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## Proof of the Week

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

As throwing a die is an experiment with a finite number of equiprobable outcomes, we can use the Classical Probability Model.

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.

$\blacksquare$

## Consequence

This is a veridical paradox.

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

## Source of Name

This entry was named for Chevalier de Méré.

He posed this problem to his friend mathematician Blaise Pascal, who solved it.