In this case I'd consider using two numbers. One reflecting the reliability and the other the utility of a particular plan, or course of action. reliability = successes / attempts. utility = successes - (attempts - successes) The numbers for the trials below would be 1. reliability = 0.5 , utility = 02. reliability = 0.5, utility = 03. reliability = 0, utility = -3 4. reliability = 0, utility = -1 Thoughts? ~PM
> Date: Fri, 26 Sep 2014 15:39:30 -0400 > Subject: Re: [agi] How many tries? > From: [email protected] > To: [email protected] > > On Fri, Sep 26, 2014 at 2:32 PM, Stanley Nilsen via AGI <[email protected]> > wrote: > > 4. probabilities > > Matt states a fairly simple equation (utility * probability < cost) that > > he calls the answer. I'm not sure it's as simple as indicated - in fact I'm > > sure it is not. > > It is not simple because estimating probabilities is not simple. > Suppose you do an experiment some number of times and observe the > following outcomes, where 1 = success and 0 = failure. For each of the > following sequences, estimate the probability of success on the next > trial: > > 1. 0110100101 > 2. 0000011111 > 3. 000 > 4. 0 > > In case 1 we observe 5 successes in 10 trials, so we guess p = 0.5. > > In case 2 we again observe 5 successes in 10 trials, but we are no > longer justified in assuming that the trials are independent. We would > assume p > 0.5 on the basis of a universal distribution, i.e. the > shortest program to generate the data is the most likely. For example, > output 5 zeros followed by ones. > > Cases 3 and 4 are examples of the zero frequency problem. This has > received a lot of study in predictive modeling for data compression. > Simply counting zeros and ones is wrong because probabilities are > never exactly 0. A LaPlace estimator adds 1 to each count so for case > 3, instead of p = 0/3 we have p = (0+1)/(3+2) = 0.2. This is > theoretically optimal if all probabilities are equally likely. But in > practice we often find that this offset is too high. For context > models trained on text, we find experimentally that offsets of 0.03 to > 0.05 are better estimators, e.g. p = (0+0.03)/(3+0.06) = 0.01. You > often have to find these values experimentally. > > -- > -- Matt Mahoney, [email protected] > > > ------------------------------------------- > AGI > Archives: https://www.listbox.com/member/archive/303/=now > RSS Feed: https://www.listbox.com/member/archive/rss/303/19999924-4a978ccc > Modify Your Subscription: https://www.listbox.com/member/?&; > Powered by Listbox: http://www.listbox.com ------------------------------------------- AGI Archives: https://www.listbox.com/member/archive/303/=now RSS Feed: https://www.listbox.com/member/archive/rss/303/21088071-f452e424 Modify Your Subscription: https://www.listbox.com/member/?member_id=21088071&id_secret=21088071-58d57657 Powered by Listbox: http://www.listbox.com
