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The Chevalier’s Dice Problem – in a nutshell.

April 4, 2019

This is a true story about New York gambling-house operator, The Butch, who made his fortune booking dice games. In 1952 he was famously challenged by a bigtime gambler known as The Brain to a simple wager. The bet was an even-money proposition that the Butch could throw a double-six in 21 rolls of two dice. We can assume symmetry – the dice were not loaded or biased in any way. All faces were equally likely to come up. So the probability of any number appearing on a given roll of either one of the dice is 1/6.

On the face of it, the edge seems to be with Butch. After all, there are 36 possible combinations that could come up when throwing two dice, from 1-1, 1-2, 1-3, to 6-4, 6-5, 6-6. Intuition might suggest, therefore, that 18 throws should give you a 50-50 chance of throwing any one of these combinations, including a double-six. In 21 throws, the chance of a double-six should, therefore, be more than 50-50. On this basis, the Butch accepted the even money bet at $1,000 a roll. After twelve hours of rolling, the Brain was $49,000 up, at which point the Butch called it a day, sensing that something was wrong with his strategy.

The Brain had in fact profited from a classic probability puzzle known as the Chevalier’s Dice problem, which can be traced to the 17th French gambler and bon vivant, Antoine Gombaud, better known as the Chevalier de Méré. The Chevalier would agree even money odds that in four rolls of a single die he would get at least one six. His logic seemed impeccable. The Chevalier reasoned that since the chance that a 6 will come up in any one roll of the die is 1 in 6, then the chance of getting a 6 in four rolls is 4/6, or 2/3, which is a good bet at even money. If the probability was a half, he would break even at even money. For example, in 300 games, at 1 French franc a game, he would stake 300 francs and expect to win 150 times, returning him 150 francs for each win with his stake returned on each occasion (total of 300 francs). With a probability of 2/3, he would expect to win 200 times, yielding a good profit.

In fact, it is straightforward to show that this reasoning is faulty, for if it were correct, then we would calculate the chance of a 6 in five rolls of the die as 5/6, and therefore the chance of a 6 in six rolls of the die would be 6/6 = 100%, and in 7 rolls, 7/6!!! Something is therefore clearly wrong here.

Still, even though his reasoning was faulty, he continued to make a profit by playing the game at even money. To see why, we need to calculate the true probability of getting a 6 in four rolls of the die. The key idea here is that the number that comes up on each roll is independent of any other rolls, i.e. dice have no memory. Since each event is independent, we can (according to the laws of probability) multiply the probabilities.

So the probability of a 6 followed by a 6, followed by a 6, followed by a 6, is: 1/6 x 1/6 x 1/6 x 1/6 = 1/1296.

So what is the chance of getting at least one six in four rolls of the die?

Since the probability of getting a 6 in any one roll of the die = 1/6, the probability of NOT getting a 6 in any one roll of the die = 5/6.

So the chance of NOT getting a 6 in four rolls of the die is:

5/6 x 5/6 x 5/6 x 5/6 = 625/1296

So the chance of getting at least one 6 is 1 minus this, i.e. 1 – (625/1296) = 671/1296 = 0.5177, which > 0.5.

So, the odds are still in favour of the Chevalier, since he is agreeing even money odds on an event with a probability of 51.77%.

This was all very well as long as it lasted, but eventually the Chevalier decided to branch out and invent a new, slightly modified game. In the new game, he asked for even money odds that a pair of dice, when rolled 24 times, will come up with a double-6 at least once. His reasoning was the same as before, and quite similar to the reasoning employed by the Butch. If the chance of a 6 on one roll of the die is 1/6, then the chance of a double-6 when two dice are thrown = 1/6 x 1/6 (as they are independent events) = 1/36.

So, reasoned the Chevalier, the chance of at least one double-6 in 24 throws is: 24/36 = 2/3.

So this is very profitable game for the Chevalier. Or is it? No it isn’t, and this time Monsieur Gombaud paid for his faulty reasoning. He started losing. In desperation, he consulted the mathematician and philosopher, Blaise Pascal. Pascal derived the correct probabilities as follows:

The probability of a double-6 in one throw of a pair of dice = 1/6 x 1/6 = 1/36.

So the probability of NO double-6 in one throw of a pair of dice = 35/36.

So, the probability of no double-6 in 24 throws of a pair of dice = 35/36 x 35/36 …  24 times = 35/36 to the power of 24, i.e. (35/36)24  = 0.5086.

So probability of at least one double-6 is 1 minus this, i.e. 1 – 0.5086 = 0.4914, i.e. less than 0.5. Under the terms of the new game, the Chevalier was betting at even money on a game which he lost more often than he won. It was an error that the Butch was to repeat almost 300 years later!

