How many coin flips are needed to reliably know a coin of weight w is unfair? I want to find out how many flips I need to flip a coin to reliably know that it is an unfair coin.
The issue is that as the coin becomes closer to 50/50, the more false-negatives you will have if you don't take dramatically more data.
I wrote some python code and numerically found out how many flips are needed to be able to confirm a weighted coin with weight w. As w gets closer to .5, exponentially more flips are needed to have a low false-negative rate, as shown in this figure I made in python:

The x-axis represents the weight of the coin, while the y-axis represents the number of flips to get a false-negative rate of 5% (while also having a 5% false-positive rate).
Is there a way of getting this answer analytically? I'm particularly interested in what approximations can be done to know what will occur as you get closer to .5.
Here's my python code attempt that was used to generate the figure. I have added comments in the code to try to explain the logic:
from scipy import stats
import numpy as np
import matplotlib.pyplot as plt



# The goal of this script is to figure out how many flips are required to
# have a low false-negative rate when testing if a coin is fair for a specific
# p-value. I use this script to understand the relation between
# the unfairness of the coin and the amount of flips needed to get a low false-negative
# rate. 
# Essentially we calculate the false-negative rate for some starting number of flips,
# and if this rate is lower than our desired threshold, then we add more number of flips,
# we continue this in a loop until we are under the false-negative rate (or until we run 
# out of iterations)
# Finally, we run this all inside the function find_numberofFlips_and_false_negatives_vs_fairness
# this function has an input of the coins fairness, and returns a number of flips as an output
# we then use this function to collect a set of points for the coins fairness vs numberofFlips 
# then we plot this

def flipAndDoPTest(numberOfFlips, weight, guessWeight):
    """
    Simulate multiple coinflips and perform a binomial test, 
    returning a pvalue.
    
    Parameters
    ----------
    numberOfFlips : int
                    The number of flips that will be simulated.
    weight        : int
                    The weight of the coin that is flipped
    guessWeight   : float 
        The hypothesized probability of success, i.e. the expected
        proportion of successes. We will typically use .5 for this.
        
    Returns
    -------
    result :
        pvalue : float
            The p-value of the hypothesis test.
    """
        
    flippedCoinSimulation = np.random.binomial(1, weight, numberOfFlips) 
    numberOfHeads = np.sum(flippedCoinSimulation==1)
    numberOfTails = np.sum(flippedCoinSimulation==0)
    pvalue = stats.binom_test(numberOfHeads, numberOfFlips, guessWeight)
    return pvalue

def accurate_pvalue_interval(numberofFlips, desiredConfidenceLevel):
    """
    Creates an new confidence interval which is closest to the desiredConfidenceLevel.
    Because of the discrete nature of coin flips, it's not possible to get any arbitrary
    confidence level. This code finds the closest confidence level that has a pvalue lower
    than the desired confidence level.
    
    Parameters    
    ----------
    numberofFlips          : int
                             The number of flips that will be done for the 
                             relevant confidence level.
    desiredConfidenceLevel : float
                             the desiredConfidenceLevel that we will find the nearest
                             confidence level for
        
    Returns
    -------
    result :
        pValUnderThreshold : float
                             the nearest valid pvalue that is smaller than the desiredConfidenceLevel
            
    """
    
    numberofIntervals = np.int(np.round(numberofFlips/2));
    intervalIndex = np.arange(numberofIntervals);
    intervalList = 2*(stats.binom.cdf(intervalIndex,numberofFlips,.5));
    pValUnderThreshold = np.max(intervalList[intervalList <= desiredConfidenceLevel])
    return pValUnderThreshold


def false_negative_calculator(numberofFlips, unfairCoinWeight, desiredConfidenceLevel):
    """
    loop through each coin and do a simulated p-test to see if for a set number of coinflips 
    if the given coin is verified to be unfair in a confidence window.
    if the coin is a fair coin, and pvalue<=closestConfidenceLevel (meaning it passes the binomial test and 
    the coin is unfair in a certain confidence window), then we add 1 to the false-positive number
    if the coin is an unfair coin, yet is not identified as one, then we add1 to the false-negative numbers
    """
    guessWeight = .5
    falsePositives = 0
    falseNegatives = 0
    numberofTrials = 1000
    
    
    #for our p-test, we want to make sure to pick an interval that fits the discretness of 
    #the problem. Therefore we input a desired confidence interval, and the function finds 
    #the nearest confidence interval that can be used.
   
    closestConfidenceLevel = accurate_pvalue_interval(numberofFlips, desiredConfidenceLevel)
    
    #see how many false negatives are obtained in a number of trials.
    for ithTrial in range(numberofTrials):
        pvalue = flipAndDoPTest(numberofFlips, unfairCoinWeight, guessWeight)     
        if pvalue>closestConfidenceLevel:
            falseNegatives += 1
    falseNegativeRate = falseNegatives/numberofTrials            
    return falseNegativeRate


def find_numberofFlips_and_false_negatives_vs_fairness(unfairCoinWeight):

    numIterations = 1000
    #maximum number of iterations used in the loop
    numberofFlips = 10
    #initial starting number of flips used in the first iteration of loop
    flipNumberIncrement = 50
    #how much the flip number increases by each iteration in loop
    numberofStandardDeviationSeparation = 4
    #decides the amount of standard deviations the unfair coin is from the fair coin that we finding the 
    #number of flips for
    falseNegativeThreshold = .05
    desiredConfidenceLevel = .05

