![\"graph1.jpg\"](\"http:\/\/www.kith.org\/journals\/vardibidian\/images\/graph1.jpg\")
\nSo. This is a math problem that I would once have been able to set up and possibly even solve, but at this point I can’t even figure out how to google the answer. So.\n
In rolling two dice (2D6) the long-run distribution is easy: for every thirty-six rolls, 2 and 12 come up essentially once each, 3 and 11 twice, 4 and 10 three times, 5 and 9 four times, 6 and 10 five, and the last six rolls come up showing 7. In the long run. More or less. You could graph it.\n
In the short term, of course, longer-odds outcomes will likely come up more often than shorter-odds outcomes. For instance, after one roll, the odds are very good indeed (five-to-one) that something that is not a seven will have come up more often than something that is a seven. After six rolls, the odds are (if I am getting this right, and perhaps I am not) even that you will have rolled a seven at all, which means that the odds are slightly better than even that you will have rolled some longer-odds number more often than a 7 (counting the outcomes where you roll, for instance, one 7 and two 6s in those six rolls.\n
Is that clear? For any given number, 7 will come up more frequently in the long term, but in the very short term (six rolls), the odds favor some<\/i> number coming up more often than the 7. If you have ever rolled dice a bunch of times, this will be instinctive, I think; if not, grab a couple of dice or a dice-rolling app and try.\n
Now, let’s see if I can generalize while remaining clear: if you roll, say, a hundred times and graph the results, you’ll probably have something that more or less approximates the ziggurat shape the distribution would predict. Something like this:\n
\n
But still there might be more, for instance, 8s than 7s:\n
![\"graph2.jpg\"](\"http:\/\/www.kith.org\/journals\/vardibidian\/images\/graph2.jpg\")
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Or even more 5s than 7s:\n
![\"graph3.jpg\"](\"http:\/\/www.kith.org\/journals\/vardibidian\/images\/graph3.jpg\")
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Because 100 rolls really is still a short run of rolls, innit? Let’s try 200:\n
![\"graph4.jpg\"](\"http:\/\/www.kith.org\/journals\/vardibidian\/images\/graph4.jpg\")
\n
Look! There are more sevens than anything else. Of course, there are just as many 5s as 6s. I’ll try again.\n
![\"graph5.jpg\"](\"http:\/\/www.kith.org\/journals\/vardibidian\/images\/graph5.jpg\")
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This time there were more 4s than 5s.\n
![\"graph6.jpg\"](\"http:\/\/www.kith.org\/journals\/vardibidian\/images\/graph6.jpg\")
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More 10s than 9s.\n
So, this is my question: It seems to me that for a reasonable number of dice rolls, the odds are in favor<\/i> of some lower-odds result coming up more<\/i> frequently than some higher-odds result. I could figure out the odds of more 10s than 9s, and more 4s than 5s, but I don’t know how to figure out the general odds of there being at least one instance of more x<\/i> than y<\/i> over n<\/i> rolls, where the odds of x<\/i> are longer than the odds of y<\/i>. And what I’m really wondering is that—it feels to me as if it would be possible to look at that as a function of n<\/i>, such that as n<\/i> increases the odds (that is, the odds of at least one such low-odds incidence occurring) decrease, and that therefore there exists an n<\/i> such that the odds are even, with lower n<\/i> having greater odds and higher n<\/i> shorter. I’d like to know how big that n<\/i> is. But I could be wrong about there being such an n<\/i>; my instincts for probability functions aren’t all that good.\n
There’s a lesson about the world in all this, that low-probability things happen all the time, and so forth. Mostly, though, it’s about actual dice: when you shoot craps or play Settlers of Catan<\/cite> or Monopoly<\/cite> or whatnot, don’t expect that over the course of the game there will necessarily be more 8s than 9s. The actual odds are that somebody will get screwed somehow.\n Tolerabimus quod tolerare debemus<\/I>, In Which Your Humble Blogger in theory produces wheat on an 5 and an 8, and has three houses on Oriental Avenue, and also ten dollars behind the pass line.<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[201],"tags":[],"class_list":["post-15174","post","type-post","status-publish","format-standard","hentry","category-navel-gazing"],"acf":[],"_links":{"self":[{"href":"https:\/\/www.kith.org\/vardibidian\/wp-json\/wp\/v2\/posts\/15174","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.kith.org\/vardibidian\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kith.org\/vardibidian\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kith.org\/vardibidian\/wp-json\/wp\/v2\/users\/7"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kith.org\/vardibidian\/wp-json\/wp\/v2\/comments?post=15174"}],"version-history":[{"count":1,"href":"https:\/\/www.kith.org\/vardibidian\/wp-json\/wp\/v2\/posts\/15174\/revisions"}],"predecessor-version":[{"id":16480,"href":"https:\/\/www.kith.org\/vardibidian\/wp-json\/wp\/v2\/posts\/15174\/revisions\/16480"}],"wp:attachment":[{"href":"https:\/\/www.kith.org\/vardibidian\/wp-json\/wp\/v2\/media?parent=15174"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kith.org\/vardibidian\/wp-json\/wp\/v2\/categories?post=15174"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kith.org\/vardibidian\/wp-json\/wp\/v2\/tags?post=15174"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
-Vardibidian.\n\n","protected":false},"excerpt":{"rendered":"