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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.

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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:

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

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

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

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

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

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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.

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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 $\frac{5}{10}$.

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