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What skills can you take 10 or 20 on?
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<blockquote data-quote="Elder-Basilisk" data-source="post: 3068192" data-attributes="member: 3146"><p>Sure, that sounds good, but if you actually want to do it that way, you should take the watcher's spot check into account.</p><p></p><p></p><p></p><p></p><p>Now, maybe you protest: but the spotter can take 20 and keep looking. OK, well, let's assume that this is possible. In that case, the usefulness would really come down to what the spotter can accomplish.</p><p></p><p>Modifying the above dialogue:</p><p></p><p></p><p></p><p>So, if the spotter takes 20 for every one roll the hider makes, it's really taking 400 (since the number of possible combinations is 400). So, it should take 40 minutes.</p><p></p><p>From a rules perspective, however, there is a problem with using the take 20 mechanic to signify this process: There is not necessarily a difference in the ability of the spotter to notice a hider who rolls well.</p><p></p><p>Taking the above example a step further:</p><p></p><p></p><p></p><p>Of course, the hider is not as well hidden as possible, he's just so well hidden that the spotter can't possibly spot him. If your reverse the bonuses, you find a similar discrepancy:</p><p></p><p></p><p></p><p>Now, that uses the take 20 mechanic which doesn't offer an opportunity to distiguish between "a roll of 20 can find the hider" and "could be hidden better" or between "a roll of 20 cannot find the hider" and "cannot possibly be hidden better." </p><p></p><p>But, you protest, surely the characters would be able to see the difference between a 10 and a 16 on the hide check. In the case of the +15 hide vs. +11 spot, a roll of 16 on the hide check means the spotter should only notice the hider 1 time in 20 and a roll of 10 would mean the spotter notices the hider 1 roll in 4. The problem is that this is not exactly true. The reason it is not true is that once the spotter succeeds, he cannot continue to try and compare the number of failures to successes. Rather, he has to say "I see you" and let the other guy decide whether or not to try and hide again. Now, it is not likely to actually take 20 rolls in order to roll a 20--it's possible that that the first roll of the dice will be a 20. So, while it is more likely that the character with the +11 spot will notice the hide check 26 attempt in shorter time than he notices the hide check 31 attempt, there is no guarantee of that at all and with the probability curve of a 20 roll, it will take quite a few "take 20s" (which aren't actually using the take 20 rules but rather rolling until the spotter notices the hider) to have an accurate assessment of how many rolls it takes the typical spotter to notice the hider.</p><p></p><p>Now, since the spotter cannot fail to spot the hider once he has succeed in a spot check--until the hider hides again, the only way to get the multiple iterations on a single hide check would be to have multiple spotters. So, maybe, you might get some mathematically reliable results taking 400 with eight or ten spotters. And maybe you'd get good enough results with three spotters. But you've still got spotters taking twenty for every one of the twenty iterations in the original hider's take 20 so it's still really taking 400. </p><p></p><p>And using more spotters also exacerbates the problem of the spot/hide differential. (In a typical adventuring party, there will be some members with great spots, but most will have lousy Spot checks. So, maybe in a 6th level party, the halfling rogue has a +20 hide (9 ranks, +5 dex, +4 size, +2 racial) and a +9 spot (9 ranks), the elf ranger has a +13 hide (9 ranks, +4 dex) and a +13 spot (9 ranks, +2 wis, +2 racial), the cleric has a -5 hide (+0 dex, -5 masterwork fullplate), and a +4 spot (+4 wisdom), and the wizard has a +2 hide (+2 dex), and a +9 spot (4 ranks cross class, +2 alertness, +3 bonus from familiar). Nobody in the party is going to be able to tell the difference between a 14 and a 20 on the rogue's hide check or the difference between a 16 and a 20 on the ranger's hide check. The elf will only fail to spot the cleric hiding on a 1 if the cleric rolls a 20 (1/400 chance). About the only place where doing a statistical analysis of the take 400 process would give the character a good idea that he is hiding as well as possible is if everyone else was trying to help the wizard or the cleric hide. (And in the cleric's case, the odds are slim enough that I suspect meaningful statistical analysis would require more than three observations).