Epsilon-Greedy Action Distribution

Problem Statement

Given action-value estimates for one state, compute the probability distribution that an epsilon-greedy policy induces over the actions. Do not sample — return the exact probabilities.

Background

Epsilon-greedy is the standard answer to the exploration/exploitation dilemma: with probability 1 - epsilon take the action that currently looks best, and with probability epsilon pick uniformly at random among all actions (the greedy action included). So for a state with n actions:

$\pi(a \mid s) = \begin{cases} 1-\varepsilon + \dfrac{\varepsilon}{n} & a = \arg\max_{a'} Q(s,a') \$6pt] \dfrac{\varepsilon}{n} & \text{otherwise} \end{cases}$

The detail that trips people up is ties. If k actions share the maximum value, the greedy mass 1 - epsilon is split evenly among all k of them, so each tied action gets (1 - epsilon)/k + epsilon/n. Handing the whole greedy mass to argmax (which returns only the first index) silently biases the policy — and later, when you write Expected SARSA, that same distribution appears inside the update target, so the bug becomes a wrong learning signal rather than just an odd action choice.

Your Task

Implement:

def epsilon_greedy_probs(q, epsilon):
    ...

• q — a list of floats, the action values Q(s, a) for the current state.
• epsilon — a float in [0.0, 1.0].

Return a list of floats of the same length as q, summing to 1.

Input / Output Format

Input is a list of floats and a float. Output is a list of floats; the grader rounds each entry to 4 decimals.

Sample

print([round(p, 4) for p in epsilon_greedy_probs([1.0, 5.0, 3.0], 0.3)])

Output:

[0.1, 0.8, 0.1]

Every action gets 0.3/3 = 0.1, and the greedy action (index 1) additionally gets 1 - 0.3 = 0.7.