DAVE THE DIVER How to Always Win the Shark Game

Select a number equal to the remainder of (cavity number minus 1) divided by 4. If this is 0, try to get to 4, and pray. When you pick a number and arrive at 4n+remainder where n is any number (including 0), you’ve already won! Now just pick 4 minus whatever number your opponent picks, and watch the magic happen.

The shark tooth game follows the same principles as the “100 Game” founded by Reverend Charles Lutwidge Dodgson, or Lewis Carroll – yeah, the Alice in Wonderland guy. Instead of selecting numbers from 1 to 10 in the “100 Game”, you can select numbers from 1 to 3. I’ll give you the short explanation, and explain the theory afterwards. Essentially, you can always win this game by following this strategy.

Locate the cavity. This is obvious. There are 20 teeth on the bottom row, so you can consider your target by it’s number from left to right – let’s say your tooth is 18, for sake of experiment.

Divide the target number minus 1 by 4 and take the remainder (or modulus, for the math nerds), as well as keeping the quotient handy. (18-1)/4 is 4 with a remainder of 1. This remainder is important, but so is the quotient, so keep 1 and 4 in mind.

Whatever your opponent picks, you can always pick the option to make the sum go to 4, or in other words: 4 minus your opponent’s number is your number. If he picks 1, you pick 3; if he picks 2, you pick 2; if he picks 3, you pick 1. In this way, you control the flow of the game, but you can also always be certain that you’re going to get the next number in the series.

However, unlike the “100 Game”, your goal is to make your opponent reach the target number, not yourself, which is why we subtracted 1 from the target number. We want to make sure that we reach the target number minus one, so that when the opponent has to go after us, he must hit the target number and therefore lose.

First, pick the remainder as an option: in this example, 1. Then, the opponent will pick a number – as discussed, subtract that number from 4 to get your number (pick 3 in response to 1, pick 2 in response to 2, pick 1 in response to 3. For our example of the target being 18, this will go (where bold is the range of an opponent’s possible choice):

1, 2-4, 5, 6-8, 9, 10-12, 13, 14-16, 17. As you pass off your choice at 17, whatever the opponent picks will reach the target number, and you win.

The exception to this, obviously, is if the remainder is 0 – you can’t pick 0, so what do you do? Let’s take 17 as our target number and think it through, much like our previous example: (17-1)/4 is quotient 4, but remainder 0.

The advantage to us is that the opponent does not do optimal play the same way we do – he will virtually always mess up. The new objective is for us, the player, to pick a number that brings us to a multiple of 4 as early as possible, so that we can make sure that we pick 16 so the opponent has to pick 17.

If the remainder is 0, get to a multiple of 4. If you start with 1, you’re farther away from that golden number of 16. If the opponent picks 1 or 2, you’re aces: pick the number that gets you to 4, bada-bing bada-boom, you’re on track for the easy win. If he picks 3 in response to your 1, not all is lost – you have several rounds to get lucky. Simply continue picking 1, wait until he does not pick 3.

However, you might ask, what if you are unlucky enough that you’re at a point where it’s at 12 and you’re to pick? Well, picking 3 and bringing it to 15 because it’s more likely he will pick 2 or 3 is still gonna be the same realistic odds as him picking 1 and 2 in response to you picking 1 and letting you get into a winning position – if he picks 3 in response to the 1, you’re screwed, obviously, but if he picks 1 in response to your 3, you’re also screwed. It’s the same odds, so it’s not like it’s a bad choice, but it overcomplicates the strategy. Likewise, picking 2 puts him in a case where if he picks 2, you lose; if he picks 1, you can counter with 1 and bring it to 16 for the definite win; and if he picks 3, you win as he gets to 17 and loses. Same odds: 2 to 1 to win.

In reality, if you can’t move into a winning position (4n+remainder, if you’re a nerd): you’re still hoping for him to pick a number that doesn’t gets it to a multiple of 4, so that you can get it to a multiple of 4, but you get more rounds. Picking a larger number will usually make the game faster, if you really care about the extra few seconds, as it gives you the likelihood that he’ll lose all on his own if you’re unlucky enough to have not gotten a winning position – but if you’re getting blocked out of picking number 4, it’s because he’s picking 1, so you’re never skipping rounds unless you’re doing suboptimal play. The odds are still always the same.