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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 :).

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    $\begingroup$ Could you change the title? Probability is a branch of mathematics. $\endgroup$ – Xi'an Apr 5 at 17:57
  • $\begingroup$ That's fair, what I meant is specifically. $\endgroup$ – ajacian81 Apr 5 at 18:34
  • $\begingroup$ Any time you draw just one ball, and 1/3 of them is winning while 2/3 are losing, your probability of success is 1/3. $\endgroup$ – Federico Tedeschi Apr 5 at 21:06
  • $\begingroup$ In an uneven fight, 1:2 vs. 10:20 does have an influence, but things like morale or whether the smaller for can hold a chokepoint resp. fighting surrounded in an open field can drastically change probabilities. $\endgroup$ – toolforger Apr 6 at 7:10
  • $\begingroup$ Also, 10:20 is equivalent to doing 10 draws of 1:2, so assuming the 10:20 are 10 matches of 1:2, the probabilities and result distribution will be the same as with drawing balls (assuming a 1:2 fight has a 1:2 win chance, which is usually not the case due to skill level). $\endgroup$ – toolforger Apr 6 at 7:11
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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.

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    $\begingroup$ I do not think n = 1 or =10. $\endgroup$ – mdewey Apr 5 at 15:30
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    $\begingroup$ $n$ is the sample size, it should be 3 in the first case and 30 in the second. This doesn't change you conclusion however, but it's not surprising that the width of the internal shrinks as $n$ increases (all else equal). Though I don't see why confidence intervals are interesting here, we all ready know the probability distribution $\endgroup$ – Repmat Apr 5 at 15:37
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    $\begingroup$ "However, we would treat the two differently in inference a out the proportion." It appears that OP doesn't know almost anything about basic probability, so explaining inference, and confidence intervals about proportions without explaining what situation you're talking about must be confusing. $\endgroup$ – JiK Apr 6 at 2:02
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    $\begingroup$ To clarify: If we have a black box random machine, and we get 1 win out of 3 draws, that leads us to different conclusions about the win probability than getting 10 wins out of 30 draws would. That has nothing to do with the 3 or 30 balls in an urn. $\endgroup$ – JiK Apr 6 at 2:04
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    $\begingroup$ Confidence intervals are irrelevant for the OP's scenario. The answer should be edited to either explain when confidence intervals become relevant, or to not have that part at all. $\endgroup$ – toolforger Apr 6 at 7:07
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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.

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  • $\begingroup$ right the fighter scenario was more along the lines of why i was thinking there'd be a possible difference. $\endgroup$ – ajacian81 Apr 5 at 18:32
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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!
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When it comes to probabilities, sometimes people use a trick '+1', which means that for every state you upgrade its count for something you have not observed. This means, you will never receive 0 probabilities for any state. If you use this trick, you get for 1 in 3 case, these probabilities:

p(a) = (1+1)/((1+1)+(2+1)) = 2/5 = 0.4

p(b) = (2+1)/((1+1)+(2+1)) = 3/5 = 0.6

and for 10 in 30 case:

p(a) = (10+1)/((10+1)+(20+1)) = 11/32 = 0.34375

p(b) = (20+1)/((10+1)+(20+1)) = 21/32 = 0.65625

the more counts you have, the closer it gets to 1/3=0.33333 and 2/3=0.66666.

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    $\begingroup$ There is nothing about inference about probabilities based on observations in the question. The question asks about the probability of drawing a green ball with known numbers of green and red balls in the urn. This answer is misleading, and using "the trick" where it doesn't apply is clearly wrong. $\endgroup$ – JiK Apr 6 at 2:06
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    $\begingroup$ To be clear, this "trick" is used when you have a good sample size, but rare enough event that you still haven't covered the sample space, so you add pseduocounts to ensure that your unobserved events don't wind up with zero probability. You generally shouldn't do this with very low sample size, since the fictitious pseduocounts will be a larger chunk of the data. In the first example, you get 3 real counts and 2 pseudocounts, so 40% of your data is completely made up - it's a poor strategy to apply this "trick" here. $\endgroup$ – Nuclear Wang Apr 6 at 12:53
  • $\begingroup$ no, this "trick" is especially useful with small amount of samples. See (Lidstone, 1920; Johnson, 1932; Jeffreys, 1948) in publication 'An Empirical Study of Smoothing Techniques for Language Modeling' (sciencedirect.com/science/article/pii/S0885230899901286). $\endgroup$ – Peter Taraba Apr 6 at 21:56

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