Probability of exactly one same-colored pair when drawing six balls from an urn consisting of five same-colored pairs.

Question

Imagine there is an urn with $10$ colored balls. There are five colors present in the urn, and two balls of each color.

What is the probability that there is exactly one pair of the same color when you draw $6$ balls from an urn in one go?

What is wrong with my appproach:

$1/10\cdot 1/1 \cdot 8/8 \cdot 6/7 \cdot 4/6 \cdot 2/5.$

Reason: For the first ball we have one out of $10$ balls -> $1/10$

the second ball is the one that has the same color than the first -> $1/1$

for the third ball we have $8$ balls left, we can choose -> $8/8$

for the fourth ball we have $7$ balls left, but we can only choose out of $6$ since we must not choose the same color -> $6/7$

for the fifth ball we have $6$ balls left, but we can only choose out of $4$ since we must not choose the same color -> $4/6$

for the sixth ball we have $5$ balls left, but we can only choose out of $2$ since we must not choose the same color -> $2/5$

I know that my solution is wrong but I don't know why. I don't want to have the right solution, but the exact reason at which point my solution is wrong and how could it been fixed. I know that in my approach the balls are drawn from an urn one after another without replacement by taking into account the order. But in the experiment the balls are drawn in one go without taking into account the order. Is that the mistake in thinking? How can I adjust my solution to come to the right result? What is my mistake in reasoning?

Answer 1

If you want to do it the multiplication way, you'd start with

• $\frac{10}{10}\frac19$ Any ball followed by $1$ from $9$ of same color for the pair
• $\frac88\frac67$Any ball of remaining $8$ and skip to next color and so on

We get $(10/10)(1/9)(8/8)(6/7)(4/6)(2/5)= \frac{8}{315}$

But there is still a hitch . We have chosen the pair together but they could actually be placed in $\binom62$ ways in the order, so multiply to get $\frac8{315}\binom62 = \frac8{21}$

---

NOTE

I have solved the problem using the path you had taken, but personally I'd solve it using combinations, as there is less chance of error

• Choose a color and choose $2$ from it

• Choose one each from the remaining colors

This gives $$\frac{\binom51\binom22\times \binom21\binom21\binom21\binom21}{\binom{10}6} = \frac8{21}$$

This could, of course, be abbreviated to just $\dfrac{5\times2^4}{\binom{10}6}$,

but the full form shows the process most transparently

Answer 2

First ball - 1/10 - this implies that the first ball has to be one specific ball. Not even just a specific color (1/5) but one ball in specific. However, it's very possible to draw 6 of 10 balls and NOT draw a specific ball, but still get exactly one matching pair.

Second ball - 1/1 says that the second ball must have the same color as the first. When you pick this, then you're not counting "ways of getting exactly one pair." You are instead counting "ways to draw the pair out first, then draw four more balls that don't match anything else." In other words, nothing in the initial question says that the first two balls must match. It could be #3 and #5 that match.

Now, if we are purposefully assuming that the pair is drawn out in the first two draws, then I believe I agree with the rest of your reasoning - each ball past there just has to not match any of the previously drawn balls.

How would you do it correctly? Well...the normal combinatorial method of calculating a probability is essentially - count the total number of possible drawings then count the number of ways to get the desired outcome and finally divide the latter by the former to get the probability. There are ${10\choose 6}=210$ total possible outcomes. Then there are ${5\choose 1}=5$ ways to choose which color gets both balls, and ${2\choose 1}^4=2^4=16$ ways to choose which of the other colors balls get chosen, so 5*16=80 total possible ways to get the desired outcome. $\frac{80}{210}\approx 0.381$.

If you really wanted to find the probability by looking at each ball in succession, then it gets pretty tedious pretty quickly.

You would first draw out a ball. Then you'd draw out a second. At this point you've got two cases - Case1 they match, case2 they don't.

In either case you draw a third ball. Case1 is good, but case 2 has to be further broken into two more cases - case2a: 1-3 maches. Case2b: Neither 1-3 nor 2-3 match.

Draw a fourth ball. Case 1 now has to be broken into two cases: Case1a: 3-4 also matches, which means you've got two matching colors. Case1b: 3-4 doesn't match. Case2a also needs to be broken into two more cases: Case2a1: 2-4 matches and we again have two matching colors. Case2a2: 2-4 doesn't match. Case2b ALSO has to be broken into two cases: Case2b1: 1-4 or 2-4 or 3-4 match. Case2b2: Still no matches.

Some of those cases are already dead ends, but many of them continue and also continue to branch off into additional cases on the 5th and 6th draws. Plus you've got to keep track of how many ways there are to draw to each case and subcase. Then once you've got it all drawn out, you've got to add up all of the various branches, both to find the total number of possible cases and the number of possible successful drawings. Alternately there's probably some statistical method, but the question was tagged Combinatorics and not Statistics, so...