</p><p></p><p>There is another part of the rules that make this process problematic for actually hiding a group of characters: concealment and cover rules. The process of hiding requires that the characters have cover or concealment from the people trying to spot them--presumably including the eventual victims of the ambush. Now, if we assume a situation where the PCs know roughly where the victims will be coming from and one PC hides, the others, presumably will be looking for positions where they also have concealment or cover from the victims' expected location. The problem is that they will not necessarily have cover or concealment from the friends that who they helped to hide in that position. Thus, in many cases, the process will go like this:</p><p></p><p>rogue, ranger, and wizard help the cleric to hide (40 minutes)</p><p></p><p>Then the rogue decides to hide, but his hiding spot does not have cover from the cleric so the cleric cannot help. The wizard and ranger help (40 minutes).</p><p></p><p>The wizard then decides to hide. Again, he does not have cover or concealment from the cleric or rogue so they cannot help try to spot him. Only the ranger can, and as we discussed above, it requires more than 1 set of spot checks for the time required to score success to accurately reflect the likelihood of succeeding. That, aside, however, the ranger how has nobody who can help him hide unless maybe he hides on the other side of the road where all three party members could attempt to spot him. And the process took three hours to complete. (Let's hope the party members don't get hungry, or need to relieve themselves and that the expected victims don't show up 30 minutes before the expected time).</p><p></p><p>From a not-strictly-mechanical perspective, there is another problem with taking 20 on hide: it assumes that hiding from the spotter is the same as hiding from the ambush's eventual victim. If the ambush's victim is taller or shorter or able to see through shadow perfectly (while the spotter has normal vision) or standing in a different place (almost a certainty at some point, if he is moving, it stands to reason that what provides concealment from a spotter standing in a certain place would not necessarily provide concealment from the eventual ambush victim standing somewhere else and walking--possibly at a different height and with different visual abilities.</p><p></p><p>That's why, in my games, you can have a spotter help you and potentially provide a circumstance bonus to your hide check (though, I don't think I'd give a circumstance bonus for a spotter who didn't actually stand a reasonable chance of spotting the hider--in he case of a +2 spot check trying to help a +20 or so hide check, I might actually assess a circumstance penalty for a false sense of security). However, I won't allow take 20 on a hide check. Hide is dependent upon the spotter and the spotter is not the same person or in the same position as the ambush's true victim.</p></blockquote><p></p>
[QUOTE="Elder-Basilisk, post: 3068192, member: 3146"] Sure, that sounds good, but if you actually want to do it that way, you should take the watcher's spot check into account. Now, maybe you protest: but the spotter can take 20 and keep looking. OK, well, let's assume that this is possible. In that case, the usefulness would really come down to what the spotter can accomplish. Modifying the above dialogue: So, if the spotter takes 20 for every one roll the hider makes, it's really taking 400 (since the number of possible combinations is 400). So, it should take 40 minutes. From a rules perspective, however, there is a problem with using the take 20 mechanic to signify this process: There is not necessarily a difference in the ability of the spotter to notice a hider who rolls well. Taking the above example a step further: Of course, the hider is not as well hidden as possible, he's just so well hidden that the spotter can't possibly spot him. If your reverse the bonuses, you find a similar discrepancy: Now, that uses the take 20 mechanic which doesn't offer an opportunity to distiguish between "a roll of 20 can find the hider" and "could be hidden better" or between "a roll of 20 cannot find the hider" and "cannot possibly be hidden better." But, you protest, surely the characters would be able to see the difference between a 10 and a 16 on the hide check. In the case of the +15 hide vs. +11 spot, a roll of 16 on the hide check means the spotter should only notice the hider 1 time in 20 and a roll of 10 would mean the spotter notices the hider 1 roll in 4. The problem is that this is not exactly true. The reason it is not true is that once the spotter succeeds, he cannot continue to try and compare the number of failures to successes. Rather, he has to say "I see you" and let the other guy decide whether or not to try and hide again. Now, it is not likely to actually take 20 rolls in order to roll a 20--it's possible that that the first roll of the dice will be a 20. So, while it is more likely that the character with the +11 spot will notice the hide check 26 attempt in shorter time than he notices the hide check 31 attempt, there is no guarantee of that at all and with the probability curve of a 20 roll, it will take quite a few "take 