# Why is not the answer to all probability questions 1/2.

Ok, I know this is wrong, but I want someone to tell me why.

Let's take a normal heads tails example of a fair coin. The probability of getting head = 1/2. And I write this is because, either it will be heads, or not. Hence two cases that makes it 1/2.

Now, I know I can't do this to all the cases. For example, the probability of getting a 2 when I roll a dice. I know the answer is 1/6 but why can't I do, either the outcome will be 2 or not. And in that case, my probability is 1/2.

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"The outcome will not be 2" has more than one option: 1,3,4,5,6 out of six , and the outcome "2" has only one. This, of course, affects the probability of each outcome, otherwise probability will be very, very boring. – DonAntonio Mar 1 '14 at 9:29
Just counting cases is combinatorics. Probability theory means assigning weights to the different cases and arguing about these in a consistent manner. – Christian Blatter Mar 1 '14 at 10:07
Either there is a dinosaur outside my window, or there isn't. So, 50/50? – Ittay Weiss Mar 1 '14 at 11:24
@IttayWeiss OMG I'll bring my camera! – David Richerby Mar 1 '14 at 12:10
OK, now that you've got some answers let's test whether you've got the concept down. You flip two coins at the same time. There are three possibilities: no heads, exactly one head, or two heads. There are three cases. Are the probabilities of each equal to one third? – Eric Lippert Mar 1 '14 at 15:26

In the case of a fair coin, the two outcomes are equally probable. $P(\text{heads}) = P(\text{tails}) = 1/2$.

In the case of a (fair) dice, the two outcomes $"2"$ and $"\text{not }2"$ are not equally probable. There are five possible ways of not getting 2, all equally probable, and only one possibility to get 2. In total there are six possible outcomes. Simply put: $$\frac{5}{6} = P(\text{not }2) > P(2) = \frac{1}{6}$$.

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Thanks, that makes sense. :) – Kraken Mar 1 '14 at 9:38
"x ways of getting A" is a dangerous and misleading intuition - it makes sense only if all the ways are equally likely. For example, if the coin is slightly biased, then still "either it will be heads, or not", but P(heads) may be 0.6 and P(tails)=0.4. It is important to note that "either it will be heads or tails" means that P(heads)+P(tails)=1, and nothing more. – Peteris Mar 1 '14 at 17:32
Thank you @Peteris - I will add some details to what I meant. – naslundx Mar 2 '14 at 15:47

To talk of probability, you need both the sample space and a probability assigned to each element of the sample space.*

So in the case of a fair coin, the probability is given by the data that the sample space is $\{H, T\}$ and that $\Pr(H) = \frac12$ (and therefore $\Pr(T) = \frac12$). It's only because of this that you can say that the probability of heads is $\frac12$; you can't conclude this just from "either it will be heads or not".

For example, you could have a biased coin, where the sample space is still $\{H, T\}$, but $\Pr(H) = 0.9$ (and $\Pr(T) = 0.1$). In that case, you cannot use the reasoning that "either it will be heads or not" to conclude that heads and tails have equal probability.

In the case of a fair coin, the full reasoning you use is "either it will be heads, or not, and both are equally likely", where the final part (in emphasis) comes from the data given to you, namely that $\Pr(H) = \Pr(T) = \frac12$.

Even in the case of dice, you can still say "either it will be $2$ or not", but you're missing the latter part of the reasoning, that both events are equally likely. (Indeed, for fair dice, the probability of rolling $2$ is only $\frac16$, whereas the probability of rolling something other than $2$ is $\frac56$. But you could have loaded dice too.)

Often, in your textbooks, you may be given just the sample space without the probabilities, leaving implicit the fact that each individual element of the sample space has the same probability. But ideally that should be specified too.

[*]: Note: The discussion above is of a discrete probability space. In general, we have a set ("sigma algebra") of events, and probabilities on them, satisfying certain axioms. But let's not worry about that now.

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You actually really can't have a biased coin; everybody uses that as an example but physically it's impossible to make. At least if you throw it, I think is possible if you spin it. – Kai Sikorski Mar 2 '14 at 5:54
@KaiSikorski You could put butter on one side of the coin; then it would always land butter side down! – Jason C Mar 2 '14 at 5:57
@JasonC You can also make the two sides the same. That way there's still two sides, but it's impossible to distinguish which one the coin landed on! (another important combinatorics concept) – Thomas Mar 2 '14 at 10:02
Real coins aren't fully symmetrical, so the probablility of heads and tails can't be exactly equal (whatever the extremely complex physical process ends up in heads or tails). Same for dice, the side with 6 dots is slightly lighter than the others, so it should end on top a bit more than 1/6 of the time. – vonbrand Mar 2 '14 at 16:27
@KaiSikorski Why can't we actually have a biased coin?? I am afraid I don't get it.. – N Unnikrishnan Aug 10 '15 at 11:39

We cannot calculate a probability without using other probabilities in the calculation.

