De Méré's Paradox

Paradox
Which of these two is more probable?


 * Getting at least one six with four throws of a die;


 * Getting at least one double six with 24 throws of a pair of dice?

The self-styled Chevalier de Méré believed the two to be equiprobable, based on the following reasoning:


 * 1) A pair of sixes on a single roll of two dice is the same probability of rolling two sixes on two rolls of one die.
 * 2) The probability of rolling two sixes on two rolls is $$1/6$$ as likely as one six in one roll.
 * 3) To make up for this, a pair of dice should therefore be rolled six times for every one roll of a single die in order to get the same chance of a pair of sixes.
 * 4) Therefore, rolling a pair of dice six times as often as rolling one die should equal the probabilities.
 * 5) So rolling 2 dice 24 times should result in as many double sixes as getting 1 six throwing 4 dice.

However, betting on getting 2 sixes when rolling 24 times, he lost consistently.

Resolution
Throwing a die is an experiment with equiprobable outcomes.

There are six sides to a die, so there is $$1/6$$ probability for a six to turn up in one throw.

That is, by Elementary Properties of Probability Measure there is a $$1 - \frac 1 6 = \frac 5 6$$ probability for a six not to turn up.

When you throw a die $$4$$ times, by Probability of Independent Events Not Happening, there is $$\left({1 - \frac 1 6}\right)^4 = \left({\frac 5 6}\right)^4$$ of a six not turning up at all.

So by Probability of Occurrence of At Least One Independent Event, there is a probability of $$1 - \left({\frac 5 6}\right)^4$$ of getting at least one six with $$4$$ rolls of a die.

Doing the arithmetic gives you a probability of $$> 0.5$$, or in favour of a six appearing in 4 rolls.

Now when you throw a pair of dice, from the definition of independent events, there is a $$\left({\frac 1 6}\right)^2 = \frac 1 {36}$$ of a pair of sixes appearing.

That is, by Elementary Properties of Probability Measure, $$\frac {35} {36}$$ for a pair of sixes not appearing.

So by Probability of Occurrence of At Least One Independent Event there is a probability of $$1 - \left({\frac {35} {36}}\right)^{24}$$ of getting at least one pair of sixes with $$24$$ rolls of a pair of dice.

Doing the arithmetic gives you a probability of $$< 0.5$$, or in favour of a pair of sixes not appearing in 24 rolls.

Consequence
This is a veridical paradox.

Counter-intuitively, the odds are distributed differently from how they would be expected to be.