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‎.nojekyll‎

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‎archive.html‎

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<div class="g-col-1" data-index="12" data-categories="cHl0aG9uJTJDYm93bGluZw==" data-listing-date-sort="1700982000000" data-listing-file-modified-sort="1775178113358" data-listing-date-modified-sort="NaN" data-listing-reading-time-sort="15" data-listing-word-count-sort="2871">
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<a href="./posts/impossible_bowling/index.html" class="quarto-grid-link">
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<div class="quarto-grid-item card h-100 card-left">
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<p class="card-img-top"><img src="posts/impossible_bowling/index_files/figure-html/5fe67f81-f378-4258-a005-1914e099b559-1-31ff655b-8ba9-40cd-a9af-7ff1538b1345.png" style="height: 150px;" class="thumbnail-image card-img"/></p>

‎index.html‎

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<div class="quarto-post image-right" data-index="12" data-categories="cHl0aG9uJTJDYm93bGluZw==" data-listing-date-sort="1700982000000" data-listing-file-modified-sort="1775178113358" data-listing-date-modified-sort="NaN" data-listing-reading-time-sort="15" data-listing-word-count-sort="2871">
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<div class="thumbnail"><a href="./posts/impossible_bowling/index.html" class="no-external">
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<p class="card-img-top"><img data-src="posts/impossible_bowling/index_files/figure-html/5fe67f81-f378-4258-a005-1914e099b559-1-31ff655b-8ba9-40cd-a9af-7ff1538b1345.png" class="thumbnail-image card-img"/></p>
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<div class="listing-author">
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Allan Farrell
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<div class="listing-reading-time">13 min</div>
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<div class="listing-reading-time">15 min</div>
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</a>
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‎index.xml‎

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<p>While hanging out at the lanes a few obvious impossible scores got thrown out: a <em>1</em>, obviously, but also a <em>449</em> – there’s no way to throw a 14 with the last ball in the tenth frame. But the question still lingered: are there any other gaps? It was not immediately obvious, to my bowling team, how you would figure that out without checking.</p>
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<p>Maybe we can brute-force this and try every conceivable bowling game? However there are <em>a lot</em> of possible bowling games. As a first pass, there are thirty balls thrown in a game and each ball has <em>up to</em> fourteen possible pinfall scores (0 through 15 excluding 1 and 14). This would give <em>14<sup>30</sup></em> possible bowling games. Even if it took a single nanosecond to evaluate each game that would take longer than the current age of the universe to work through.</p>
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<p>But that’s not a great upper bound, it doesn’t take into account the rules of bowling: you can only knock down up to five pins in any given frame, for example if the first ball scores a <em>13</em> then the second ball doesn’t get to choose from fourteen possibilities, it gets to chose from two: 0 and 2. Still, it is going to be a large number. The vast majority of those games are going to be completely redundant, since we are only looking for scores from 2 to 450.</p>
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<div class="callout callout-style-default callout-note callout-titled">
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<div class="callout-header d-flex align-content-center">
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<div class="callout-icon-container">
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<i class="callout-icon"></i>
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</div>
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<div class="callout-title-container flex-fill">
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Note
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</div>
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</div>
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<div class="callout-body-container callout-body">
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<p>I was thinking about this some more and there is a different way of looking at this that gives a better estimate for the number of possible games. First let’s consider a single frame (within the first 9 frames). We can count the number of ways to knock down any subset of pins (including all and none):</p>
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<p><img src="https://latex.codecogs.com/png.latex?%0A%5Cbinom%7B5%7D%7B0%7D%20+%20%5Cbinom%7B5%7D%7B1%7D%20+%20%5Cbinom%7B5%7D%7B2%7D%20+%20%5Cbinom%7B5%7D%7B3%7D%20+%20%5Cbinom%7B5%7D%7B4%7D%20+%20%5Cbinom%7B5%7D%7B5%7D%20=%2032%0A"></p>
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<p>For a single frame there are three possibilities:</p>
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<table class="caption-top table">
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<thead>
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<tr class="header">
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<th style="text-align: left;">Frame</th>
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<th style="text-align: left;">Number of Possibilities</th>
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<th style="text-align: left;">description</th>
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</tr>
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</thead>
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<tbody>
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<tr class="odd">
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<td style="text-align: left;">? ? ?</td>
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<td style="text-align: left;">32</td>
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<td style="text-align: left;">All three balls are thrown</td>
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</tr>
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<tr class="even">
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<td style="text-align: left;">X</td>
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<td style="text-align: left;">1</td>
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<td style="text-align: left;">A single ball is thrown, a strike</td>
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</tr>
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<tr class="odd">
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<td style="text-align: left;">? \</td>
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<td style="text-align: left;">31</td>
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<td style="text-align: left;">Two balls are thrown, a spare</td>
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</tr>
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</tbody>
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</table>
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<p>So there are 64 possible ways of bowling a single frame of bowling <em>that is not the tenth frame</em>.</p>
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<p>The tenth frame has some extra rules, so lets go through it:</p>
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<table class="caption-top table">
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<thead>
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<tr class="header">
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<th style="text-align: left;">Frame</th>
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<th style="text-align: left;">Number of Possibilities</th>
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<th style="text-align: left;">description</th>
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</tr>
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</thead>
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<tbody>
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<tr class="odd">
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<td style="text-align: left;">? ? ?</td>
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<td style="text-align: left;">32</td>
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<td style="text-align: left;">Anything but a strike or spare in the first two balls</td>
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</tr>
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<tr class="even">
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<td style="text-align: left;">X ? ?</td>
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<td style="text-align: left;">32</td>
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<td style="text-align: left;">First ball a strike, two non strikes</td>
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</tr>
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<tr class="odd">
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<td style="text-align: left;">X X ?</td>
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<td style="text-align: left;">32</td>
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<td style="text-align: left;">Two strikes, one anything</td>
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</tr>
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<tr class="even">
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<td style="text-align: left;">? \ ?</td>
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<td style="text-align: left;">992</td>
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<td style="text-align: left;">A spare (31 options) and anything (32)</td>
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</tr>
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</tbody>
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</table>
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<p>Which gives 1088 possible ways of bowling the 10th frame.</p>
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<p>So this gives a total count of possible 5 pin bowling games of <img src="https://latex.codecogs.com/png.latex?1088%20%5Ctimes%2064%5E9"> which is about <img src="https://latex.codecogs.com/png.latex?2%20%5Ctimes%2010%5E%7B19%7D"></p>
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</div>
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</div>
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</section>
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<section id="nothing-fancy" class="level2 page-columns page-full">
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<h2 class="anchored" data-anchor-id="nothing-fancy">Nothing fancy</h2>