When we say that a coin has $P(H)=1/2$, or a die has $P(2)=1/6$ that is not something we learn using probability theory, it is an assumption about physics.

We assume that the die can only land with a face up, and we assume that all faces are equal in geometry and weight distribution, and therefore that the sum of probability for the faces is 1, and that the probability is the same for all 6 faces. Pure physics.

If I roll two dice I believe that each of them exert the same probabilities as a single die, and that they do this independently of one another. This is not something I can learn using probability theory either, it is an assumption about physics that the dice do not coordinate their rolls.

Once we have all these assumptions about die physics we can do all the cool die equity calculations using probability theory.

Unlike the rest of the world, dice are made for producing independent stochastic variables, the assumptions about die rolls have a solid foundation in physics that is pretty much undisputed. But when we try to figure the probabilities of something with real world relevance simple physics will not suffice, rather we might have to rely on psychology, sociology, economics or palaeontology. Putting probabilities on different outcomes is often guesswork, and different variables have all sorts of odd correlations, meaning that basic probability theory (which always assume independent variables) won't work.

When the coin argument seems to be universally applicable it is actually because it is a false argument, it just happens to produce the right result for fair coins.

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Probability is the study of "making the most out of insufficient information". If you knew all relevant information about a situation, then the probability of an outcome is always 100% or 0%.

The reason your answer isn't 1/2 for rolling the dice is because you aren't using all information available. In fact, if you gain more information (weights of the dice, staring position, initial direction of the throw) then the probability goes from 1/6 towards 100% or towards 0%.

There is no such thing as "the probability of an event". It is "the probability of an even given what we know". Two different people can get 2 different probabilities just because they have differing degrees of knowledge.

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Two different people can get 2 different estimates of probability. No more than one will be right, and experiment (or Monte-Carlo simulation) will tell which. – Ruslan Mar 2 '14 at 15:51

Imagine you call your friend on the phone and ask him to give you an answer that is two or not two. Your friend decides to make the answer based on the throw of a dice and he will answer 2 or not two depending on the dice. You do not know that your friend is rolling a dice. After the first answer You can stick to the assumption that this is a 50/50 chance. If you ask your friend for a few more answers of two or not two you may still believe that the chances are 50/50, but if you ask your friend a great number of times and carefully write down the results You will see that you only get a 2 in approximately $\frac{1}{6}$ of the cases. Now You might ask your friend if he is using a dice since the outcome looks like the distribution you would expect from a dice. This is why $\frac{1}{6}$ is used as the chance for a 2 from one dice, $\frac{4}{52}$ from two's from a deck of cards and $\frac{1}{2}$ for the flip side for a coin and from that the calculation of combined odds have been developed - given rules that will be used when adding dices or having more coins ( given an even distribution of outcome ).

You might want to rethink your opinion of "always 50/50 " before you start betting on the outcome - it will be pretty close on red/black in the casino but far from correct if you play a single number on the roulette.

Henrik

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It's important to note that sometimes the wrong steps will give the right answer, by chance.

Your statement of "The probability of getting heads = 1/2. And I write this is because, either it will be heads, or not." is faulty, even though it produces the correct answer. A more correct statement would be "either it will be heads, or it will be tails, with an equal chance for each. Therefore, 1/2."

Compare to a die with 'heads' written on one face, and 'tails' written on the other five. In this case, still, either it will be heads or it will not -- and either it will be tails or it will not -- but because they are not equal probabilities, you have a 1/6 chance of heads and 5/6 chance of tails. It should be obvious that if five times as many sides say 'tails' then you have a five-times greater chance of getting tails.

Which is a lengthy way of saying that "Either it's A or it's not A" doesn't properly weigh the two possibilities. Just because there are two possibilities doesn't mean each has an equal chance of being chosen.

Otherwise, every day I'd have a 50% chance of becoming the supreme omnipotent ruler of all.

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Update: Still not the supreme omnipotent ruler of all. – Trevel Aug 17 '15 at 13:04
Further update: STILL not the supreme omnipotent ruler of all. – Trevel Apr 11 at 12:36

## protected by Zev ChonolesMar 2 '14 at 15:41

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