This Question stems from Ronald Fishers Tea problem in how to design experiments which I have reworded.

Anyway, suppose you have 2 different items in front of you, 5 of each which we can just call red and blue. So you have 10 items total, and you want to randomly select the 5 red ones. The probability of getting this correct is $$\frac{1}{10 \choose 5}$$ but I can't shake the intuition that it would be $$\frac{5}{10}$$ and I can't reason with myself as to why it wouldn't be. Can someone explain intuitively and not mathematically why it wouldn't be because I am able to conclude mathematically that it's not.

The probability of randomly drawing one red item is $$\frac{5}{10}$$. But, after drawing that first one, the probability of drawing another red item is now $$\frac{4}{9}$$, since 9 items remain in total, and of them, 4 are still red. The next one is $$\frac{3}{8}$$, then $$\frac{2}{7}$$, and the final one has probability $$\frac{1}{6}$$.

The probability of all five of these events occurring is their product:

$$p = \frac{5}{10}*\frac{4}{9}*\frac{3}{8}*\frac{2}{7}*\frac{1}{6}$$

$$= \frac{5*4*3*2*1}{10*9*8*7*6}$$

$$= \frac{5!}{\frac{10!}{5!}}$$

$$= \frac{1}{\frac{10!}{5!5!}}$$

$$= \frac{1}{10 \choose 5}$$

Here is an intuitive answer: by the same reasoning, if there are $$100 \, 000$$ balls, $$50 \, 000$$ red and $$50 \, 000$$ blue, then the chance of picking $$50 \, 000$$ reds would be $$50 \, 000/100 \, 000 = 1/2$$. But this is absurd! If you make $$50 \, 000$$ selections, then the chance of all of them being red would be tiny.

You might be confusing this with a different scenario. If there are $$10$$ balls and $$5$$ reds, then the chance of picking a red is equal to $$5/10 = 1/2$$. However, this is different from making $$5$$ selections, and considering the probability that all $$5$$ of them are red.

If the probability of getting $$5$$ reds were $$\frac{5}{10}=\frac{1}{2}$$, then by symmetry, we would expect the probability of getting $$5$$ blues to also be $$\frac{1}{2}$$. From this, we would conclude that the probability of getting a mix of reds and blues would be $$1-\frac{1}{2}-\frac{1}{2}=0$$. However, we know that the probability of getting a mix of reds and blues is non-zero, leading to a contradiction. Therefore, the probability of getting $$5$$ reds cannot by $$\frac{5}{10}$$.