The potato paradox is a mathematical calculation that has a counter-intuitive result. The Universal Book of Mathematics states the problem as such:

White potatoes are actually around 79% water;[1] agar is 99% water.[2]

Fred brings home 100 kg of potatoes, which (being purely mathematical potatoes) consist of 99% water (being purely mathematical water). He then leaves them outside overnight so that they consist of 98% water. What is their new weight?

The surprising answer is 50 kg.[3]

The misunderstanding that causes the paradox stems from the overlap between the amount of water in the potatoes, and the proportion of the potatoes that is water.

The initial 100kg of potatoes is described as 99% water. That means that the 100kg of potatoes are made up of 99kg of water and 1kg of dry mass. The percentage water and the mass of water are the same number, which can make it appear that removing 1kg of water is all that is necessary.

If the 1kg of water could be replaced with something else, so that the potatoes remained at a weight of 100kg, then replacing 1kg of water would indeed make them 98% water; but the problem does not allow this. If the water is not replaced with something else, then removing 1kg of water also reduces the total mass of the potatoes by 1kg, to 99kg. That means that the proportion of water is then 98kg out of 99kg, which is 98.98%. Removing 1kg mass of water reduces the proportion of water by only 0.02%.

Because removing water lowers the total mass of the potatoes as well as the mass of water, the proportion of water is reduced much more slowly than the mass of water.

The difficulty in mentally calculating 98/99 can also have an effect on the problem. 98/99 is 0.98989898.. (recurring), so a reader who estimates the result may wrongly believe that it is closer to 98%.

### Method 1

If the potatoes are 99% water, the dry mass is 1%. This means that the 100kg of potatoes contains 1kg of dry mass. This mass will not change, as only the water evaporates.

In order to make the potatoes be 98% water, the dry mass must become 2% of the total weight - double what it was before. The amount of dry mass - 1kg - cannot be changed, so this can only be achieved by reducing the total mass of the potatoes. Since the proportion that is dry mass must be doubled, the total proportion of the potatoes must be halved, giving the answer 50kg.

### Method 2

A visualization where blue boxes represent kg of water and the orange boxes represent kg of solid potato matter. Left, prior to dehydration: 1 kg matter, 99 kg water (99% water). Middle: 1 kg matter, 49 kg water (98% water).

In the beginning (left figure), there is 1 part non-water and 99 parts water. This is 99% water, or a non-water to water ratio of 1:99. To double the ratio of non-water to water to 1:49, while keeping the one part of non-water, the amount of water must be reduced to 49 parts (middle figure). This is equivalent to 2 parts non-water to 98 parts water (98% water) (right figure).

In 100 kg of potatoes, 99% water (by weight) means that there is 99 kg of water, and 1 kg of non-water. This is a 1:99 ratio.

If the percentage decreases to 98%, then the non-water part must now account for 2% of the weight: a ratio of 2:98, or 1:49. Since the non-water part still weighs 1 kg, the water must weigh 49 kg to produce a total of 50 kg.

## Explanations using algebra

### Method 1

After the evaporating of the water, the remaining total quantity, ${\displaystyle x}$ , contains 1 kg pure potatoes and (98/100)x water. The equation becomes:

{\displaystyle {\begin{aligned}1+{\frac {98}{100}}x&=x\\\Longrightarrow 1&={\frac {1}{50}}x\end{aligned}}}

resulting in ${\displaystyle x}$  = 50 kg.

### Method 2

The weight of water in the fresh potatoes is ${\displaystyle 0.99\cdot 100}$ .

If ${\displaystyle x}$  is the weight of water lost from the potatoes when they dehydrate then ${\displaystyle 0.98(100-x)}$  is the weight of water in the dehydrated potatoes. Therefore:

${\displaystyle 0.99\cdot 100-0.98(100-x)=x}$

Expanding brackets and simplifying

{\displaystyle {\begin{aligned}99-(98-0.98x)&=x\\99-98+0.98x&=x\\1+0.98x&=x\end{aligned}}}

Subtracting the smaller ${\displaystyle x}$  term from each side

{\displaystyle {\begin{aligned}1+0.98x-0.98x&=x-0.98x\\1&=0.02x\end{aligned}}}

Which gives the lost water as:

${\displaystyle 50=x}$

And the dehydrated weight of the potatoes as:

${\displaystyle 100-x=100-50=50}$

### Method 3

After the potatoes are dehydrated, the potatoes are 98% water.

This implies that the proportion of non-water weight of the potatoes is ${\displaystyle (1-.98)}$ .

If x is the weight of the potatoes after dehydration, then:

{\displaystyle {\begin{aligned}(1-.98)x&=1\\.02x&=1\\x&={\frac {1}{.02}}\\x&=50\end{aligned}}}

## Implication

The answer is the same as long as the concentration of the non-water part is doubled. For example, if the potatoes were originally 99.999% water, reducing the percentage to 99.998% still requires halving the weight.