p}}{p}$$ As long as $$ 0 \le p \le 1 $$ the following is true: $$\sqrt{1-p} \ge 1-p$$ $$<n> - std(n) = \frac{1 - \sqrt{1-p}}{p} \le \frac{1 - (1-p)}{p} = 1$$ The scores s in 'Pixelcopter' are the number of blocks passed decreased by 5 (for crash). So the average is lower by 5 and the std is the same. No matter how small p is, our 'least score' is: $$ (<n> - 5) - std(n) = <n> - std(n) - 5 \le - 4$$ But as we use only 10 episodes to calculate statistics and episode duration is limited, we can still achieve the goal, better agent, more chances. But understanding this is disappointing