Simon has tried to find out whether each of the players have equally fair chance to win a NIM game. In many games that’s exactly the case, but not in the game of NIM, as Simon proves in the course of these two videos. If you just want to see the math behind it, feel free to only watch the second video:

The game of NIM is about flipping two coins and playing with 12 more coins. Depending on what a player throws (two heads, heads-tails, tails-heads or two heads) she can claim a number of coins from the 12 (corresponding to the binary value of the throw, that can vary from 1 to 4 coins). The person who claims the last coin wins all the 12 of them. Simon has proved that the person starting the game (according to the rules of the game, that’s the person who brought in the money) has better chances of winning!

Simon learned about the game of NIM from Matt Parker’s Stand Up Math channel, but worked out the proof himself.