The flawed argument claims to be based on mathematical induction, and proceeds as follows.
Suppose that we have a set of five horses. We wish to prove that they are all the same colour. Suppose that we had a proof that all sets of four horses were the same colour. If that were true, we could prove that all five horses are the same colour by removing a horse to leave a group of four horses. Do this in two ways, and we have two different groups of four horses. By our supposed existing proof, since these are groups of four, all horses in them must be the same color. For example, the first, second, third and fourth horses constitute a group of four, and thus must all be the same colour; and the second, third, fourth and fifth horses also constitute a group of four and thus must also all be the same colour. For this to occur, all five horses in the group of five must be the same colour.
But how are we to get a proof that all sets of four horses are the same colour? We apply the same logic again. By the same process, a group of four horses could be broken down into groups of three, and then a group of three horses could be broken down into groups of two, and so on. Eventually we will reach a group size of one, and it is obvious that all horses in a group of one horse must be the same colour.
By the same logic we can also increase the group size. A group of five horses can be increased to a group of six, and so on upwards, so that all finite sized groups of horses must be the same colour.
The argument above makes the implicit assumption that the two subsets of horses to which the induction assumption is applied have a common element. This is not true when n = 1, that is, when the original set only contains 2 horses.
Indeed, let the two horses be horse A and horse B. When horse A is removed, it is true that the remaining horses in the set are the same colour (only horse B remains). If horse B is removed instead, this leaves a different set containing only horse A, which may or may not be the same colour as horse B.
The problem in the argument is the assumption that because each of these two sets contains only one colour of horses, the original set also contained only one colour of horses. Because there are no common elements (horses) in the two sets, it is unknown whether the two horses share the same colour. The proof forms a falsidical paradox; it seems to show something manifestly false by valid reasoning, but in fact the reasoning is flawed. The horse paradox exposes the pitfalls arising from failure to consider special cases for which a general statement may be false.