As I alluded to in my previous post, simple probability is really a matter of observation, to produce a model. If you roll a six sided die over and over again, you’ll see that each face comes up pretty much equally often, and so you model it as 1\/6 probability for each face. There’s not really a huge amount of principle behind the basic models: it’s really just whatever works. This is the root of the distinction between interpretations: a frequentist starts with an experiment, and builds a descriptive model based on it, and says that the underlying phenomena being tested has the model as g, a property; a Bayesian does almost the same thing, but says that the model describes the state of their knowledge.<\/p>\n
Where probability starts to become interesting in when you combine<\/em> things. I know the probability of outcomes for rolling one die: how can I use that to see what happens when I roll five dice together? I know the probability of drawing a specific card from a deck: what are the odds of being dealt a particular poker hand?<\/p>\n We’ll start with the easiest part: combining independent probabilities. The probability of two events are independent when there’s no way for the outcome of one to influence the outcome of the other. For example, if you’re flipping a coin several times, the result of one coin flip has no effect on the result of a subsequent flip. On the other hand, dealing 10 cards from a deck is a sequence of dependent events: once you’ve dealt one card, the next deal must come from the remaining<\/em> cards: you can’t deal the same card twice.<\/p>\n If you know the probability space of your trials, then recognizing an independent situation is easy: if the outcome of one trial doesn’t alter the probability space of other trials, then they’re independent.\n<\/p>\n Look back at the coin flip example: we know what the probability space of a coin flip looks like: it’s got two, equally probable outcomes. If you’ve flipped a coin once, and you’re going to flip another coin, the result of the first flip can’t do anything that alters the probability space of a subsequent flip.\n<\/p>\n But if you think about dealing cards, that’s not true. With a standard deck of cards, the initial probability space has 52 outcomes, each of which is equally likely. So the odds of being dealt the 5 of spades is exactly 1\/52.<\/p>\n Now, suppose that you got lucky, and you did get dealt the 5 of spades on the first card. What’s the probability of being dealt the 5 of spades’s on the second? If they were independent events, it would still be 1\/52. But once you’ve dealt one card, you can’t deal it again. The probability of being dealt the 5 of spades as the second card is 0: it’s impossible. The probability space only has 51 possible outcomes, and the 5 of spades is not<\/em> one of them. The space has changed. That’s the definition of a dependent event.<\/p>\n When you’re faced with dependent probabilities, you need to figure out how the probability space will be changed, and incorporate that into your computation. Once you’ve incorporated the change in the probability space of the second test, then you’ve got a new independent probability, and you can combine them. Figuring out how to alter the probability space can be extremely difficult, but that’s what makes it interesting.<\/p>\n When you’re dealing with independent events, it’s easy to combine them. There are two basic ways of combining event probabilities, Suppose you’re looking at two test with independent outcomes. I know that the probability of event e<\/em> is P(e)<\/em>, and the probability of event f<\/em> is P(f)<\/em> Then the outcome of e & f<\/em> – that is, of having e<\/em> as the outcome of the first trial, and f<\/em> as the outcome of the second, is P(e)×P(f)<\/em>. The odds of rolling HTTH on a coin is (1\/2)*(1\/2)*(1\/2)*(1\/2)=(1\/16). <\/p>\n If you’re looking at independent alternatives – that is, the probability of e<\/em> OR F<\/em>, you combine the probabilities of the event with addition: P(e) + P(f)<\/em>. So, the odds of drawing any heart from a deck: for each card, it’s 1\/52. There are thirteen different hearts. So the odds of drawing a red are 1\/52 + 1\/52 + … = 13\/52 = 1\/4. <\/p>\n That still doesn’t get us to the really interesting stuff. We still can’t quite work out something like the odds of being dealt a flush. To get there, we need to learn some combinatorics, which will allow us to formulate the probability spaces that we need for an interesting probability.<\/p>\n","protected":false},"excerpt":{"rendered":" As I alluded to in my previous post, simple probability is really a matter of observation, to produce a model. If you roll a six sided die over and over again, you’ll see that each face comes up pretty much equally often, and so you model it as 1\/6 probability for each face. There’s not […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[53],"tags":[],"class_list":["post-2192","post","type-post","status-publish","format-standard","hentry","category-probability"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_shortlink":"https:\/\/wp.me\/p4lzZS-zm","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/posts\/2192","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/comments?post=2192"}],"version-history":[{"count":0,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/posts\/2192\/revisions"}],"wp:attachment":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/media?parent=2192"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/categories?post=2192"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/tags?post=2192"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nand they should be familiar from logic: event1 AND event2, and event1 OR event2.\n<\/p>\n