The Dice Probability Puzzle of the 17^th Century

One of the most-known probability problems, which partly laid the ground of modern probability theory, is dated back to the year 1654 and is connected with the name of the Flemish Renaissance gambler – Chevalier de Mere. He systematically tried his luck with two dice games.

1^st Dice Game of Chevalier de Mere

The first game was played with a single dice. Chevalier de Mere accepted bets that he would roll at least one six in four consecutive attempts. He knew that the probability to roll six was 1/6 in a single roll. That is why he supposed that his chance to roll six in four tries was (1/6) × 4 = 2/3.

2^nd Dice Game of Chavelier de Mere

In the second game two dice were rolled. Chevalier de Mere was successful if he managed to roll 2 sixes   in 24 consequent rolls. He knew again that the probability to roll two sixes in a single roll was 1/36 (only one possibility out of 6 times 6 events). So he presumed that his chance to win was (1/36) × 24 = 2/3 and thus equal to the first dice game.

Wrong Presumption

Despite that he suffered financial losses with the second dice game. Is the chance to win in both games really the same? Why was his presumption wrong?

Chevalier de Mere, desperately seeking to reveal the reason of his failure, turned sight to his friend – the excellent French mathematician and physicist Blaise Pascal (1623–1662). After a careful analysis Pascal truly spotted the right solution of the problem, which needed to be seen on from another angle.

Correct Solution of Chevalier de Mere's Problem

The essence of the correct solution consists in indirect calculation of favorable events based on the calculation of losing events towards all possibilities. The house edge of all casino games is evaluated on the similar principle (or universally on the basis of the expected value). Let us closely follow Pascal's ideas.

There can be these possibilities in the first game, when one dice is rolled in four attempts:
6 × 6 × 6 × 6 = 64 = 1 296
The total volume of the losing possibilities when six is not rolled (1, 2, 3, 4, 5) is:
5 × 5 × 5 × 5 = 54 = 625
Now we can figure out the number of winning events:
1296 − 625 = 671.

We can see that 671 > 625, that means the number of winning events is higher than the number of losing events. Chevalier de Mere really had a small advantage.

Why did we (or Blaise Pascal) proceed this way? Since it is clear that the direct calculation of winning events would lead us nowhere, because 1 × 1 × 1 × 1 = 14 = 1(!). Let us analyze the second game.

With two dice and 24 rolls the number of total possibilities is:
36 × 36 × ... × 36 = 3624 = 22 452 257 707 354 600 000 000 000 000 000 000 000
Possibilities to lose:
35 × 35 × ... × 35 = 3524 = 11 419 131 242 070 600 000 000 000 000 000 000 000
Possibilities to win then:
3624 − 3524 = 11 033 126 465 284 000 000 000 000 000 000 000 000

Now we can see that the presumption of Chevalier de Mere, as regards the equality of the winning chances in both dice games, was wrong, since the number of losing possibilities exceeds the number of winning possibilities in the second dice game.

Mathematical Calculation of the Probability to Win

We can calculate the probability to win in the first game as a supplement towards the number of losing possibilities:

P(A) = 1 − (5/6)4 = 0.517746914.

Purely from the mathematical point of view the probability ranges from 0 (impossible event) to 1 (certain event). However we can find also percentage stipulation in practice. After multiplication by 100 and rounding the chance to win is 51.77%. It means that Chevalier de Mere would probably win approximately 52 out of 100 games (rolls).

Analogously we can calculate the probability of a win in the second dice game:

P(A) = 1 − (35/36)24 = 0.491403876.

The winning chances do not even reach one half(!). Any player would probably lose 51 games and win 49 games only.

The Summary of Chevalier de Mere's Probability Problem

Both calculation methods of winning probability can convince us of three findings:

• The winning probability is not the same for both dice games; the second game is disadvantageous;
• The calculations confirm the correctness of Pascal’s solution;
• The whole probability problem illustrates and underlines the importance of knowledge of true mathematical odds in various gambling games and their bets.

Chevalier de Mere made a mistake in his presumption that the winning probability was the same for both dice games, which was far from it. Nevertheless if he had stuck to the first game only he would have really had the edge – approximately 1.77%.

Ask yourself whether it is high or low. For instance the house edge in the French Roulette is 2.7% when betting on numbers and 1.35% in case of even-money bets. You can have a look at the calculation based on the expected value.

You might be interested:

→ Three Door Puzzle – The Monty Hall Problem;
→ Two Beagles Probability Riddle;
→ Overview of Articles on Probability.