20s" (which aren't actually using the take 20 rules but rather rolling until the spotter notices the hider) to have an accurate assessment of how many rolls it takes the typical spotter to notice the hider. Now, since the spotter cannot fail to spot the hider once he has succeed in a spot check--until the hider hides again, the only way to get the multiple iterations on a single hide check would be to have multiple spotters. So, maybe, you might get some mathematically reliable results taking 400 with eight or ten spotters. And maybe you'd get good enough results with three spotters. But you've still got spotters taking twenty for every one of the twenty iterations in the original hider's take 20 so it's still really taking 400. And using more spotters also exacerbates the problem of the spot/hide differential. (In a typical adventuring party, there will be some members with great spots, but most will have lousy Spot checks. So, maybe in a 6th level party, the halfling rogue has a +20 hide (9 ranks, +5 dex, +4 size, +2 racial) and a +9 spot (9 ranks), the elf ranger has a +13 hide (9 ranks, +4 dex) and a +13 spot (9 ranks, +2 wis, +2 racial), the cleric has a -5 hide (+0 dex, -5 masterwork fullplate), and a +4 spot (+4 wisdom), and the wizard has a +2 hide (+2 dex), and a +9 spot (4 ranks cross class, +2 alertness, +3 bonus from familiar). Nobody in the party is going to be able to tell the difference between a 14 and a 20 on the rogue's hide check or the difference between a 16 and a 20 on the ranger's hide check. The elf will only fail to spot the cleric hiding on a 1 if the cleric rolls a 20 (1/400 chance). About the only place where doing a statistical analysis of the take 400 process would give the character a good idea that he is hiding as well as possible is if everyone else was trying to help the wizard or the cleric hide. (And in the cleric's case, the odds are slim enough that I suspect meaningful statistical analysis would require more than three observations). There is another part of the rules that make this process problematic for actually hiding a group of characters: concealment and cover rules. The process of hiding requires that the characters have cover or concealment from the people trying to spot them--presumably including the eventual victims of the ambush. Now, if we assume a situation where the PCs know roughly where the victims will be coming from and one PC hides, the others, presumably will be looking for positions where they also have concealment or cover from the victims' expected location. The problem is that they will not necessarily have cover or concealment from the friends that who they helped to hide in that position. Thus, in many cases, the process will go like this: rogue, ranger, and wizard help the cleric to hide (40 minutes) Then the rogue decides to hide, but his hiding spot does not have cover from the cleric so the cleric cannot help. The wizard and ranger help (40 minutes). The wizard then decides to hide. Again, he does not have cover or concealment from the cleric or rogue so they cannot help try to spot him. Only the ranger can, and as we discussed above, it requires more than 1 set of spot checks for the time required to score success to accurately reflect the likelihood of succeeding. That, aside, however, the ranger how has nobody who can help him hide unless maybe he hides on the other side of the road where all three party members could attempt to spot him. And the process took three hours to complete. (Let's hope the party members don't get hungry, or need to relieve themselves and that the expected victims don't show up 30 minutes before the expected time). From a not-strictly-mechanical perspective, there is another problem with taking 20 on hide: it assumes that hiding from the spotter is the same as hiding from the ambush's eventual victim. If the ambush's victim is taller or shorter or able to see through shadow perfectly (while the spotter has normal vision) or standing in a different place (almost a certainty at some point, if he is moving, it stands to reason that what provides concealment from a spotter standing in a certain place would not necessarily provide concealment from the eventual ambush victim standing somewhere else and walking--possibly at a different height and with different visual abilities. That's why, in my games, you can have a spotter help you and potentially provide a circumstance bonus to your hide check (though, I don't think I'd give a circumstance bonus for a spotter who didn't actually stand a reasonable chance of spotting the hider--in he case of a +2 spot check trying to help a +20 or so hide check, I might actually assess a circumstance penalty for a false sense of security). However, I won't allow take 20 on a hide check. Hide is dependent upon the spotter and the spotter is not the same person or in the same position as the ambush's true victim. [/QUOTE]
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