LogFAQs > #916860263

It's time for a decision game! 10 lockers are beside the door... Within one is an oxygen mask. Which one will we open? Ah, here's our Monty Hall. But we have to just choose one...let's try 7. Sure enough, we're going to get the Monty Hall. If we chose correctly, 8 of the 9 wrong ones will randomly open. If we chose wrong, the other eight wrong ones will open. 4 is the only other one that stays closed. The exit door won't unlock for 20 minutes. Junpei shows his lack of understanding of the Monty Hall problem, which Akane referenced by name upon being told of the rules. Akane tries to explain, but passes out again, just like with the poison.

Now, having spent the better part of a post explaining why switching is inherently a better option, and explaining that the more doors there are, the chances that switching is a better option increase...I stick with 7. One, because I know these games like to screw with the odds, and I suspect no matter which option I take, I'll have to reload to get the correct one. Two, because the door that stayed closed was Door 4, and since this game was made in Japan, that signifies Death. The question is, since this series of games works on Schrodinger's Cat principles...what are the actual odds that I chose correctly? Are they 10%? 50%? 0%?

...It's not in 7, of course. But what will happen if I choose 4 next time?

It cuts to Sigma and Diana opening the X-Door, with all seven other names listed as deceased. But Sigma tries to convince Diana that they're all alive, so this can't follow the version where Phi is incinerated. There was one other that opened past midnight...but that was the rec room, which was another C-team fragment! So this follows something I don't know yet... I naturally jump right back to Monty Hall. To see if switching is always correct, I take a different initial choice to both 4 and 7. Namely, q--er, 9. Sure enough, 10 is the other one that stays closed. So at the very least, the identity of the correct door doesn't stay constant. Now, for the question. Does the game randomly pick the correct one each time, and the odds that switching is the correct option are the expected 90%? Or does the game only choose where the mask is after you've made your second choice? I chose red, and had to go back and choose blue to get the game started properly. Another person in the topic initially chose blue, and had to go back to choose red. No matter which choice you make, the same thing happens.

Except here, there's a third option. What if the choice of correct one is made after your first choice? Then the odds would actually be 50%. I find that unlikely, though.

Sure enough, when I switch to 10, it's empty. 9 was correct here. I believe it's likely that this time, no matter which choice I make, it'll be correct. Just like the dice, where it's a 1 in 216 chance, but the game always rigs it to happen on the third try. Or at least, I assume it's always the third try. I did the math to figure out how many tries you'd have to do to "expect" to get it, assuming that it would in fact be lower than 128 (basing it on the "birthday problem", wherein it only takes a mere 25 people to be in a randomly selected group for the odds that two of them share a birthday to exceed 50%), but honestly that's foolish, because even an infinite number of tries doesn't guarantee it. Care to guess how many times it actually is?

It's 150. Exactly 150, which is a bizarrely round number, though only by the arbitrary fact that our numbering system favors 2's and 5's above all other prime factors (though 150 still has a 3 in it as well). To 4 digits,
215/216 to the 149th is .5009, and to the 150th it's .4985. So it would take 150 trials for "at least one success" to be more likely than "all failures".
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Also known as Cyberchao X.