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?