Answer 3

I know that in my approach the balls are drawn from an urn one after another without replacement by taking into account the order. But in the experiment the balls are drawn in one go without taking into account the order. Is that the mistake in thinking? How can I adjust my solution to come to the right result? What is my mistake in reasoning?

I think the main issue is that you seem to be bouncing around a bit too freely between counting states and propagating probabilities... when really you should adhere to one-or-the-other.

The all-at-once versus one-by-one-without-replacement accounting strategies are both perfectly valid if properly accounted (though in most problems one or the other tends to be much more straightforward).

---

Let's first define the balls by the following: $$\{A,a,B,b,C,c,D,d,E,e\}$$

Hopefully, this uppercase-lowercase system will help us keep track of the pairs (if/when enumerating states Aa-versus-aA ever comes up), while alphabetical is slightly easier than remembering some problem-specific choice of color names.

Now, let's look at your approach and see where we need to make an alteration or two. $$\Big(\frac{1}{10}\Big)\Big(\frac{1}{1}\Big)\Big(\frac{8}{8}\Big)\Big(\frac{6}{7}\Big)\Big(\frac{4}{6}\Big)\Big(\frac{2}{5}\Big)$$

What does 1/10 represent? You describe it as the probability for picking one ball from ten...but is it the odds of "any ball" or "a (specific) ball"? This is actually an very important distinction when it comes to propagating probabilities (or counting states).

If I asked you the odds for "Pick a capital letter." from a bag of $\{A,a,B,b,C,c,D,d,E,e\}$ in a single draw, I'm sure you could tell me those odds are 50%. I'm sure you could also tell me the odds of drawing $D$ in particular are 10%... but what does that mean we were enumerating in your solution? It means we were looking at the odds of drawing one ball in particular. (Let's say we were talking about drawing $A$ in particular, just so we can keep going...)

Next, what does that mean $1/1$ represents? You say it represents drawing the one possible success from the only remaining single match left for that color/letter... but this feels to me like you're confusing a $\binom{1}{1}$ (i.e. "one-choose-one") enumeration-of-states with a probability value. Had I asked, "What are the odds of drawing '$a$' from the bag (after you've already drawn '$A$')?" you would simply told me 1/9. So why are we now looking at a 1/1 = 100% probability for drawing specifially '$a$'? Clearly something has become confused.

If we fix it by assuming that we were looking at odds to draw specifically $Aa$ (by switching to a $(1/9)$ probability), then the rest of the work (probably?) looks fine for calculating the odds of drawing specifically $AaWXYZ$... but that still means that we need to calculate the $(AWaXYZ=\frac{1}{10}\frac{8}{9}\frac{1}{8}\frac{6}{7}\frac{4}{6}\frac{3}{5})$, $AWXaYZ$, $AWXYaZ$, and $AWXYZa$ states carefully.... and then extrapolate that for all possible $aA$, $Bb$, $bB$, ... etc. cases and all without double-counting.

Any way that we look at it... there is much more work to be done in accounting for all the possible probabilities for the possible range of assumed states. (Though we can be sure that a simple 'multiply by 5! to account for different orderings' will be incorrect.)

---

Finally, if I myself had to try to do all of this accounting for all the possible outcomes, I'd probably elect to simplify using the usual toolkit of $\binom{n}{k}$ (i.e. 'n-choose-k').

What is the probability of success? Well, that would be best described as the ratio of successful possible outcome states to total possible outcome states:

$$ P(success) = \frac{\sum S_{good}}{\sum S_{all}} $$

but how many possible different outcome states are there? Well, we are choosing six things (without repetition) from a mix of ten, so...

$$ P(success) = \frac{\sum S_{good}}{\binom{10}{6}} $$

but what are the possible successful outcomes? Well, we need to account for the "Two-A's-One-B-One-C-One-D-One-E" scenario, as well as the "One-A-Two-B's-One-C-One-D-One-E" scenario, etc. which means...

$$ P(success) = \frac{\binom{2}{2}\binom{2}{1}\binom{2}{1}\binom{2}{1}\binom{2}{1} + \binom{2}{1}\binom{2}{2}\binom{2}{1}\binom{2}{1}\binom{2}{1} \ + \ ... \ + \ \binom{2}{1}\binom{2}{1}\binom{2}{1}\binom{2}{1}\binom{2}{2}}{\binom{10}{6}} $$

which should give... $$ P(success) = \frac{5 \cdot 2^4}{\binom{10}{6}} \approx 38.1\% $$