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Title / Subject - or probabilities are hard
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<blockquote data-quote="CapnZapp" data-source="post: 7265643" data-attributes="member: 12731"><p>In my ToA campaign, I've reused the guts of the AL adventure Over the Edge to add some needed initial motivation for striking out "randomly" in the jungle - in my scenario the party is looking for some way to get rid of the heavy armor penalties, and Pozanna tells the party a gold statue they can trade for a solution is in one out of five sites in exchange for them mapping the sites.</p><p></p><p>They only get two (random) site coordinates, because Pozanna says other adventurers have already been given the other site locations. What they don't know is that both these other parties will have failed and died on site (somewhat like the original story). Only if they haven't found the gold statue after four sites will the fifth and final site NPC party emerge victorious, and the heroes lose out.</p><p></p><p>Anyway. Instead of already from the start rolling 1d5 to place the gold statue, I thought I make the rolls as I go along. I thought it would be boring to know already from the start if the heroes' search would be futile, and to increase the odds by giving the party "first shot" at rolling for each site.</p><p></p><p>That is, I would instead roll 1d5 when the party is visiting the first site, a 1 means the statue is there. If not, then they need to roll 1 on 1d4 when visiting the second site, and so on.</p><p></p><p>It turns out I needn't have bothered - <em>the probabilities are exactly the same</em>.</p><p></p><p>Probability using method #1: 20% the gold statue is in any given site, and thus it's 20% it'll be in the last site they look (or won't look as it were).</p><p></p><p>Probability using method #2: the probability it hasn't been found after visiting four sites is (1-1/5)*(1-1/4)*(1-1/3)*(1-1/2) which is... drum-roll... 20%!</p><p></p><p>So, well, anyway... just wanted to share that bit of insight.</p></blockquote><p></p>
[QUOTE="CapnZapp, post: 7265643, member: 12731"] In my ToA campaign, I've reused the guts of the AL adventure Over the Edge to add some needed initial motivation for striking out "randomly" in the jungle - in my scenario the party is looking for some way to get rid of the heavy armor penalties, and Pozanna tells the party a gold statue they can trade for a solution is in one out of five sites in exchange for them mapping the sites. They only get two (random) site coordinates, because Pozanna says other adventurers have already been given the other site locations. What they don't know is that both these other parties will have failed and died on site (somewhat like the original story). Only if they haven't found the gold statue after four sites will the fifth and final site NPC party emerge victorious, and the heroes lose out. Anyway. Instead of already from the start rolling 1d5 to place the gold statue, I thought I make the rolls as I go along. I thought it would be boring to know already from the start if the heroes' search would be futile, and to increase the odds by giving the party "first shot" at rolling for each site. That is, I would instead roll 1d5 when the party is visiting the first site, a 1 means the statue is there. If not, then they need to roll 1 on 1d4 when visiting the second site, and so on. It turns out I needn't have bothered - [I]the probabilities are exactly the same[/I]. Probability using method #1: 20% the gold statue is in any given site, and thus it's 20% it'll be in the last site they look (or won't look as it were). Probability using method #2: the probability it hasn't been found after visiting four sites is (1-1/5)*(1-1/4)*(1-1/3)*(1-1/2) which is... drum-roll... 20%! So, well, anyway... just wanted to share that bit of insight. [/QUOTE]
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