The odds are only 1/15. It's kind of a trick question, because you only have 15 chances to pick the right key., assuming you only get one key with each attempt. Factorial math assumes you have 15 chances on the first attempt, then 14 on the 2nd, 13th on the 3rd and so on.
@@Jokefish Take fifteen playing cards (including an ace of spades) and shuffle them. Then pick the top card one by one and put it aside until you get the ace of spades. What's the chance that you pick the ace of spades last ? You may think 'but what if you shuffle the remaining cards every time?' but that doesn't matter, as the remaining cards are already randomized. If you want the full calculation: The chance you miss the correct key first time is 14/15, the chance you miss it again the second time is 13/14, third time it's 12/13 et cetera. So the whole thing is 14/15 * 13/14 * 12/13 * ... * 2/3 * 1/2 and all the numbers from 2 to 14 cancel each other out so you're left with 1/15
Thank you, that makes it make more sense to me 😄 I can't imagine though that 1 in 15 people playing this would get the key as last, that doesn't sound right does it?
@@Jokefish what makes it feel wrong, like it should be something higher, is the length of time it takes to reach the room for that one chance to draw a key before starting over again
I heard what you said about the 4 doors and i think i might understand? Lets use 100 doors to make it more clear that theres an advantage in changing our guess and then see if we can math our way to a clearer understanding. Choose 1 door out of the 100. Youre probably wrong yeah? Its ok, its just a low 1 percent guess. Im now gonna remove 98 wrong doors. Your guess of course did not get removed, and also one other door remains as well. Would you like to change your guess or stay with it? Think about why your current guess still has a 1% chance of being right, whereas the other door has a 99% chance of being right (despite looking now like a mere 50/50). This of course is an extreme example taken much further than you had stated, but I think it begins to shed light. What if I only removed 95, leaving your initial guess (1% chance still) and 4 others? The other 4 come from a "group of 99 doors which collectively have a 99% chance of containing the right one" as opposed to your initial guess. We still know the 99% chance of it being "not our guess", which is now only divided among 4 doors. That's 24.75% chance of it being any one of the other 4. The benefit is much smaller than when I removed 98, but it's mathematically still more likely that the correct door is in the "not your initial guess" group. So in your thing, the initial guess is a 25% chance with 1 out of 4. Now the group of 3 you didnt guess collectively has a 75% chance of containing the correct door. That group retains 75% even if we remove 1, because your guess was made at a true 1 out of 4 chance and therefore doesn't magically become a better guess no matter what. The other group will always split that 75%, even if we removed 2 of them. It would then be 75% likely that the other 1 is correct (as its guarenteed the 2 that were removed are incorrect). If we only remove 1 from "the other group" as in your example, I believe they would each have a 37.5% chance of being the right one vs your 25% chance initial guess door. I am just spitballing this and I might be crazy afterall.
I think you could be very well correct.. I think my idea was a tad bit simpler; picking 1 in 4 is 25% chance, picking 1 in 3 is 33.3% chance, no further thoughts 😂
The odds are only 1/15. It's kind of a trick question, because you only have 15 chances to pick the right key., assuming you only get one key with each attempt. Factorial math assumes you have 15 chances on the first attempt, then 14 on the 2nd, 13th on the 3rd and so on.
It stills feels weird! I can't imagine that 1 in 15 Dungeon Clawler players would have ended up getting the key as the last, that just feels weird 😄
@@Jokefish Take fifteen playing cards (including an ace of spades) and shuffle them. Then pick the top card one by one and put it aside until you get the ace of spades. What's the chance that you pick the ace of spades last ?
You may think 'but what if you shuffle the remaining cards every time?' but that doesn't matter, as the remaining cards are already randomized.
If you want the full calculation:
The chance you miss the correct key first time is 14/15, the chance you miss it again the second time is 13/14, third time it's 12/13 et cetera.
So the whole thing is 14/15 * 13/14 * 12/13 * ... * 2/3 * 1/2 and all the numbers from 2 to 14 cancel each other out so you're left with 1/15
Thank you, that makes it make more sense to me 😄
I can't imagine though that 1 in 15 people playing this would get the key as last, that doesn't sound right does it?
@@Jokefish what makes it feel wrong, like it should be something higher, is the length of time it takes to reach the room for that one chance to draw a key before starting over again
Huh, perhaps yes, I had not thought about that at all.. that definitely makes it feel a lot more effort than it really is
I heard what you said about the 4 doors and i think i might understand? Lets use 100 doors to make it more clear that theres an advantage in changing our guess and then see if we can math our way to a clearer understanding. Choose 1 door out of the 100. Youre probably wrong yeah? Its ok, its just a low 1 percent guess. Im now gonna remove 98 wrong doors. Your guess of course did not get removed, and also one other door remains as well. Would you like to change your guess or stay with it? Think about why your current guess still has a 1% chance of being right, whereas the other door has a 99% chance of being right (despite looking now like a mere 50/50). This of course is an extreme example taken much further than you had stated, but I think it begins to shed light. What if I only removed 95, leaving your initial guess (1% chance still) and 4 others? The other 4 come from a "group of 99 doors which collectively have a 99% chance of containing the right one" as opposed to your initial guess. We still know the 99% chance of it being "not our guess", which is now only divided among 4 doors. That's 24.75% chance of it being any one of the other 4. The benefit is much smaller than when I removed 98, but it's mathematically still more likely that the correct door is in the "not your initial guess" group. So in your thing, the initial guess is a 25% chance with 1 out of 4. Now the group of 3 you didnt guess collectively has a 75% chance of containing the correct door. That group retains 75% even if we remove 1, because your guess was made at a true 1 out of 4 chance and therefore doesn't magically become a better guess no matter what. The other group will always split that 75%, even if we removed 2 of them. It would then be 75% likely that the other 1 is correct (as its guarenteed the 2 that were removed are incorrect). If we only remove 1 from "the other group" as in your example, I believe they would each have a 37.5% chance of being the right one vs your 25% chance initial guess door. I am just spitballing this and I might be crazy afterall.
Wtf just happened I think I just learned math somehow while watching you use a new character and i only have the demo 😂
I think you could be very well correct.. I think my idea was a tad bit simpler; picking 1 in 4 is 25% chance, picking 1 in 3 is 33.3% chance, no further thoughts 😂
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