Your backtest is a random number generator

Search a large enough space of rules and something will clear any threshold you set. The Sharpe ratio you are admiring is a maximum over many trials, and the distribution of a maximum is not the distribution of a draw.

What the number actually is

If you test $m$ independent strategies whose true edge is zero, the expected best observed Sharpe grows roughly like $\sqrt{2 \log m}$. Twenty variations buys you about 2.4 standard deviations of pure noise.

The deflated ratio

Bailey and López de Prado's deflated Sharpe adjusts for the number of trials and for the skew and kurtosis of returns. It is not a cure, but it converts a number you cannot interpret into one you can.

from scipy.stats import norm

def deflated_sharpe(sr, n, m, skew, kurt):
    """Probability the observed Sharpe exceeds a zero-skill benchmark."""
    sr0 = expected_max_sharpe(m)
    se = ((1 - skew * sr + (kurt - 1) / 4 * sr**2) / (n - 1)) ** 0.5
    return norm.cdf((sr - sr0) / se)

The honest version of a backtest reports how many things you tried.