# Intuition for probability of drawing first ball = probability of drawing second ball

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?

• This is a slight generalization of math.stackexchange.com/questions/1189349/… – David K Aug 3 '17 at 5:48
• In fact, the probability of any ball being green is the same as the probability of the first ball being green, for which I have given a short intuitive explanation, explained more elaborately earlier by Graham Kemp. – true blue anil Aug 3 '17 at 18:25
• Wait... You place the first ball back into the jar? – Malandy Aug 3 '17 at 21:48
• I think the key phrase here is "dependent, identically distributed", as opposed to "independent, identically distributed" (i.i.d.) which you may have heard before. Specifically the "identical" part. – Mehrdad Aug 3 '17 at 22:51
• @Malandy In this case the second draw is not conditional on the first. Say there are 3 balls, one red and two green. The probability the first ball is green is $2/3$. There are six ways to draw two balls while preserving order of the draw: G1 G2, G2 G1, G1 R, G2 R, R G1, and R G2. Of those six possible draws, four have a green ball as the second ball drawn. The probability the second ball is green is also $2/3$. Proof for other values of $r$ and $g$ is left as an exercise. – Todd Wilcox Aug 4 '17 at 2:18

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.

• I don't think this actually conveys any intuition. Somebody who suspects that the probability of the second ball being green will be different from the first will reinforce their wrong answer with this thought experiment. You seem to be arguing that the colour of the second ball isn't affected by my ignorance of the colour of the first one, but that's completely obvious. The question is, why doesn't it depend on the actual colour of the first ball? (E.g., what you buy for lunch today obviously isn't affected by whether or not I know your salary, but it certainly depends on your actual salary.) – David Richerby Aug 3 '17 at 8:23
• @DavidRicherby - it does depend on the actual color of the first ball. – kludg Aug 3 '17 at 9:00
• @kludg Under that interpretation, where the first ball's color is relevant, the claim is false. The claim's true only if you assume that they meant the probabilities without knowledge of the first ball's color. It's circular. – Nat Aug 3 '17 at 9:08
• @kludg Of course it does. Doh! I guess I meant to say something like, "How do we know we're in a case where drawing the first ball but not knowing its colour doesn't change the probability of the second ball's colour?" Because certainly not all situations behave that way -- for example, if somebody adds extra green balls to the urn if the first ball was red, then the "theorem" fails but it's not clear why the intuition no longer applies. My underlying objection is that the claimed intuition doesn't seem to use any of the properties of the system it purports to explain, and only people who... – David Richerby Aug 3 '17 at 10:12
• OP asked for intuition on the problem. I think this is a reasonable way of seeing that the second ball is just as likely to be green as the first. Of course after the actual first ball is drawn, (knowledge of) its specific color will change (one's assessment of) the probability of the second ball's color being green. But of course it works out so that the added probability when the first ball is red and the lost probability when the first ball is green gives overall the same probability that the second ball is green as the first. – paw88789 Aug 3 '17 at 13:24

Colors don't have preferences for positions, so a green ball is equally likely to occupy any spot.

Probability of (red&green) is equal to probability of (green&red), therefore probability of second green is equal to probability of first green.

Algebraically: $$1=\underbrace{P(R_1R_2)+P(R_1G_2)}_{=P(R_1)}+\underbrace{\overbrace{P(G_1R_2)+P(G_1G_2)}^{=P(R_1G_2)+P(G_1G_2)=P(G_2)}}_{=P(G_1)}$$

• That is true, but drawing without replacement it does not look intuitive obvious to me. A misleading little voice says "surely it is more likely to draw the more common colour first which has the side effect of increasing the probability of drawing the less common colour second" – Henry Aug 3 '17 at 12:46

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.)

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.

There are multiple ways to interpret the claim, and its truth depends on that interpretation. The three interpretations are:

1. The probabilities are the same, period.

2. The probabilities are the same when evaluated before drawing either ball.

3. 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.

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.

• It doesn't matter when you calculate the probability, what matters is which probability you calculate. In scenario 1) you aren't calculating P( second ball green) but rather P(second ball green | first ball red). – DRF Aug 3 '17 at 18:33
• My point is that if you have already drawn a ball and know what the color of that ball is when you calculate probability, then all that is relevant is the number of balls remaining. Really there's a third scenario where a green ball has already been drawn. Honestly I think the question (not OPs question, but the one he's quoting) is poorly framed. – aslum Aug 3 '17 at 18:36

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

$$P(G_2) = P(G,G) + P(R,G) = \frac{g}{g+r} \cdot \frac{g-1}{g+r-1} + \frac{r}{g+r} \cdot \frac{g}{g+r-1}$$ $$P(G_2) = \frac{g(g-1)}{(g+r)(g+r-1)} + \frac{rg}{(g+r)(g+r-1)} = \frac{g(g - 1) + rg}{(g+r)(g+r-1)} = \frac{g}{g+r}$$