11
$\begingroup$

If I were to roll a die, what would ​be the probability of getting $2$? Certainly it would be $\dfrac 16$ (because there are $6$ numbers and sample space contains 6 numbers) But I think we can look at it another way. We need $2$ right? and from $1-6$ there is only one number which is $2$ i.e. $2$. If I were to roll there would essentially be two types of numbers ie. one which is $2$ and one which is not ie $1,3,4,5,6$. So there are two basic outcomes. P(E) = number $2$÷total outcomes ie $2$. P(E)= $\dfrac 12$ So why is the probability $\dfrac 16$(when it is clear that the outcome we desire, can either happen or not happen.) Shouldn't the probability be $\dfrac 12$ instead, why not?

$\endgroup$
  • 6
    $\begingroup$ The probability of getting 2 may be $1/6$ or $1/2$ depending on the chosen probability model. $\endgroup$ – kludg Oct 24 '17 at 16:15
  • 3
    $\begingroup$ You need to decompose these into simple events. What exactly happens, and when? A simple event is one that cannot be broken down further. The event $\overline E$ can be broken down further since there are $5$ possibilities where you don't get $2$. $\endgroup$ – Decaf-Math Oct 24 '17 at 16:15
  • 24
    $\begingroup$ This is a perfect chance to remind that $Pr(E)=\frac{|E|}{|S|}$ is only valid when all outcomes in the sample space $S$ are equally likely to occur. In situations where the outcomes are not known to be equally likely then you may not use the above and must instead approach more carefully. $\endgroup$ – JMoravitz Oct 24 '17 at 16:21
  • 2
    $\begingroup$ “Crisis”? ${}{}$ $\endgroup$ – gen-ℤ ready to perish Oct 24 '17 at 16:51
  • 3
    $\begingroup$ Either you're going to win the lottery, or you're not going to win the lottery. Is your theory that you win the lottery 50% of the times you play it? Here, here's 100 lottery tickets, all with different numbers on them. Either every number wins, or not every number wins. Is your theory that 50% of the time, every one of those hundred tickets is a winner? $\endgroup$ – Eric Lippert Oct 24 '17 at 20:53
45
$\begingroup$

The probability is $\frac{1}{6}$ because there are $6$ elements in the sample space that are all equally likely.

If we take a sample space of $2$ and not $2$, then there are $2$ elements in the sample space, but they are not equally likely.

The classic example is that you don't have a $50\%$ chance of winning the lottery because the two options are "win" and "don't win".

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ "you don't have a 50% chance of winning the lottery" what a bad news... :( But the example is very good :). $\endgroup$ – Surb Oct 25 '17 at 10:24
25
$\begingroup$

This is actually a good instance of a really common logical fallacy. I'm not sure if the fallacy has any particular name -- one might call it the "equal likelihood fallacy," perhaps. The fallacy is assuming that events in a probability space have equal likelihood, even though they may not.

In general, one requires an affirmative reason to state that events have equal likelihood. When I roll a die, I can say that the chance of it rolling on any particular side is $1/6$ because (a) it's symmetric, so it shouldn't have any geometric bias in favor of one side over another, and (b) my experience confirms (or rather, doesn't dispute) this assumption. Similarly, when I draw a card from a well-shuffled deck, the word "well-shuffled" means there shouldn't be any particular favoritism toward any card; that is, all cards are equally likely.

But absent a reason like that, it's dangerous to say that events are equally likely, and it's very often false. If I flip 4 coins, then the number of those coins that could come up heads will be 0, 1, 2, 3, or 4. However, it is not true that each of those has a 20% chance of occurring; instead, having 2 coins come up heads is significantly more likely than the other possibilities, while 0 and 4 are the least likely.

For a good example of this fallacy in action, I'll refer you to this clip from the Daily Show: http://www.cc.com/video-clips/hzqmb9/the-daily-show-with-jon-stewart-large-hadron-collider The fallacy can be seen at about the 2:38 mark, and the proper rebuttal is at about the 5:30 mark. I'll warn you that the video is perhaps mildly vulgar and not safe for work (strong language). However, the video gives some hilarous examples of mistake made in the argument you supplied.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks and don't worry I'm still in tenth standard. Too early for office. $\endgroup$ – user167920 Oct 24 '17 at 17:01
  • $\begingroup$ Perhaps not safe for class, then. :) I'm also not sure if it will be regionally-locked; it wouldn't shock me if it's for US viewers only, so my apologies if it is. $\endgroup$ – Aaron Montgomery Oct 24 '17 at 17:11
  • 2
    $\begingroup$ @AaronMontgomery LOL. Thanks for the video! $\endgroup$ – Bram28 Oct 24 '17 at 17:33
  • 1
    $\begingroup$ A related fallacy: When the Fields medal was awarded to Perelman for proving the Poincaré conjecture, Steven Colbert complained about not getting one for himself: "They gave out four of these! So I had a one in four chance!" $\endgroup$ – Matthew Leingang Oct 25 '17 at 15:22
5
$\begingroup$

The probability of not getting a $2$ is far more likely than getting a $2$. Hence, you can't just say that since getting a $2$ is one of two possible outcomes, its probability is $\frac{1}{2}$, for that assumes each of the possible outcomes is equally likely.

If you want to think about this problem in terms of the two possible outcomes of getting a $2$ versus not getting a $2$, you need to take their respective probabilities into account:

$$P(2) = \frac{P(2)}{P(2) + P(not \ 2)} = \frac{\frac{1}{6}}{\frac{1}{6} + \frac{5}{6}} = \frac{\frac{1}{6}}{1} =\frac{1}{6}$$

Crisis averted!

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Assume an omnipotent creature approaches me and tells me that it had decided the following: it will either kill me, or all the other $7.5$ billion people in earth. And it will decide what course of action to take, by flipping a fair coin, i.e. a coin where its side has the same probability of landing, $1/2$. If it lands heads, it will take my head. Otherwise, it will take $7.5$ billion heads. Here the probability of me dying is $1/2$.

The "elementary events" here are indeed just two, and we have decided to give to each equal probability, even though their physical reality is rather non-comparable (surely, even for an omnipotent creature killing $7.5$ people will take more energy and time than killing just one person).

But assume now that the omnipotent creature has created a fair die with $7.5 \text{bn} +1$ sides, and it will roll this die in order to determine whether it will kill me or it will kill the other $7.5$ billion. If the die comes up $1$, it will kill me. If it does not come up $1$, it will kill all the others. Here the probability of me dying is $1/(7.5 \text{bn} +1)$, a bit lower than in the previous situation.

The physical actions that may take place in both situations are the same: either I will die, or everybody else will.

But the allocation of probability to these two events is quite different in the above two scenarios, and it should be obvious that any other "probability scenario" could be constructed by the omnipotent creature... what does that tells us about the concept of probability?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "Bit" lower you say? $\endgroup$ – user167920 Oct 25 '17 at 0:26
  • $\begingroup$ @user167920 Yeah, it's all a matter of perspective I guess :). Rather it is somewhat more than just "bit" lower. $\endgroup$ – Alecos Papadopoulos Oct 25 '17 at 0:30
-1
$\begingroup$

P(E) = number 22÷total outcomes

The problem in the construction you have created is that the selection part (choose a side out of 6, the set {1,2,3,4,5,6}) doesn't match the total outcomes part (choose from the set {2, not 2}).

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.