A movie, a tragedy and its odds.

jmb

New member
Joined
May 3, 2018
Messages
2
Inspired by "some" recent movie I saw, imagine the following scenario:
I sit with my friends Adam (A) and Ben (B) in a restaurant, when some accident happens that kills (0) everyone with the probability of 1/2. I am aware of this probability.
Luckily I survive (1), but hurt my eyes during accident so I cannot see who of my friends made it as well. Chances are 1/4 for each outcome A1B1, A1B0, A0B1, A0B0. [Knowledge state (KS) 1]
After a few minutes, I stumble over a person on the floor, which I can conclude to be definitely one of my friends and being alive, but I do not know who exactly. The person on the floor is either Adam or Ben. This rules out outcome A0B0.
If I now think about the chances that both of my friends survived this accident. I conclude it to be 1/3 as I have to be in one of the possible 3 remaining states of the world: A1B1, A1B0, A0B1 [KS 2]

Upon touching the person on the floor I am able to identify the friend as either Ben, who has a long beard, or Adam, who always shaves properly. The moment I am able to identify the person I know I am in either of two states:
1. The guy on the floor is Adam, so we have to be in state: A1B1 or A1B0 [KS 31]. The chances that we all made it is 1/2.
2. The guy on the floor is Ben, so we have to be in state: A1B1 or A0B1 [KS 32]. The chances that we all made it is 1/2.
No matter who it is on the floor, the moment I can identify him as either Adam or Ben, the chances that we all made it changes from 1/3 to 1/2, even though I know with 100% certainty in NS 2 that I have to be in either in KS 31 or 32.

I am not so sure why this somewhat hurts my head. In KS2 we simply ask: given that we know that at least one of two coins show heads, what is the probability of both showing heads. In KS 31/32 I ask after the first coin showed heads, whats the probability that the second coin shows heads as well.

It is however just weird to think about this scenario where I find a guy, knowing it has to be either Adam or Ben, and in the moment I know who it is exactly, I can be somewhat less concerned about our little group of friends.
At face value, that makes little sense to me and hence I expect something to be wrong, but I can't see what it is and I was unable to google this question. :confused:

If anyone could enlighten me I'd be most grateful.

Thanks a lot in advance.
Jan
 
Last edited:
Hi Jan,

I'm sorry to see that no one got to your question until now. The first part of your problem, about the probabilities changing with additional information, can be understood in terms of conditional probabilities. It's not that you're computing the same probability and it's magically changing. You're computing three different probabilities.

To illustrate this, we first need to know some concepts from probability theory. Suppose you have two events X and Y. We'll use the overbar to denote the negation of an outcome (\(\displaystyle \bar{X}\) means "NOT X") so that we have the notation

\(\displaystyle P(X) \): Probability that event X occurs
\(\displaystyle P(\bar{X}) \): Probability that event X does NOT occur

likewise for Y

In general the probability that both X AND Y will occur is given by the expression

\(\displaystyle \displaystyle P(X~\mathrm{and}~Y) = P(X|Y)P(Y) \)


where P(X|Y) is read as "probability of X given Y" and is called the conditional probability of X given Y. In other words, it expresses what the probability of X occurring is, under the given condition that Y has occurred. The multiplication on the righthand side makes perfect sense, because suppose that P(X|Y) = 0.5 and P(Y) = 0.1. What this is saying is that in 50% of the cases where Y occurs, X also occurs. And, Y occurs 10% of the time. Therefore, in 50% of that 10% of cases, both X and Y will occur. So that's 50% times 10% or 0.5*0.1 = 0.05.

We can take a rearrangement of the above equation to be a definition of conditional probability:

\(\displaystyle \displaystyle \frac{P(X~\mathrm{and}~Y)}{P(Y)} = P(X|Y) \)


Now, if X and Y are independent events, then whether or not Y occurs has absolutely no bearing on whether X occurs. The two outcomes don't influence each other. Therefore, in that scenario, \(\displaystyle P(X|Y) = P(X)\) and hence the two probabilities simply multiply:

\(\displaystyle \displaystyle P(X~\mathrm{and}~Y) = P(X)P(Y)~~\mathrm{(independent~events)} \)

We can also ask what is the probability of either X OR Y occurring. To solve this imagine a Venn diagram with two overlapping circles, one representing event X occurring and the other representing event Y occurring. The areas of the circles are proportional to the probabilities of of the respective events. The probability of X OR Y is given by the area of the union of the two circles (the total area enclosed by both together). To get this area, you could simply sum the areas of the two circles (sum the probabilities) but then you will have double counted the portion in the middle where the two circles overlap (which corresponds to X AND Y occurring) so you have to subtract that out. Therefore:

