Today’s problem is a mathematical treatment of an idiom. The straw that broke the camel’s back is a saying about how a camel near capacity buckles upon the trivial weight of an additional straw. The proverb refers to situations where a seemingly small final act precipitates a large reaction or failure. The moral is that the final straw was really no different from any others, but it just happened to have extra significance due to the accumulated weight.

The puzzle investigates this claim: is it really true that the last straw is no different from any others?

Let’s make the problem precise. One by one straws are placed on a mathematical camel’s back. The straws have random weight, being drawn independently from the uniform distribution 0 to 1. The camel’s back breaks when the total weight of the straws exceeds 1.

Is the final straw really the same as any other straw? What’s the expected weight of the straw that breaks the camel’s back?

Give it a try before reading the solution below. If you’re stuck, I offer an intuitive explanation before getting into the mathematical calculation.

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"All will be well if you use your mind for your decisions, and mind only your decisions." Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon. .

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Intuitive answer to The Straw That Broke the Camel’s Back

It turns out the idiom is wrong: the straw that breaks the camel’s back will tend to be heavier than average.

The intuition is that a typical straw has an average weight of 0.50. The last straw, however, is something that broke the camel’s back. Since heavier straws are more likely to break the camel’s back, this would imply the last straw will be heavier than average.

Let’s do a concrete example. Suppose you keep drawing randomly between 1 cent and 5 cent coins until you exceed 1 dollar. The first coin is equally likely to be 1 or 5 cents–for an average of 3 cents. What about the last coin? Note that if you’re at 99 cents, you will exceed a dollar with either coin. But if you’re at 98, 97, 96, or 95 cents, you will only exceed a dollar if you draw the 5 cent coin. Therefore, the distribution of the last coin shifts towards the coin of larger denomination.

Or think about it in terms of the casino game blackjack. You receive cards valued from 1 to 11 and you bust if your total exceeds 21. Considering only hands where you bust, what’s the average value of the last draw you received? Note that the typical card drawn is between 1 and 11 (several face cards are 10), with an average between 6 and 7. But you won’t bust on certain cards. For instance, the only way you can bust with a 1 is if you were at 21 and you hit. That’s not going to happen, so you’ll never bust on a 1. Similarly, the only way you can bust with a 2 is if you were at 20 and took a hit. This is almost never going to happen, so you don’t bust on a 2. It’s also less likely you will bust on a 3 or a 4. Since low valued cards are less likely to cause you to bust, this means that bust cards will have a higher average value.

(There is a more refined discussion of the discrete version here.)

Now that we have the intuition of why the final straw is heavier, let us solve the problem with the uniform distribution.

Numerical Analysis

A problem like this is easy to simulate in a spreadsheet. Use the function RAND() to generate a random draw from 0 to 1. In the next column have a running total, and whenever the sum exceeds 1, zero the running total and place that draw into the third column.

From 10,000 draws, there are approximately 3,000 draws that make the sum exceed 1. Here is a box and whisker plot comparing the weight of a typical straw to the weight of a straw that broke the camel’s back (made the sum exceed 1).

The typical straw is exactly what we’d expect: it’s evenly spread out from 0 to 1. The last straw is definitely skewed towards larger weights, with a median of 0.68.

The average straw has a weight of 0.50 while the last straw is about 0.64. The exactly value, which we will derive below, is equal to 2 – e/2–and suddenly the constant e appears like magic!

(I thank this video for pointers on making a box and whisker plot. Basically you need to create a table of the quartile values and make it a stacked chart with cleverly hacked error bars.)

Analytic solution

There is a closely related puzzle which is: what’s the expected number of draws for the uniform sum to exceed 1? This is a delightful problem too with an unexpected answer–you should check it out.

I thought I could translate that problem into a solution for the current problem. However, tried as I would, I got stuck. I decided to ask for help and posted the problem on Math StackExchange. To my surprise, I received two replies with answers in less than a day! I thank users Jason and Did for their help. I will present a slightly modified version of Did’s solution.

First some notation. Let X 1 , X 2 , …, X n be n independent draws from the uniform distribution. Denote the sum as S n = X 1 + X 2 + … + X n .

What we wish to do is find the expected value of X k , where k is the first value where the sum exceeds 1. In other words, we would have S k – 1 < 1 and S k ≥ 1.

We will take a step back and derive another useful function. For x between 0 and 1, define g(x) as follows

The function g(x) is the sum of the probabilities that the uniform sum does not exceed x. We can explicitly calculate this from the probability distribution of the uniform sum, which is known as the Irwin-Hall distribution.

The function g(x) is the sum of the probabilities that the uniform sum does not exceed x. In other words, this is the probability that X 1 + X 2 + … + X n ≤ x for 0 < x ≤ 1 and each X i is between 0 and 1. Geometrically this shape is a simplex whose volume equals x/n!

(Alternately, we can explicitly calculate the probability from the distribution function of the uniform sum, which is known as the Irwin-Hall distribution. The probability distribution function for the uniform sum, where 0< y < 1, is given by the nth draw is yn – 1/(n – 1)! We need to take the integral of this between 0 and x to find the cumulative probability that S n ≤ x. The result is x/n! and the answer agrees with the derivation of the volume of the simplex method.)

Now we can derive g(x) since we know the nth term of the summation is x/n!

What a happy coincidence! We started out with a probability sum and that lead us into an infinite series which happens to be related to the constant e.

We’re halfway there. We now need to compute the expectation of the final draw. We need to compute the probability density function f(x) which will be the first draw where the uniform sum exceeds 1. So our expression is the following.

Now we do a neat trick. We can write an equivalent expression for the inequality. First we will subtract S n from each term, then we will multiply through by -1 (which switches the direction of the signs), and then finally we add 1 to each term. The steps are the following.

S n < 1 ≤ S n + x 0 < 1 - S n ≤ x 0 > -1 + S n ≥ –x 1 > S n ≥ 1 – x

Now we can re-write f(x) in terms of g(x), which we already have solved for.

Now that we have an expression for f(x), we can compute the expectation of the draw that makes the sum exceed 1.

So the final draw, the one that makes the sum exceed 1, is about 28 percent larger than the average draw of 0.50.

The moral

While the math got a bit involved, the intuition nudged us to the qualitative lesson. The straw that breaks the camel’s back is a wrong idiom. In our mathematical example, it was true that the final straw was identical and independently drawn from 0 to 1–so in one sense the last straw really is the same as any other. However, the fact that it broke the camel’s back means something–it means the final straw is heavier than average.

I think this provides a lesson for interpersonal conflicts. When a relationship turns sour, the offending party often replies that they were acting exactly the same way as before, and they blame the other person for over-reacting. In fact, this might very well be true: if one’s behavior is drawn from a uniform distribution, then the final act was a realization of exactly the same stochastic process. However, it is not true that the person is over-reacting: the final straw tends to be worse behavior than average.

In fact, the situation for personal relationships is likely even more skewed. Imagine you allow each friend to offend you up to a limiting point. Each time they betray you or harm you adds to the weight of the relationship. Now a small offense is quickly forgiven and forgotten. However it is the larger acts that stay in your memory. So small acts are unlikely to break up a relationship–the final straw will be something more egregious than average.

In conclusion, the straw that breaks the camel’s back is a wrong idiom. It is not simply the same inconsequential act that overloads a system; the final straw is worse than average, precisely because it is the final straw.