Mathematically, 1 in 3 and 10 in 30 are equal. What about in probabilities? Assume I am in a draw for a prize. There are 3 rubber balls, 2 red ones (loser) and 1 green ball (winner). If the green ball is drawn, I win.
Now let's have a second draw (simultenously). It has 20 red balls and 10 green ones – same thing, 1 draw, if I get a green ball, I win. 
Assume equal random distribution of the balls. Is there any difference in the win probabilities between the two scenarios? 

Side note: In an uneven fight, 2 v 1 vs 20 v 10, I believe the 1 fighter vs 2 has a better probability of winning than 10 over 20 (it's easier to overcome a 2:1 advantage once than it is 20 times). Not sure if it plays into this at all :).
 A: The probabilities are the same! However, we would treat the two differently in inference about the proportion.
Let’s construct a 95% confidence interval for the proportion in each case. The usual formula for this is:
$$
\hat{p} \pm 1.96\sqrt{\hat{p}(1-\hat{p})/n}
$$
For 1 in 3 draws, we get:
$$
0.33 \pm 1.96\sqrt{0.33(1-0.33)/3}
$$
For 10 in 30 draws, we get:
$$
0.33 \pm 1.96\sqrt{0.33(1-0.33)/30}
$$
The second situation will give the narrower confidence interval. Likewise, the p-value for a proportion test will be lower in the second case.
However, both cases give the same $1/3$ probability as the $\hat{p}$ proportion.
A: For the ball problem, the probabilities are the same and $1/3$. For truly random draws, it is neither harder nor smaller to draw one green ball in an urn consisting of 1 green & 2 red vs 1K green & 2K red. Your fighter example doesn't reflect the same situation though. 
A: For the ball problem the probabilities are the same because every ball is equally likely to be picked.
In the 3 ball bag there are 3 possible (equally likely) balls. Any ball gets picked on average 1/{number of balls} of the time.  in other words, the probability of getting the green ball is 1/3.
In the 30 ball bad there are 30 possible (equally likely) balls.  Any ball gets picked on average 1/30 of the time.  Suppose all the green balls were numbered 1,2,...,10.
Green1 gets picked 1/30 of the time.
Green2 gets picked 1/30 of the time.
...
Green10 gets picked 1/30 of the time.
So Green gets picked 1/30 + 1/30 + .. 1/30 = 10/30 = 1/3 of the time.
But in the fight


*

*Not all the fighters are equal

*Even if the fighters were equal (e.g. they were robots) they could gang up.

*Even if the fights were separate i.e. just 10 lots of 1vs2, 1vs2 isn't a random draw. 
If all the fighters are equal then I would expect 2 ALWAYS to beat 1!