\(\displaystyle \displaystyle P(X~\mathrm{or}~Y) = P(X)+P(Y)-P(X~\mathrm{and}~Y) \)

Notice that if the two events are mutually exclusive (i.e. it is not possible for both of them to occur at the same time) then \(\displaystyle P(X~\mathrm{and}~Y)=0\) and the two probabilities simply add together:

\(\displaystyle \displaystyle P(X~\mathrm{or}~Y) = P(X)+P(Y)~~\mathrm{(mutually~exclusive~events)} \)

Let me know if there is anything you haven't understood here. In my next post, we'll see how to apply these concepts to your problem.
 
So, in your problem, we have the following events:

Probability that Alan survives = \(\displaystyle P(A) = \frac{1}{2}\)
Probability that Alan dies = \(\displaystyle P(\bar{A}) = \frac{1}{2}\)
Probability that Ben survives = \(\displaystyle P(B) = \frac{1}{2}\)
Probability that Ben dies = \(\displaystyle P(\bar{B}) = \frac{1}{2}\)

Case 1:
The first question you asked was, in the absense of any information, what is the probability that both Alan AND Ben survive? By assumption, their survival is independent of each other (it's simply a coin flip whether each one lives). So A and B are independent events and:

\(\displaystyle \displaystyle P(A~\mathrm{and}~B) = P(A)P(B) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4} \)

Case 2:
The second question you asked was, given that we know one of the friends has survived, what is the probability that both of them have survived? So the given information is that A OR B has occurred (either one or the other or both have survived). So the question you are asking above amounts to the conditional probability of (A and B) given (A or B):

\(\displaystyle \displaystyle P((A~\mathrm{and}~B) | (A~\mathrm{or}~B)) = \frac{P((A~\mathrm{or}~B)~\mathrm{and}~(A~\mathrm{and}~B))}{P(A~\mathrm{or}~B)} \)

where the righthand side has come simply from the definition of conditional probability given in my first post, with X = (A and B) and Y = (A or B). At first glance, the numerator of this may seem tricky to evaluate. But, we can re-write it using the defintion of conditional probability (again) as

\(\displaystyle \displaystyle P((A~\mathrm{and}~B) | (A~\mathrm{or}~B)) = \frac{P((A~\mathrm{or}~B)|(A~\mathrm{and}~B))P(A\,\mathrm{and}\,B)}{P(A~\mathrm{or}~B)} \)

Now, we can think this through: the events (A and B) and (A or B) are not independent. The reason is, if I know that both A and B have occurred, then I know with 100% certainty that (A or B) has occurred. Both have survived, so it is definitely true that one or the other or both have survived. Therefore \(\displaystyle P((A~\mathrm{or}~B)|(A~\mathrm{and}~B)) = 1\) and the equation above becomes:

\(\displaystyle \displaystyle \begin{aligned}P((A~\mathrm{and}~B) | (A~\mathrm{or}~B)) &= \frac{1 \cdot P(A\,\mathrm{and}\,B)}{P(A~\mathrm{or}~B)} \\
&= \frac{1 \cdot P(A)P(B)}{P(A) + P(B) - P(A\,\mathrm{and}\,B)}\\
&= \frac{1 \cdot \frac{1}{2}\cdot\frac{1}{2}}{\frac{1}{2} + \frac{1}{2} - \frac{1}{4}}\\
&= \frac{1/4}{3/4} = \frac{1}{3} \end{aligned}\)

Case 3:
Fortunately, case 3 is a bit easier. In this case, you know that Ben, specifically, has survived. This is the given condition. You would like to know the conditional probability that both have survived, given that Ben has survived. So the conditional probability you are trying to compute is:

\(\displaystyle \displaystyle P((A~\mathrm{and}~B) | B) \)

Now we can think this through again. The events (A and B), and B are not independent. If we know B has survived, there is a greater chance that both have survived. In fact, if B has survived, then that's half the battle, and all that is now required for both to survive is for A to also have survived. The probability that both survive given B has survived is then simply the probability that A survives:

\(\displaystyle \displaystyle P((A~\mathrm{and}~B) | B) = P(A) = \frac{1}{2} \)

If you're not happy with me just asserting this, you can derive it using the exact same reasoning as we did in Case 2. Try it!

