Fair vs Loaded
tap a face to select event A · compare P(A) across both dice✓ naive works✗ naive failsFair Die11/621/631/641/651/661/6P(A) = 2/6Loaded Die13/1021/1031/1042/1051/1062/10P(A) = 5/10equal weightvariable weightselected (event A)
What do you think?
I flip a coin ten times and it lands heads every single time. What's the probability the next flip is heads?

The hidden assumption

Set the bias slider below and flip 100 times. Watch how observed frequency changes with bias.

Fair vs Loaded
tap a face to select event A · compare P(A) across both dice✓ naive works✗ naive failsFair Die11/621/631/641/651/661/6P(A) = 0/6Loaded Die13/1021/1031/1042/1051/1062/10P(A) = 0/10equal weightvariable weightselected (event A)
Biased Coin Simulator
00.51
?
What do you think?
With bias set to 0.7, approximately what fraction of 100 flips will be heads?
%
Enter a whole number from 0 to 100

The formula everyone knows

You've probably seen this before:

P(A)=number of favorable outcomestotal number of outcomesP(A) = \Large\frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}

There are two outcomes, one heads and one tails, so the probability is one half. Except it's often wrong.

Naive Definition of Probability

If all outcomes are equally likely, then: P(A)=ASP(A) = \Large\frac{|A|}{|S|} where A|A| is the number of outcomes in event AA and S|S| is the size of the sample space.

A fair die has 6 equally likely faces. What's P(rolling a 3)? (fraction, e.g. 2/7)
What's P(rolling an even number) on a fair die? (fraction, e.g. 2/7)

The equally likely assumption

Notice the phrase "equally likely" in the definition. It determines whether the formula gives you the right answer.

What do you think?
Which scenario violates the 'equally likely' assumption?

Where naive probability works

The formula works when outcomes are genuinely symmetric:

  • Fair dice rolled on a flat surface
  • Coins manufactured to exacting standards
  • Cards shuffled by a machine
  • Names drawn from a hat

In these cases, counting works. Count the ways to win, count the total, divide.

Where naive probability fails

What do you think?
A coin worn thin on one side lands heads 60% of the time. Using the naive formula P(heads) = 1/2 would give...

The formula fails when symmetry doesn't hold:

  • A coin worn thin on one side
  • Dice with microscopic bubbles
  • Weather patterns shaped by geography
  • Elections influenced by countless factors

Can you tell the difference? Sort these real-world scenarios into whether the naive formula works or fails:

Symmetry Check
1 / 10

A player making a free throw

The counting problem

Even when the naive definition applies, you still have to count correctly.

A standard deck has 52 cards. How many hearts are there? (whole number)
What's P(drawing a heart) from a shuffled deck? (fraction, e.g. 2/7)
How many face cards (J, Q, K) are in a deck? (whole number)

For a coin, counting is easy. For two dice, it gets subtle. For passwords or poker hands, counting is where most of the effort goes.

Before using the naive formula, ask yourself: are these outcomes actually symmetric? Is there any reason nature would favor one over another? If you're not sure, you need different tools.

Test your understanding

True or False: The naive formula works for any probability problem. (true or false)
A bag has 3 red and 5 blue marbles. What's P(red)? (fraction, e.g. 2/7)
I flip a bent coin 100 times and get 80 heads. Using the naive formula gives P(heads) = ? (fraction, e.g. 2/7)
What key phrase must be true for the naive formula to work? (two words)

What's next

The next lesson is about the rules of probability: what can we actually do with P(A)?