Can you solve the monster duel riddle? - Alex Gendler

3,629,660 views ・ 2020-12-08

TED-Ed


Please double-click on the English subtitles below to play the video.

00:07
You’ve come a long way to compete in the great Diskymon league
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and prove yourself a Diskymon master.
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Now that you’ve made it to the finals, you’re up against some tough competition.
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As you enter the arena, the referee explains the rules.
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There are three Diskydisks you can use.
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Disk A will always summon a level 3 Burgersaur.
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Disk B summons a Churrozard that has a 56% chance of being level 2,
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a 22% chance of being level 4, and a 22% chance of being level 6.
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Disk C will summon a level 5 Wartortilla 49% of the time,
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and a level 1 Wartortilla 51% of the time.
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All Diskymon fully heal between battles, and the higher level Diskymon always wins,
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no matter what type it is.
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In round one, you’ll face a single opponent
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and get to choose your disk before she picks from the remaining two.
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Which one gives you the best chance of winning?
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Pause here to figure it out yourself
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Answer in 3
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Answer in 2
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Answer in 1
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Before you start calculating probabilities,
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take a look at the disks themselves.
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Disks B and C each have a more than 50% chance
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of summoning a level 2 or a level 1 Diskymon, respectively.
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This means that disk A’s guaranteed level 3 Burgersaur
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will always have better than even odds of winning.
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If you choose B or C, your opponent could pick A and gain an advantage over you.
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And C fares worst of all, being more than 50% likely to lose to any opponent.
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So you choose A, hoping for the best, and sure enough,
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your level 3 Burgersaur triumphs over the level 2 Churrozard.
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Now it’s time for round two, and while you’ve prepared for trouble,
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you didn’t anticipate they’d make it double.
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You get to choose any one of the three disks again,
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but this time, you’ll be in a battle royale against two opponents,
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each using one of the other disks.
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Whoever summons the highest level Diskymon wins.
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Should you stick with A, or switch?
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Pause now to figure it out yourself
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Answer in 3
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Answer in 2
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Answer in 1
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For many Diskymon trainers,
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it seems intuitive that if A is the best at beating B or C,
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it should also be the best at beating B and C.
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02:48
Strangely enough, that couldn’t be further from the truth.
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Let’s calculate the odds.
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For A to win, B has to summon a level 2 Diskymon,
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and C has to summon a level 1.
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Those are independent events, so their odds are 56% times 51%, or 29%.
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For disk B, a level 2 Churrozard would automatically lose to the Burgersaur.
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But you’d have two ways to win.
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The 22% chance of summoning a level 6 would give you an outright win,
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while a level 4 could still win if C summons a level 1.
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Adding up those mutually exclusive possibilities gives you odds of about 33%.
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Finally, C will win with a level 5 Wartortilla
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as long as B doesn’t summon its level 6, giving C a 38% chance overall.
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So while disk A’s middling consistency was an advantage in a single matchup,
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multiple fights increase the odds
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that one of the other disks will summon something better.
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And although C was the worst first-round option,
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its decent chance of summoning a strong level 5 gives it an advantage
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when facing two opponents simultaneously.
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This sort of counterintuitive result is why misleading statistics
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are a favored tool of unscrupulous politicians
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and nefarious Diskymon trainers alike.
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Fortunately, your Wartortilla comes out level 5 and makes short work of its foes.
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You’re about to celebrate when your rivals capture the referee
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and announce a surprise third round.
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You’ll have to repeat each of the previous matches in succession,
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with all the same rules except for one: you must keep the same disk throughout.
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Which should you choose to give yourself the best chance
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at becoming that which no one ever was?
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