3

Whenever I encounter a probability question, especially those with pulling objects out of a box, I think, "Wait, are they all being selected at the same time (combinations), or are they being chosen one by one without replacement?"

And unfortunately, I still don't have an answer for that.

For example, let there be $10$ cards laid face down on table, $5$ red, $5$ blue.

I pick up two cards. What is the probability that both of them are red?

Well, I could do combinations, $\displaystyle \frac{\binom{5}{2}}{\binom{10}{2}}= \frac{2}{9}$.

Alternatively, we have $\displaystyle \frac{5}{10} \times \frac{4}{9}=\frac{20}{90}=\frac{2}{9}$.

In this case the probabilities were the same, regardless of whether they were picked up at different moments in time or simultaneously.

Does this hold for all probability questions? Is there something to be assumed when it is not clear how the objects are being selected?

Saketh Malyala
  • 13,637
  • It's a known fact that probability questions often are somewhat ambiguous, resulting in some known paradoxes, too. The conclusion is obvious, though: as long as that ambiguity is present, you should refuse to consider the problem. –  Jun 18 '17 at 22:25
  • Insightful. Thank you @ProfessorVector – Saketh Malyala Jun 18 '17 at 22:28
  • Erm... but there was that "as long as"! They aren't always ambiguous (and you gave an example yourself), and the trick is to recognise those situations. Of course, the clear cases are mostly some combinatorics, and boring like hell. –  Jun 18 '17 at 22:46
  • There are many ways to do the same type of selection. If the possible outcomes are the same, and they are all equally likely in both cases, then it doesn't matter how you do it. This is the key to many probability questions: if you think of doing the selection in the right way it is easy. – Robert Israel Jun 19 '17 at 00:29

3 Answers3

4

If you don't replace, then it doesn't matter whether you pick things up all at the same time or one by one, assuming all you are interested in is how many out of how many have a certain property (i.e. If order does not matter).

This is really nice, since that means that there are two different ways to think about and solve problems like this, as your post nicely demonstrates.

Of course, if you do use replacement, you will have to use a different method.

Ans finally, if it is not clear whether replacement is used or not, then you should point that out and ask for clarification.

Bram28
  • 100,612
  • 6
  • 70
  • 118
3

Assume your red cards are numbered R1 through R5 and the blue ones B1 through B5.

Your first method counts unordered pairs. All 45 unordered pairs are equally likely, and of them, 10 correspond to successful cases.

In the second method, you're essentially counting ordered pairs. Again, they are all equally likely, and there are 90 of them, or twice as many. Of these, 20 correspond to successful cases, also twice as many.

Because in going from unordered to ordered pairs, you're multiplying both the numerator and denominator by $2$, the probability $\frac{10}{45} = \frac{20}{90}$ doesn't change.

Of course, it's not a coincidence that both numbers are multiplied by the same number, 2. This is how many ways there are to order an unordered pair.

user49640
  • 2,704
1

For some reason this is often not emphasized, but in probability problems there is an important step at the beginning where you have to introduce a sample space and introduce some assumptions about the probability law in order to model the real world ​scenario that was described in the problem. And there may be more than one way to model the same real world scenario.

For the problem you mentioned we could number the cards 1 through 10 (with $1$ through $5$ corresponding to red cards) and let the sample space $S$ be the set of all two-element subsets of $\{1,\ldots,10\}$. We make the modeling assumption that all outcomes have equal probability.

Alternatively, we could choose to let the sample space be the set of all ordered pairs $(a,b)$ where $a$ and $b$ are integers between $1$ and $10$ and $a \neq b$. We again make the modeling assumption that all outcomes have equal probability.

Different models can lead to different counting problems when computing various probabilities.

So why don't they tell you which model to use when they give you the problem? Probably because they realize that in this case it doesn't matter. (If it did matter, the question was not posed precisely enough.)

A lot of famous probability "paradoxes" result from introducing different and conflicting models for the same real world scenario.

littleO
  • 51,938