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Big balance beats big edge. The Gambler’s Ruin Problem.

April 13, 2017

The famed correspondence between two titans of 17th century French intellectual thought, Blaise Pascal (Pascal’s Wager) and Pierre Fermat (Fermat’s Last Theorem) was to mark the foundation of modern probability theory. But it was sparked off by a question posed to Pascal by legendary French gambler of the time, Antoine Gombaud, better known as the Chevalier de Mere.

The question related to a new dice game the Chevalier had invented. According to the rules of the 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 seemed impeccable. If the chance of a 6 on one roll of the die = 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, he reasoned, the chance of at least one double-6 in 24 throws is: 24/36 = 2/3. So this should be a profitable game for the Chevalier. When it didn’t turn out that way, he asked the great philosopher and mathematician, Blaise Pascal to look into it, as you do.

Pascal derived the correct probabilities as follows:

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

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

So, 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 = 1 – 0.5086 = 0.4914

So the Chevalier was betting at even money on a game which he lost (albeit marginally) more often than he won, which is why he was losing over time.

What if he 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.

I will examine just one of the conclusions of this correspondence today, and it relates to the infamous ‘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?

Example: 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 the ‘Gambler’s Ruin’ problem, and many gamblers over the years have been ruined because of their unawareness of it.

So how can we avoid falling victim to the problem of ‘Gambler’s  Ruin?’

‘Never bet more than you can afford to lose’.

‘When the Fun Stops, Stop!’

Now that’s a start.

 

Further Reading and Links

https://selectnetworks.net/

Letters between Fermat and Pascal on Probability:  https://www.york.ac.uk/depts/maths/histstat/pascal.pdf

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