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#1




Question on probability involving handicapping.
Let's say I'm playing a game and there is a certain probability of me winning p(win). But to complicate things there are multiple factors affecting my chance to win which (luckily) can be quantified. For example
Standing on one foot => 20% chance to win Being 5 feet closer => +15% chance to win How do I use those to calculate my chance to win? The method I use is to treat them as multipliers. new p(win) = (1.15)(0.80) x p(win). My logic is that suppose there is a factor making it impossible to win then the multiplier would be zero and it doesn't matter what the other numbers are since new p(win) = 0. Is this the right approach & logic? 
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#2




Essentially, you're asking "What do I mean by '+15% chance to win'?" (Or, what does someone who says such a thing mean by it?) This is something that the person saying such things ought to be clear about, but IMHO yours is the most natural interpretation.

#3




Assuming these are all independent factors, your approach is reasonable. If going 5 feet closer is somehow correlated with standing on one foot, it won't work.

#4




Quote:
First suppose I initially had a 50% chance to win, then had one factor that gave a +20% chance and a second that gave a 20% chance. Ideally I should get back to a 50% chance but instead I have a 0.5*0.8*1.2=48% chance. Also suppose I have a 90% chance to win and then step 5 feet closer to the basket. Using the4 OP's method I end up with a (0.9*1.15)=103.5% chance to win. This can't happen even if I'm willing to give %110. So what statisticans do instead is use the logistic function. P(win)=e^X/(1+e^X) Where X is the sum of your positive and negative factors. So stepping closer to the basket my add 0.3 to X while standing on one foot may subtract 0.4 from X. When X=0 you have a 50% chance of winning. As X gets very large, the probability of winning goes to 1, and as X gets very negative the probability goes to zero. Also factors are additive so a postive factor if I have a postive factor of +0.3 and a negative factor of 0.3 they will cancel out. Also notice that P(lose)=1P(win)=e^X/(1+e^X) , so positive factors that increase your chance to win, decrease your opponents chance to win by the exact same amount. Unfortunately it no longer becomes quite so simple to describe what an increase or decrease of a particular amount means. Starting from a 50% chance to win, adding 0.3 to X, will decrease my chances of winning to 42.6% (so according to the OPs method about 15.8%) , but if my chances of winning was at 42.6%, adding another 0.3 will change my chances to 0.3543 or (according to the OPs method by 18.7%). Last edited by Buck Godot; 11032016 at 11:54 AM. 
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