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From 2011 Stat 110:
A jar contains r red balls and g green balls, where r and g are fixed positive integers. A ball is drawn from the jar randomly (with all possibilities equally likely), and then a second ball is drawn randomly.
Explain intuitively why the probability of the second ball being green is the same as the probability of the first ball being green.
I can show this is true algebraically, but what are some intuitive explanations?
Let us just keep drawing balls and line them up as we do so; thus forming a line of $r$ red and $g$ green balls in order of withdrawal.
I now point to any ball in the line. What is the probability that it is red?
Should it matter at all where in the line I have pointed? The first? The second? The last? Anywhere in between?
Every ball has the same chance of being the first ball drawn, and $r$ of the $r+g$ balls are red.
Every ball has the same chance of being the second ball drawn, and $r$ of the $r+g$ balls are red.
$\vdots$
Every ball has the same chance of being the last ball drawn, and $r$ of the $r+g$ balls are red.
The counterinuition is that the colour of the first ball drawn influences the probability for the second ball, and this is true but only when you have knoweldge of what the first may be. Without that conditional knowledge, each ball that was in the jar has the same chance to be the second ball drawn.
Draw both balls with your eyes shut. Then look at the second ball before looking at the first ball.
Colors don't have preferences for positions, so a green ball is equally likely to occupy any spot.
Instead of a jar of balls, imagine a deck of red and green playing cards.
Shuffle the deck, then (optionally) "cut" it by taking the top card and putting it on the bottom. Reveal the new top card, and consider the probability of that card being green.
Intuitively, that probability shouldn't depend on whether you actually cut the deck or not.
(Cutting the deck corresponds to drawing ball #2, and not cutting the deck corresponds to drawing ball #1.)
Probability of (red&green) is equal to probability of (green&red), therefore probability of second green is equal to probability of first green.
Algebraically: $$1=\underbrace{P(R_1R_2)+P(R_1G_2)}_{=P(R_1)}+\underbrace{\overbrace{P(G_1R_2)+P(G_1G_2)}^{=P(R_1G_2)+P(G_1G_2)=P(G_2)}}_{=P(G_1)}$$
There are multiple ways to interpret the claim, and its truth depends on that interpretation. The three interpretations are:
The probabilities are the same, period.
The probabilities are the same when evaluated before drawing either ball.
The probabilities are the same when evaluated before drawing each ball.
Only the second claim is true; the first and third claims are false.
The weird thing is that, if you assume that the claimant is correct, then you conclude that they were making Claim 2. But if you don't add in that caveat as you read the problem statement, it probably sounds like Claim 1, which is false. Or it could be read as Claim 3, which is also false.
For all we know, the author of the book actually did mean Claim 1 or Claim 3 and was therefore wrong in their thinking. Presumably someone who wrote a Statistics textbook wouldn't make such a silly mistake, but the only way to conclude that they didn't is to assume it.
I'd guess that it's this circular reasoning that makes the claim seem non-intuitive. In order to see the claim as true, the caveat of "when evaluated before drawing either ball" has to be inferred.
Here's a claim to focus on:
Before you draw any balls, the probability of drawing a green ball is the same for any draw:
$$p_{\text{green}}=\frac{n_{\text{green}}}{n_{\text{green}}+n_{\text{red}}}$$
For example, if there're 10 balls and only 1 of them is green, then you have a 10% chance of drawing the green ball on any round. So 10% odds on Round 1, and 10% odds on Round 2.
However your estimated odds $p_{\text{green}}$ will change as the balls are drawn because your knowledge of $n_{\text{green}}$ and $n_{\text{red}}$ are changing. For example, on Round 10, there's only 1 ball left and you know what color it'll be.
This is similar to paw88789's answer - draw out both balls with one hand, but mark the right-hand ball as the second ball, before you pull your hand out of the bag.
Ordering the balls by the sequence in which they were drawn, is equivalent to assigning the numbers $1, 2, 3, \dots, r+g$ to the balls.
Inscribe these numbers upon the balls in black ink.
Now inscribe the same sequence of numbers upon the ball in white ink, except that we write the white "one" on the ball with the black "two", and vice versa.
Note that for each black-ink sequence, there is precisely one corresponding white-ink sequence.
The probability that the "first" ball is red (where "first" is decided by the white-ink ordering) must be the same as the probability that the "second" ball is red (where "second" is decided by the black-ink ordering).
Intuitiviely we can see the probabilities are the same, because this approach is identical to pulling out two balls at the same time. Even if we pulled out 5 balls at once, each of them would have the same chance of being green, what is confusing our intuition is the assumption that we've seen (and evaluated) the first ball when taking out the second one. Until we evaluate them, removing balls in sequence is the same as removing balls all at once.
The main assumption here is that we do not know the color of the first ball when picking the second. As a result, the probability of getting any ball at any pick is the same, as there aren't any balls which are biased towards the first or the second pick. It is also the same as if the two balls were picket at the same time, or with different hands.
Let the green balls be labelled from $1,..,g$ and the red balls be labelled from $g+1,...,g+r$, then the sample space is the set of all pairs $(a,b) \in \{1,...,g+r\}$ such that $a \neq b$. Each of these pairs is equally likely so we can use the Naive definition of probability.
The probability of the first being green is $P(G_1) = \frac{g}{g+r}$. The probability of the second being green is
$$ P(G_2) = P(G,G) + P(R,G) = \frac{g}{g+r} \cdot \frac{g-1}{g+r-1} + \frac{r}{g+r} \cdot \frac{g}{g+r-1} $$ $$ P(G_2) = \frac{g(g-1)}{(g+r)(g+r-1)} + \frac{rg}{(g+r)(g+r-1)} = \frac{g(g - 1) + rg}{(g+r)(g+r-1)} = \frac{g}{g+r}$$
It matters when you calculate the probability.
If r and g are equal, and you know the color of the first ball drawn, then when you calculate the probability for the second draw is not the same as if you calculate the probability of the second draw before the first ball is drawn.
Say r and g=10.
Scenario 1) I draw a red ball. If we then calculate probability, the chances for the second draw are 9/19 red and 10/19 green.
Scenario 2) We calculate the probability that the second ball will be green before having drawn any balls. The chances are 50/50 for both draws... More precisely there's a 25% chance the balls will be GG or RR or GR or RG.
