Can you solve the monster duel riddle? - Alex Gendler
You’ve come a long way to compete in the great Diskymon league and prove yourself a Diskymon master. Now that you’ve made it to the finals, you’re up against some tough competition. As you enter the arena, the referee explains the rules. There are three Diskydisks you can use. Disk A will always summon a level 3 Burgersaur. Disk B summons a Churrozard that has a 56% chance of being level 2, a 22% chance of being level 4, and a 22% chance of being level 6. Disk C will summon a level 5 Wartortilla 49% of the time, and a level 1 Wartortilla 51% of the time. All Diskymon fully heal between battles, and the higher level Diskymon always wins, no matter what type it is.
In round one, you’ll face a single opponent and get to choose your disk before she picks from the remaining two. Which one gives you the best chance of winning? Pause here to figure it out yourself. Answer in 3. Answer in 2. Answer in 1. Before you start calculating probabilities, take a look at the disks themselves. Disks B and C each have a more than 50% chance of summoning a level 2 or a level 1 Diskymon, respectively. This means that disk A’s guaranteed level 3 Burgersaur will always have better than even odds of winning.
If you choose B or C, your opponent could pick A and gain an advantage over you. And C fares worst of all, being more than 50% likely to lose to any opponent. So you choose A, hoping for the best, and sure enough, your level 3 Burgersaur triumphs over the level 2 Churrozard. Now it’s time for round two, and while you’ve prepared for trouble, you didn’t anticipate they’d make it double. You get to choose any one of the three disks again, but this time, you’ll be in a battle royale against two opponents, each using one of the other disks. Whoever summons the highest level Diskymon wins. Should you stick with A, or switch? Pause now to figure it out yourself. Answer in 3. Answer in 2. Answer in 1.
For many Diskymon trainers, it seems intuitive that if A is the best at beating B or C, it should also be the best at beating B and C. Strangely enough, that couldn’t be further from the truth. Let’s calculate the odds. For A to win, B has to summon a level 2 Diskymon, and C has to summon a level 1. Those are independent events, so their odds are 56% times 51%, or 29%. For disk B, a level 2 Churrozard would automatically lose to the Burgersaur. But you’d have two ways to win. The 22% chance of summoning a level 6 would give you an outright win, while a level 4 could still win if C summons a level 1. Adding up those mutually exclusive possibilities gives you odds of about 33%.
Finally, C will win with a level 5 Wartortilla as long as B doesn’t summon its level 6, giving C a 38% chance overall. So while disk A’s middling consistency was an advantage in a single matchup, multiple fights increase the odds that one of the other disks will summon something better. And although C was the worst first-round option, its decent chance of summoning a strong level 5 gives it an advantage when facing two opponents simultaneously. This sort of counterintuitive result is why misleading statistics are a favored tool of unscrupulous politicians and nefarious Diskymon trainers alike.
Fortunately, your Wartortilla comes out level 5 and makes short work of its foes. You’re about to celebrate when your rivals capture the referee and announce a surprise third round. You’ll have to repeat each of the previous matches in succession, with all the same rules except for one: you must keep the same disk throughout. Which should you choose to give yourself the best chance at becoming that which no one ever was?