    #Loop explained in the function deviation_separation_calculator()
    for stepi in range(numIterations):
        falseNegativeRate = false_negative_calculator(numberofFlips, unfairCoinWeight, desiredConfidenceLevel)
        if falseNegativeRate <= falseNegativeThreshold:
            print("Completed in", stepi, "total increments and requires", numberofFlips, "number of flips", "with a false negative rate of:", round(falseNegativeRate*100, 4), "%")
            break
        else:
            numberofFlips += flipNumberIncrement
    if falseNegativeRate > falseNegativeThreshold:
        print("Could not find solution in the given number of iterations:", numIterations)
    return numberofFlips



    
weightList = np.array([.9, .8, .7, .6, .55, .54, .53, .525, .520, .515, .510])
#list of weights of the unfair coin; as this gets closer to .5 the more numberofFlips will be needed for
#our condition to be satisfied

numofFlipsList = np.arange(len(weightList))
for i in range(len(weightList)):
    numofFlipsList[i] = find_numberofFlips_and_false_negatives_vs_fairness(weightList[i])
    print("For fairness weight of:", weightList[i], "number of flips required is",numofFlipsList[i])

plt.plot(weightList, numofFlipsList)   


There are various ways you could examine this problem analytically, but a typical way is to frame the problem as a hypothesis test for a stipulated probability for the coin.  Suppose we let $X_1,X_2,X_3, ... \sim \text{IID Bern}(\theta)$ denote the outcomes of the coin-flips where $\theta$ is the probability of flipping a head (here denoted by a one).  We can use a classical binomial proportion test to test the hypotheses:
$$H_0: \theta = \tfrac{1}{2}
\quad \quad \quad \quad \quad 
H_\text{A}: \theta \neq \tfrac{1}{2}.$$
There are various types of binomial test in statistical analysis, but the simplest is the Wald test, which uses the normal approximation to the binomial proportion.  If you want to know how many flips you need to reliably detect an unfair coin, the usual thing to do would be to find out how many flips you need to obtain some minimum stipulated power against a specified value for the parameter that is close to the null value.
To do this you will need to specify three things: (1) the significance level for your hypothesis test; (2) the parameter value at which you want to compute the power (presumably a value close to your null value); and (3) the minimum power you will consider to be sufficient to constitute a "reliable" test.  In the section below I give an example of this using the power function for the Wald test.

Computing sample size via power of the Wald binomial test: The two-sided Wald test for the null hypothesis $H_0: \theta = \tfrac{1}{2}$ uses the test statistic an approximate null distribution:
$$Z_n \equiv \sqrt{n} \cdot \frac{p_n - \tfrac{1}{2}}{\sqrt{p_n (1-p_n)}} \overset{\text{approx}}{\sim} \text{N}(0,1).$$
At significance level $0 < \alpha < 1$, the test has acceptance-region $-z_{\alpha/2} \leqslant Z_n \leqslant z_{\alpha/2}$, which can be rewritten as:
$$\bigg( p_n - \frac{1}{2} \bigg)^2 \leqslant z_{\alpha/2}^2 \cdot \frac{p_n (1-p_n)}{n},$$
which can be shown to be equivalent to $L(\alpha,n) \leqslant n p_n \leqslant U(\alpha,n)$ with the lower and upper bounds:
$$L(\alpha,n) \equiv \frac{n}{2} \Bigg[ 1 - \sqrt{\frac{z_{\alpha/2}^2}{n + z_{\alpha/2}^2}} \Bigg] 
\quad \quad \quad \quad \quad 
U(\alpha,n) \equiv \frac{n}{2} \Bigg[ 1 + \sqrt{\frac{z_{\alpha/2}^2}{n + z_{\alpha/2}^2}} \Bigg] .$$
Consequently, the exact power function for the test is:
$$\begin{align}
\text{Power}_\alpha(\theta)
&= 1 - \mathbb{P} ( \text{Accept } H_0 | \theta ) \\[6pt]
&= 1 - \mathbb{P} ( L(\alpha,n) \leqslant n p_n \leqslant U(\alpha,n) | \theta ) \\[6pt]
&= 1 - \sum_{L(\alpha,n) \leqslant x \leqslant U(\alpha,n)} \text{Bin} (x|n, \theta). \\[6pt]
\end{align}$$
We can program this power function in R as follows (we have vectorised this function with respect to the input n to make the next step easier).
#Create power function for the Wald binomial test
power.binom.test <- function(n, prob, alpha = 0.05) {
  z2    <- qnorm(1-alpha/2)^2
  OUT   <- rep(0, length(n))
  for (i in 1:length(n)) {
    nn    <- n[i]
    TERM  <- sqrt(z2/(nn+z2))
    LOWER <- ceiling((nn/2)*(1-TERM))
    UPPER <- floor((nn/2)*(1+TERM))
    OUT[i] <- 1 - sum(dbinom(LOWER:UPPER, size = nn, prob = prob)) }
  OUT }

To compute the required sample size we need to specify the three elements discussed above.  For illustrative purposes, let's stipulate that we are using a test with a 5% significance level and we want to compute the power at the point $\theta = 0.51$ and we require that the power at this point must be at least 90%.
#Set parameters for the computation
ALPHA     <- 0.05
THETA.ALT <- 0.51
MIN.POWER <- 0.9

#Compute required sample size
POWER     <- power.binom.test(n = 1:30000, prob = THETA.ALT, alpha = ALPHA)
SAMP.SIZE <- min(which(POWER >= MIN.POWER))

#Show required sample size
SAMP.SIZE
[1] 26226

In this case we see that we require a minimum sample size of $n = 26,226$ to have 90% power in detecting the alternative value $\theta_1 = 0.51$ using a Wald test with 5% significance level.  This is just one example of this type of calculation, and you could use different numbers if you prefer.
