But if you put your wand away, how are you going to smack that spoony bard upside the head?
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Those damn bards; their skills and stuff are a scattered mess yet somehow, they get OP at damn near everything except eXtreme wizardy damage.
Never underestimate powerful charisma and sideburns, I guess.
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The solution given is dependent on EVERYONE knowing not only which wand their opponents have, but the exact odds of every possible wand working at a given moment.
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Appearing weak so that your opponents will target each other instead of you is a solid strategy no matter which wand you choose and has nothing to do with math. (It’s a bad riddle, in that it lays out the math and expects a mathematical solution, but then throws in a key component that makes the math not matter) BUT, while it is clearly stated that you can CHOOSE to miss, the assumption is that you CAN miss.
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This COMPLETELY throws the math out of whack with the official solution, in that nobody is considering the odds of an opponent actually hitting their target. The 100% effective wand is only 50% effective in the hands of a wizard who misses half the time, while the 70% wand is 63% effective in the hands of a wizard who hits his target 90% of the time.
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The riddle also assumes that YOU will be the only one to employ the missed shot tactic. It is entirely possible that we end up with a complete standoff, where the last person to act almost invariably wins.
Also-
A) How long does the vine spell last?
B) Is this a “last man standing” or “first blood” finish?
C) Is a wizard eliminated when they are hit, regardless of the spell effect itself?
This is relevant because the vines are the only spell which could theoretically incapacitate someone without permanently removing them from play. This significantly changes the number of strategies available.
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Let’s ignore all the wand stuff (I heard this question with guns a long time ago) and just doing the odds of hitting math. The miss on purpose strategy being right might depend on the order of the shots and who has the best aim. I might be missing some easy shortcut, but the math seems really complicated. Obviously if you all had the same chance of hitting, you’d definitely all want to pass. But when the odds of hitting are different, I’m not sure that’s the case.
I don’t know about the values in this video, I think I originally heard it as playing 1 has a 30% to kill, player 2 has a 70%, player 3 has a 90% chance of hitting. So I made a simulation because the markov chains were getting too complex for napkins. I compared two strategies - fire at the person with the highest chance to hit vs. pass unless there is only one other person.
If all three go “highest” then the odds of winning at 29%, 58%, 13% for players 1, 2, 3.
If player one switches their strategy to “pass unless 2” then the odds of winning change to 36%, 45%, 19% - a nice improvement for player 1.
But if player two follows suit then the odds of winning change to 32%, 0%, 68%.
That’s not rounding, it’s literally zero. Because player three will always kill player two, no matter how many rounds it takes. Basically you hand it off to the third player and say, “Do you prefer a 100% chance of drawing or a 68% chance of winning”. In a game to the death that decision will be made by whether a draw means you die or live (but if a draw means you live then obviously no one should shoot anyone else, you don’t even need math to figure that out). But usually in any game I wouldn’t think someone would take a draw over a 68% play to win.
I tried with some different numbers and orders and it came out as I expected: It’s right for the person with the lowest chance of success to pass until there are two players, but for the middle and high player, they should do the “obvious” thing and shoot at the other player with the highest chance of winning.
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