What if the Chevalier had changed the game to give himself 25 throws?

Now, the probability of throwing at least one double-6 in 25 throws of a pair of dice is:

1 – (35/36)25 = 0.5055.

These odds, at even money, are in favour of the Chevalier, but this probability is still lower than the probability of obtaining one ‘6’ in four throws of a single die.

In the single-die game, the Chevalier has a house edge of 51.77% – 48.23% = 3.54%.

In the ‘pair of dice’ game (24 throws), the Chevalier’s edge =

49.14% – 50.81% = -1.72%

In the ‘pair of dice’ game (25 throws), the Chevalier’s edge =

50.55% – 49.45% = 1.1%

A better game for the Chevalier would have been to offer even money that he could get at least one run of ten heads in a row in 1024 tosses of a coin. The derivation of this probability is similar in method to the dice problem.

First, we need to determine the probability of 10 heads in 10 tosses of a fair coin.

The odds are: ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½

Odds = (1/2)10 = 1/1024, i.e. 1023/1.

Based on this, what is the probability of at least one run of 10 heads in 1024 tosses of the coin? Is it 0.5? No, because although you can expect ONE run of 10 heads on average, you could obtain zero, 2, 3, 4, etc.

So what is the probability of NO RUN of 10 heads in 1024 tosses of the coin?

This is: (1-1/1024)1024

The probability of NO RUNS OF TEN HEADS = (1023/1024)1024 = 37%

So probability of AT LEAST one run of 10 heads = 63%.

Now assume you have tossed the coin already 234 times out of 1024, without a run of 10 heads, what is your chance now of getting 10 heads?

Probability of NO RUNS OF TEN HEADS in remaining 790 tosses = (1023/1024)790 = 46%

So probability of at least one success = 54%.

The Chevalier could have played either of these games and expected to come out ahead. But the game would have taken a long time. He preferred the shorter game, which produced the longer loss.

Until he was put right by Monsieur Pascal.

Most importantly, though, the Chevalier’s question led to a correspondence, most of which has survived, which led to the foundations of modern probability theory.

Out of this correspondence emerged quite a few jewels, one of which has become known as the ‘Gambler’s Ruin’ problem.

This is an idea set in the form of a problem by Pascal for Fermat, subsequently published by Christiaan Huygens (‘On reasoning in games of chance’, 1657) and formally solved by Jacobus Bernoulli (‘Ars Conjectandi’, 1713).

One way of stating the problem is as follows. If you play any gambling game long enough, will you eventually go bankrupt, even if the odds are in your favour, if your opponent has unlimited funds?

For example say that you and your opponent toss a coin, where the loser pays the winner £1. The game continues until either you or your opponent has all the money. Suppose you have £10 to start and your opponent has £20. What are the probabilities that a) you and b) your opponent, will end up with all the money?

The answer is that the player who starts with more money has more chance of ending up with all of it. The formula is:

P1 = n1 / (n1 + n2)

P2 = n2   / (n1 + n2)

Where n1 is the amount of money that player 1 starts with, and n2 is the amount of money that player 2 starts with, and P1 and P2 are the probabilities that player 1 or player 2, your opponent, wins.

In this case, you start with £10 of the £30 total, and so have a 10/ (10+20) = 10/30 = 1/3 chance of winning the £30; your opponent has a 2/3 chance of winning the £30. But even if you do win this game, and you play the game again and again, against different opponents, or the same one who has borrowed more money, eventually you will lose your entire bankroll. This is true even if the odds are in your favour. Eventually you will meet a long-enough bad streak to bankrupt you. In other words, infinite capital will overcome any finite odds against it. This is one version of the ‘Gambler’s Ruin’ problem, and many gamblers over the years have been ruined because of their unawareness of it.

Exercise

  1. What is the probability of throwing at least one double-six in 26 throws of a pair of dice?
  2. You and your opponent toss a coin, where the loser pays the winner £10. The game continues until either you or your opponent has all the money. Suppose you have £100 to start and your opponent has £400. What are the probabilities that a) you and b) your opponent, will end up with all the money?

References and Links

DeMere’s Paradox. ProofWiki. https://proofwiki.org/wiki/De_M%C3%A9r%C3%A9%27s_Paradox

One gambling problem that launched modern probability theory. Introductory Statistics. https://introductorystats.wordpress.com/2010/11/12/one-gambling-problem-that-launched-modern-probability-theory/

deMere’s Problem. WolframMathWorld. http://mathworld.wolfram.com/deMeresProblem.html

Gambler’s Ruin. WolframMathWorld. http://mathworld.wolfram.com/GamblersRuin.html

 

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