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<blockquote data-quote="Cadence" data-source="post: 6035884" data-attributes="member: 6701124"><p>I think the easiest way to visualize this problem is a tree diagram. A nice explanation is at: <a href="http://www.mathsisfun.com/data/probability-tree-diagrams.html" target="_blank"><span style="color: #0066cc">http://www.mathsisfun.com/data/probability-tree-diagrams.html</span></a> </p><p> </p><p>Google also has a variety of videos on the subject. One that seemed particularly reasonable is at:</p><p> </p><p>[ame="http://www.youtube.com/watch?v=PVF5QBMF4lk"]<span style="color: #0066cc">http://www.youtube.com/watch?v=PVF5QBMF4lk</span>[/ame] </p><p> </p><p> </p><p>The tree diagram for your particular problem would first branch into ten values for the first die, 0 to 9 (just like the example on the web-link branches into two values - head and tail for the first flip). Each of those then branches into ten values of 0 to 9 (like the second column of heads and tails for the second flip). That gives you 100 end branches, each with a probablity of 0.01 (since each one has 1/10 x 1/10 when you multiply them out...just like each branching in the web-page example above is 0.5x0.5=0.25). Now you wouldn't want to draw out all of those branches, but you can still count up how many of them would be exactly a one on the first roll and a one on the second roll (only one end branching out of the 100 possible ones). So now you add up the probabilities of all of the end-points you are worried about (just the 1 and 1 in this case) and get 0.01=1% (in one example on the web-page they wanted the chance of "at least one head", which went with three values of 0.25, so 0.75=75% total) . A tree diagram that had three sets of branches (instead of two like this example) would help you figure out the chances of getting each possible ability score on 3d6 (like the first example in the video, but values of 1 to 6 instead of just H and T).</p><p> </p><p>The more mathy ways of calculating probabilities like these are usually the same as what the tree diagram represents graphically, they just assign symbols to everything in sight and use the "multiplication rule", "conditional probabilities", the "addition rule", and "combinatorics".</p></blockquote><p></p>
[QUOTE="Cadence, post: 6035884, member: 6701124"] I think the easiest way to visualize this problem is a tree diagram. A nice explanation is at: [URL="http://www.mathsisfun.com/data/probability-tree-diagrams.html"][COLOR=#0066cc]http://www.mathsisfun.com/data/probability-tree-diagrams.html[/COLOR][/URL] Google also has a variety of videos on the subject. One that seemed particularly reasonable is at: [ame="http://www.youtube.com/watch?v=PVF5QBMF4lk"][COLOR=#0066cc]http://www.youtube.com/watch?v=PVF5QBMF4lk[/COLOR][/ame] The tree diagram for your particular problem would first branch into ten values for the first die, 0 to 9 (just like the example on the web-link branches into two values - head and tail for the first flip). Each of those then branches into ten values of 0 to 9 (like the second column of heads and tails for the second flip). That gives you 100 end branches, each with a probablity of 0.01 (since each one has 1/10 x 1/10 when you multiply them out...just like each branching in the web-page example above is 0.5x0.5=0.25). Now you wouldn't want to draw out all of those branches, but you can still count up how many of them would be exactly a one on the first roll and a one on the second roll (only one end branching out of the 100 possible ones). So now you add up the probabilities of all of the end-points you are worried about (just the 1 and 1 in this case) and get 0.01=1% (in one example on the web-page they wanted the chance of "at least one head", which went with three values of 0.25, so 0.75=75% total) . A tree diagram that had three sets of branches (instead of two like this example) would help you figure out the chances of getting each possible ability score on 3d6 (like the first example in the video, but values of 1 to 6 instead of just H and T). The more mathy ways of calculating probabilities like these are usually the same as what the tree diagram represents graphically, they just assign symbols to everything in sight and use the "multiplication rule", "conditional probabilities", the "addition rule", and "combinatorics". [/QUOTE]
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