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File: AnOddityInTennis ''' <link rel="alternate" type="application/rss+xml" ''' href="/rss.xml" title="RSS Feed"> ********> width="25%" |>> ''' <a title="Subscribe to my feed" ''' rel="alternate" ''' href="https://www.solipsys.co.uk/rss.xml"> ''' <img style="border-width: 0px;" ''' src="https://www.feedburner.com/fb/images/pub/feed-icon32x32.png" ''' align="middle" ''' alt="" />Subscribe!</a> _ ''' <a href="https://twitter.com/ColinTheMathmo"> ''' <img src="https://www.solipsys.co.uk/new/images/TwitterButton.png" ''' title="By: TwitterButtons.net" ''' width="212" height="69" ''' alt="@ColinTheMathmo" ''' /></a> <<| ---- My lastest posts can be found here: * ColinsBlog ---- Previous blog posts: * DecisionTreeForTennis * DecisionTreesInGames * AMatterOfConvention * DoYouNourishOrTarnish * BinarySearchReconsidered * TwoEqualsFour * TheLostPropertyOffice * TheForgivingUserInterface * SettingUpRSS * WithdrawingFromHackerNews ---- Additionally, some earlier writings: * RandomWritings. * ColinsBlog2010 * ColinsBlog2009 * ColinsBlog2008 * ColinsBlog2007 * ColinsBlogBefore2007 ******** !! Decision Trees in Games (Part 3) - 2011/05/24 [[[> https://www.solipsys.co.uk/new/images/Tennis_1.png _ |>> Fig 1: Probability of winning a game <<| ---- https://www.solipsys.co.uk/new/images/DecisionTree30_15.png _ |>> Fig 2: Decision tree from 30:15 <<| ---- https://www.solipsys.co.uk/new/images/Tennis_2.png _ |>> Fig 3: Probability of winning from 30:15 <<| ---- https://www.solipsys.co.uk/new/images/Tennis_3.png _ |>> Fig 4: Close up on the high end <<| ]]] In DecisionTreesInGames we looked at how to compute probabilities in trees of possibilities, and then in DecisionTreeForTennis we extended this to "Directed Acyclic Graphs" and beyond so we could analyse the probability of winning a game of tennis. We ended up calculating that if your probability of winning a given point is /p,/ then your probability of winning a game is: |>> ''' <i>P = p⁴+4qp⁴+10q²p⁴+20p³q³.D</i> <<| where /D/ is the probability of winning from Deuce and is given by: |>> ''' <i>D = p²/(1-2pq)</i> <<| Figure 1 shows a graph of this. ---- Graphs curtesy of Wolfram Alpha: |>> [[[ ''' <a href="http://www.wolframalpha.com/"> ''' <img src="https://www.solipsys.co.uk/new/images/WolframAlpha2.png" ''' alt="Wolfram Alpha"> ''' </a> ]]] <<| ---- It's interesting to see that once your chances of winning each point gets above 80% you're pretty much certain to win the game, and increasing the per-point percentage is pretty much, well, pointless (if you see what I mean). So that's interesting, but hardly surprising. Using exactly the same techniques we can now compute how our chances change through a game. Suppose, for example, we win two points and lose another, so we are 30:15 up. We're ahead, we're winning, so you'd suppose our chances of winning the game have increased. Figure 2 shows the decision tree (well, graph) and so we can compute our chances of winning are now: |>> ''' <i>P = p²+2qp²+3pq².D</i> <<| where, as before, /D/ is the probability of winning from Deuce. Figure 3 shows this new probability in red, and, as we can see, it has gone up, just as expected. No, but wait! Let's have a closer look at the high percentage portion of the graph. Figure 4 shows a close up view of the section where our chances of winning each point are around 90% or more, and there's something of a surprise. Our chances of winning, even though we're already ahead, have actually gone down. How can that be true? It's hard to be convincing about this without resorting to the equations. However, one intuitive way of thinking about it is this. When your chances of winning each point are so large, more than 90%, the chances of losing a single point are small. So small, in fact, that losing a single point in the game is enough to swing the chances of the whole game against you. Not wholesale, obviously, since you're still overwhelmingly likely to win the game, but by some. Having two points is almost irrelevant - it's to be expected. So being 30:15 up in the game actually is a /bad/ thing, because you've lost a point you were expected to win. The cross-over happens at ''' <i><b>(1+&radic;7)/4</b></i> which a shade over 91.1% as the graph shows. And that's the other slightly unexpected thing. If your chances of winning each point independently are less than 90%, the cross-over doesn't happen, because losing a point isn't sufficiently surprising. Which I find slightly surprising. ---- |>> | |>> <<<< Prev <<<< ---- DecisionTreeForTennis <<| | : | |>> >>>> Next >>>> ---- DissectingASquare <<| | ---- ********> ''' <a href="https://twitter.com/ColinTheMathmo">You should follow me on twitter</a> ******** ''' <a href="https://twitter.com/ColinTheMathmo"> ''' <img src="https://www.solipsys.co.uk/new/images/TwitterButton.png" ''' title="By: TwitterButtons.net" ''' width="212" height="69" ''' alt="@ColinTheMathmo" ''' /></a> ********< <<| ---- !! Comments ''' <div id="disqus_thread"></div> ''' <script type="text/javascript"> ''' /* * * CONFIGURATION VARIABLES: EDIT BEFORE PASTING INTO YOUR WEBPAGE * * */ ''' var disqus_shortname = 'solipsyslimited'; ''' ''' // The following are highly recommended additional parameters. Remove the slashes in front to use. 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