Coin toss - win if $0.4 \leq $ heads $ \leq 0.6$, should it be tossed 100 or 10 times

Question

A coin will be tossed, you win a dollar if the percentage of heards is between $40$% and $60$%. Which is better: 10 tosses or 100 tosses.

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I feel as though the answer should be 100 tosses. My reasoning for this is that as the number of tosses increases the percentage of heads will approach 0.5

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A follow up to this

A coin is tossed, and you win a dollar if there are more than $60$% heads. Which is better: 10 tosses or 100 tosses

Here I feel as though 10 tosses is better, using the reasoning from before, that the proportion of heads will approach 0.5 as the number of tosses is increased.

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If anyone could provide a bit of a sounder reasoning than my intuitions here, or explain why my intuitions are false, that would be appreciated.

Edit

Here is a simulation for the tosses:

results = []
N       = 10

for t in range(trials):
    heads = 0
    for n in range(N):
        f = flip()
        if f == 1:
            heads += 1
    result = heads/N
    results.append(result)
    if result <= 0.6 and result >= 0.4:
        plt.scatter(t, result, color = "green", s = 45)
    else:
        plt.scatter(t, result, color = "red", s =45)


r = [x for x in results if x <= 0.6 and x >= 0.4 ]
print(len(r)/trials)

plt.axhline(y=0.6, alpha = 0.3)
plt.axhline(y=0.4, alpha = 0.3)
plt.title("Outcome of a series of 10 simulated coin tosses")
plt.show()

edit 2

In [3]: from math import factorial as f

Compute for 100 rolls, and 61 heads

In [1]: n = 100
In [2]: k = 61

In [4]: ( f(n) * (0.5)**n ) / (f(k) * f(n - k))
Out[4]: 0.00711073226992655

Compute for 10 rolls, and  heads

In [5]: n=10
In [7]: k=6

In [8]: ( f(n) * (0.5)**n ) / (f(k) * f(n - k))
Out[8]: 0.205078125

Answer 1

Summarizing from comments: Your intuition is correct, though there may be a bit of a trick in the 2nd question (I'm not sure if it was meant): To get more than 60% would require 70% of 10 tosses, but only 61% of 100 tosses. I doubt the intent of this problem, if you need "proof", would be in tossing a coin-- in fact there is a chance that the randomness of 10 and 100 tosses may lead you to the wrong answer.

You pointed out that you can use the binomial formula to calculate the probability of any number of heads in any number of tosses. If you evaluate that for 100 and 10 tosses you should expect the results to show that it is less likely to have a deviation from 50% at a higher toss number.

If the probability of getting 60% is higher for 10 tosses than 100%, all of the subsequent probabilities will be be higher. Intuitively it is very to understand that 10 of 10 heads is much more likely than 100 of 100 heads. As you increase the percentage heads it gets less and less likely that 100 tosses will obtain it as compared to 10 tosses.