It is however just weird to think about this scenario where I find a guy, knowing it has to be either Adam or Ben, and in the moment I know who it is exactly, I can be somewhat less concerned about our little group of friends.
At face value, that makes little sense to me and hence I expect something to be wrong, but I can't see what it is and I was unable to google this question. :confused:

So, what you seem to be saying is that it's surprising to you that given more data/information, you can be a little more certain about an outcome. But, this is something that had darn well better be true! If having more information didn't decrease our uncertainty about a hypothesis, then scientists would not be able to inform their hypotheses by doing observations and measurements!
 
Last edited:
Thank you so much for the introduction into proper notation and the elaborate explanation.
I think it got me a bit further. If the following is true:

\(\displaystyle \displaystyle \begin{aligned}P((A~\mathrm{and}~B) | (A~\mathrm{or}~B)) &= \frac{1 \cdot P(A\,\mathrm{and}\,B)}{P(A~\mathrm{or}~B)} \\
&= \frac{1 \cdot P(A)P(B)}{P(A) + P(B) - P(A\,\mathrm{and}\,B)}\\
&= \frac{1 \cdot \frac{1}{2}\cdot\frac{1}{2}}{\frac{1}{2} + \frac{1}{2} - \frac{1}{4}}\\
&= \frac{1/4}{3/4} = \frac{1}{3} \end{aligned}\)

given that \(\displaystyle \displaystyle P(A) = P(B) = \frac{1}{2} \),
is the following true as well?
\(\displaystyle \displaystyle P(A | (A~\mathrm{or}~B)) \) = \(\displaystyle \displaystyle P(B | (A~\mathrm{or}~B)) =\frac{2}{3} \) as we know that one made it for sure


\(\displaystyle \displaystyle \begin{aligned} P(A | (A~\mathrm{or}~B)) &= \frac{P((A~\mathrm{or}~B)|A) P(A)}{P(A~\mathrm{or}~B)} \\
&= \frac{P((A~\mathrm{or}~B)|A) P(A)}{P(A) + P(B) - P(A\,\mathrm{and}\,B)} \\
{with~} P((A~\mathrm{or}~B)|~A) = 1 \\
&= \frac{1 \cdot \frac{1}{2} }{\frac{3}{4} } \\
&= \frac{1/2}{3/4} = \frac{2}{3} \end{aligned}\)

But wouldn't that somehow make these independent events somehow dependent on each other?
 
Thank you so much for the introduction into proper notation and the elaborate explanation.
I think it got me a bit further.

Hi jmb! Apologies for the delay in responding. I'm glad that my posts were helpful. :)

is the following true as well?
\(\displaystyle \displaystyle P(A | (A~\mathrm{or}~B)) \) = \(\displaystyle \displaystyle P(B | (A~\mathrm{or}~B)) =\frac{2}{3} \) as we know that one made it for sure


\(\displaystyle \displaystyle \begin{aligned} P(A | (A~\mathrm{or}~B)) &= \frac{P((A~\mathrm{or}~B)|A) P(A)}{P(A~\mathrm{or}~B)} \\
&= \frac{P((A~\mathrm{or}~B)|A) P(A)}{P(A) + P(B) - P(A\,\mathrm{and}\,B)} \\
{with~} P((A~\mathrm{or}~B)|~A) = 1 \\
&= \frac{1 \cdot \frac{1}{2} }{\frac{3}{4} } \\
&= \frac{1/2}{3/4} = \frac{2}{3} \end{aligned}\)

Yes, it is true. You can check it by counting outcomes. We know that A or B is true. The three possibilities that satisfy that are
\(\displaystyle \displaystyle AB\),

\(\displaystyle \displaystyle A\bar{B}\),

\(\displaystyle \displaystyle\bar{A}B\)

where I've used AB to mean "A and B". So, in two out of three of these possibilities, A survives.

But wouldn't that somehow make these independent events somehow dependent on each other?

Sorry, but I don't follow. Wouldn't that somehow make which two independent events dependent on each other?

One thing that is important to know, that I didn't state explicitly, but that we have been using above, is the theorem that allows you flip around which event is the condition and which event is the one dependent on the condition. You derive this theorem as follows:

\(\displaystyle \displaystyle \begin{aligned} P(X~\mathrm{and}~Y) &= P(Y~\mathrm{and}~X)\\
\therefore P(X|Y)P(Y) &= P(Y|X)P(X)\\
P(X|Y) &= \frac{P(Y|X)P(X)}{P(Y)}
\end{aligned} \)

We call this result Bayes' Theorem, and it is crucially important and widely used in probability theory and statistics.
 
Last edited:
Top