Why is p for 8 times heads out of 21 flips not 8/21?

Flipping a fair coin 21 times. The probability of getting 8 heads in any order is

$$p = \frac{21!}{8!(21-8)!}0.5^8(1-0.5)^{21-8} = 0.097$$

I get that the Binomial formula takes into account the different sequences in which the heads can occur during the 21 flips. But in any case, I would at least had expected the Binomial formula to give a value a bit closer to $0.5$.

The question is: What am I then calculating by $8/21 = 0.38$?

• So should the probability for 21 heads out of 21 be 1 ? :) Jul 31, 2014 at 10:26
• Sure not but still, the question is, what is 8/21 giving me? Jul 31, 2014 at 10:28
• 8/21 is giving you the numbers to plug into the equation. Jul 31, 2014 at 10:33
• Isn't that the Binomial formula? (though a Binomial RV is a sum of n Bernoulli RVs)
– BCLC
Jul 31, 2014 at 10:52

8/21 is the proportion of heads in the result.

Instead of calculating the probability of 8 heads, you can calculate the probability that the proportion of heads in 21 coin flips will be 8/21. They both turn out 0.097 (assuming your calculation is correct)

• I see my flaw in my understanding now. I usually think of probability in terms of $accepted/possible$ cases. But there are many more than just 8 ways of having 8 heads in 21 flips. Jul 31, 2014 at 11:34
• Btw @TMOTTM, I think prob questions are more relevant for math.stackexchange.com "In probability theory we consider some underlying process which has some randomness or uncertainty modeled by random variables, and we figure out what happens. In statistics we observe something that has happened, and try to figure out what underlying process would explain those observations."
– BCLC
Jul 31, 2014 at 11:45
• @BCLC: Nevertheless we invite questions on probability theory: See "What topics can we ask about here?". If you think we shouldn't you can raise the issue on Cross Validated Meta. Aug 1, 2014 at 9:56

Think of one fair die with $21$ sides, of which $8$ have the letter $H$ inscribed, and the other $13$ have the letter $T$ inscribed.

Throw the die once. What is the probability that you will get an $H$? It is $8/21$. Now compact the $21$ dimensions into just $2$, taking into account how many time each letter appeared in the $21$-dimensional world: it appears that the $T$ dimension should have a higher probability of occurring than the $H$ dimension, in one throw in the $2$-dimensional world, if one wanted to keep the correspondence with the $21$-dimensional world...

...which tells us that the sample proportions of a sequence of results from independent throws of a coin, estimate the probability distribution that holds for each throw.

What am I then calculating by $8/21=0.38$?

You're calculating an estimate of the probability that one future coin flip will turn up heads.

Of course, by doing so you're ignoring everything else you know about the coin, including your expectation that the coin is fair, and treating it as a binomial experiment with an unknown probability of success. Which is kind of a silly thing to do when you know you're dealing with a fair coin. (If you don't have any reason to believe the coin should be fair, this may be a reasonable thing to do.)