After three successful draws, must the next compensate for the streak? Probability specifies a model for one trial; frequency describes an observed sequence. A short streak creates no debt for chance to repay. Use inequalities and counting.

Start with outcomes

An experiment is a procedure with an outcome unknown beforehand; an elementary outcome is one possible result, and an event is a set of results. Our model selects one of four equally accessible cards:

Ω = {A,B,C,D}, E = {A,B}, P(E) = (2)/(4) = (1)/(2).

“Favorable / all” applies to a finite space of equally likely outcomes. Without that condition counting alone does not determine probability: unequal card chances must be added as weights. OpenStax

The impossible event ∅ has probability 0, the certain event Ω has 1. The model rests on nonnegativity, total probability 1 and addition for pairwise disjoint events. General axioms also require addition for countable families, whose members can be numbered.

P(Ω) = 1, P(∅) = 0, 0 ≤ P(E) ≤ 1.

Union E∪F means at least one, intersection E∩F means both, and complement means not E, with probability 1−P(E). Subtract the shared part to avoid counting it twice:

P(E∪F) = P(E) + P(F) − P(E∩F).

F = {B,C}: P(E∪F) = (1)/(2) + (1)/(2) − (1)/(4) = (3)/(4).

Axioms and consequences, MIT.

What a sequence shows

If E occurs mₙ times in n trials, its relative frequency is:

ν_(n) = (m_(n))/(n).

The diagram uses an authored sequence A,B,A,C,A,D,C,B,D,C,A,D with replacement. Its first three outcomes belong to E, and six of all twelve do. This is a constructed illustration, not a measurement log.

Equally likely outcomes

Ω={A,B,C,D}; E={A,B}

ABCD01/4outcomeP

Cumulative frequency of E

A B A C A D C B D C A D; -- p=1/2

03691201/21nνₙ
Top: categories A–D, each chance 1/4; A,B form E. Bottom: authored sequence, not measurements. n counts trials, νₙ is dimensionless. Lines only connect discrete points; frequency at n=0 is undefined.

Independence here means previous outcomes do not change the chance of a new trial. With independent trials of the same probability p, the law of large numbers says: the probability that frequency differs from p in absolute value by more than any prescribed positive threshold tends to zero as n grows. It does not guarantee monotone improvement or balance in short sequences. In our independent model, replacing the card and reshuffling uniformly preserves the next chance 1/2 even after three successes. Law of large numbers, MIT.

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Areas and Monte Carlo

For a point chosen uniformly on a segment, a subsegment has probability equal to its length fraction; in a uniform planar region use the area fraction. Uniformity means equal probabilities for equal lengths or areas, not positive equal probabilities for individual points. A point can be a possible outcome with probability 0, so “P=0 implies impossible” is not universal. Continuous uniform model.

In the square 0≤x,y≤1, triangle x+y≤1 occupies half the area: the diagonal divides the square into equal parts. Monte Carlo estimates that fraction as k/N by generating N independent uniform points and counting k inside the triangle. The estimate fluctuates and does not replace the geometric proof of 1/2.

Compare frequency with a model, but check the assumptions first: equal chances, replacement and independence. Otherwise exact arithmetic may describe a different experiment.