Why does the probability of an event change in a binomial experiment with the proportional change of successes and failures?

Question

Let us assume the following two binomial experiments, assuming coin tosses with a fair coin $(p = 0.5)$:

General: $\binom{n}{k}p^{k}(1-p)^{n-k}$

$\binom{10}{9} \cdot 0.5^{9} \cdot 0.5^{1} = 0.009766$

$\binom{20}{18} \cdot 0.5^{18} \cdot 0.5^{2} = 0.0001812$

Why is it that in the second case the probability of the event decreases, although the successes and failures have been proportionally (doubled) changed here? That seems counterintuitive to me. Is there an intuitive explanation for this?

Many thanks in advance!

Note: I‘m an undergraduate economics student.

Answer 1

Think about an extreme case (often a useful strategy).

If you flip just twice the probability of an equal number of heads and tails is $1/2$. That is clearly not the case if you flip $1000$ times - exactly $500$ heads would be very surprising. What you do know is that the probability of a ratio near $1/2$ is high.

In your case, if you preserved all the probabilities going from $10$ to $20$ coin flips you would have no probability left for $19$ heads out of $20$.

Answer 2

You may find it helpful to think of the second experiment, i.e. achieving two successes in $20$ coin tosses, as two sequences of the first experiment i.e. tossing the coin $10$ times twice.

You have calculated the probability of achieving one success in $10$ throws - let's call this $p$. The probability $q$ of achieving exactly $2$ successes in $10$ tosses is obviously slightly more than $p$, but of similar order of magnitude. Likewise, the probability $r$ of exactly no successes in $10$ tosses is less than $p$ but of similar order of magnitude.

However, the probability of exactly $2$ successes in $20$ tosses is now $$p^2+2qr$$.

So this is roughly $3p^2$ and therefore very much smaller than $p$.

Answer 3

There are two contributing factors in your case. The first logic is explained by Ethan Bolker above (all specific outcomes gets less likely as n increases, simply since there are more potential outcomes and the total probability needs to sum up to unity).

The second logic has to do with your specific example of a "tail outcome" (i.e. an unlikely outcome). There is a law (the law of large numbers) saying that the tails get thinner as n increases. The intuition is basically that it is much less likely to get 5 sixes throwing 5 dice compared to getting one six when throwing one dice.