What Does Taking The Definite Integral Of This Function Mean?

73 Views Asked by At

I want to try to understand the game below a bit better.


The game is this: there are three locked doors, and at the beginning of the game you get either 1 or 2 keys. You use the keys to unlock the dooor, open it, and collect object O from behind the door. There are 4 possible events, listed below.

  • Event A = get 1 key, object O is behind a door, I pick the door out of three with object O
  • Event B = get 2 keys, object O is behind a door, I pick 2 out of 3 doors, object O is behind one of the doors I pick
  • Event C = get 2 keys, there are two object O's behind 2/3 doors, I pick two doors and both have object O behind them
  • Event D = don't get object O, either from some combination of events of it not appearing behind the door(s), or I don't pick the chest containing it.

There is no event where object O is behind all 3 doors.

Event A, B, C, and D are mutually exclusive

  • K1 = P(getting 1 key)
  • K2 = P(getting 2 keys)
  • P(Object O "deciding" to be behind a door) = 0.08

set K2 = x; K1 = (1-K2); 1 ≥ K2 ≥ 0

  • $P(A) = (1-x)(0.08)(1/3) = (1-x)(2/75)$
  • $P(B) = (x)(0.08)(2/3) = (2/75)x$
  • $P(C) = (x)(0.08)(0.08)(2/3) = (8/1875)x$

$P(A U B U C) = P(winning)$

enter image description here enter image description here

Shortcut shown by /u/sasha

WolframAlpha short way (above) input/output

WolframAlpha long way: P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+P(ABC)

Comparing equations, they are the same.


I graphed P(winning), where the x-axis is the probabilty of getting two keys (1 ≧ x ≧ 0), and the y-axis is the probability of winning. P(winning) = {0.0266 for x = 0 ; 0.0308 for x = 1}.

$$\int_{0}^{1} P(winning) = 0.028624$$

Although the number of keys I can receive (0 or 1) is descrete, the probability of getting 2 keys is continous and random. Intution tells me I would get an average of 1.5 keys per game. If I let $X$ be 0.5, P(winning) = 0.028566.

From my understanding, taking the integral of a probability function yeilds probability density. Yet, it makes no sense that I would have a 2.86% chance of getting 2 keys 0 to 100% of the time (it should be 1). The area under the curve (from 0 to 1 nor from -infinity to infinity) converges to a number.

Exactly what, if anything, does that integral tell me?