‎posts/impossible_bowling/index.html‎

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<p>While hanging out at the lanes a few obvious impossible scores got thrown out: a <em>1</em>, obviously, but also a <em>449</em> – there’s no way to throw a 14 with the last ball in the tenth frame. But the question still lingered: are there any other gaps? It was not immediately obvious, to my bowling team, how you would figure that out without checking.</p>
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<p>Maybe we can brute-force this and try every conceivable bowling game? However there are <em>a lot</em> of possible bowling games. As a first pass, there are thirty balls thrown in a game and each ball has <em>up to</em> fourteen possible pinfall scores (0 through 15 excluding 1 and 14). This would give <em>14<sup>30</sup></em> possible bowling games. Even if it took a single nanosecond to evaluate each game that would take longer than the current age of the universe to work through.</p>
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<p>But that’s not a great upper bound, it doesn’t take into account the rules of bowling: you can only knock down up to five pins in any given frame, for example if the first ball scores a <em>13</em> then the second ball doesn’t get to choose from fourteen possibilities, it gets to chose from two: 0 and 2. Still, it is going to be a large number. The vast majority of those games are going to be completely redundant, since we are only looking for scores from 2 to 450.</p>
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<div class="callout callout-style-default callout-note callout-titled">
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<div class="callout-header d-flex align-content-center">
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<div class="callout-icon-container">
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<i class="callout-icon"></i>
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</div>
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<div class="callout-title-container flex-fill">
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Note
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</div>
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<div class="callout-body-container callout-body">
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<p>I was thinking about this some more and there is a different way of looking at this that gives a better estimate for the number of possible games. First let’s consider a single frame (within the first 9 frames). We can count the number of ways to knock down any subset of pins (including all and none):</p>
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<p><span class="math display">\[
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\binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} = 32
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\]</span></p>
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<p>For a single frame there are three possibilities:</p>
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<table class="caption-top table">
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<thead>
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<tr class="header">
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<th style="text-align: left;">Frame</th>
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<th style="text-align: left;">Number of Possibilities</th>
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<th style="text-align: left;">description</th>
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</tr>
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</thead>
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<tbody>
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<tr class="odd">
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<td style="text-align: left;">? ? ?</td>
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<td style="text-align: left;">32</td>
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<td style="text-align: left;">All three balls are thrown</td>
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</tr>
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<tr class="even">
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<td style="text-align: left;">X</td>
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<td style="text-align: left;">1</td>
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<td style="text-align: left;">A single ball is thrown, a strike</td>
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</tr>
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<tr class="odd">
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<td style="text-align: left;">? \</td>
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<td style="text-align: left;">31</td>
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<td style="text-align: left;">Two balls are thrown, a spare</td>
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</tr>
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</tbody>
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</table>
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<p>So there are 64 possible ways of bowling a single frame of bowling <em>that is not the tenth frame</em>.</p>
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<p>The tenth frame has some extra rules, so lets go through it:</p>
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<table class="caption-top table">
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<thead>
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<tr class="header">
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<th style="text-align: left;">Frame</th>
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<th style="text-align: left;">Number of Possibilities</th>
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<th style="text-align: left;">description</th>
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</tr>
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</thead>
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<tbody>
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<tr class="odd">
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<td style="text-align: left;">? ? ?</td>
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<td style="text-align: left;">32</td>
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<td style="text-align: left;">Anything but a strike or spare in the first two balls</td>
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</tr>
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<tr class="even">
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<td style="text-align: left;">X ? ?</td>
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<td style="text-align: left;">32</td>
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<td style="text-align: left;">First ball a strike, two non strikes</td>
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</tr>
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<tr class="odd">
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<td style="text-align: left;">X X ?</td>
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<td style="text-align: left;">32</td>
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<td style="text-align: left;">Two strikes, one anything</td>
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</tr>
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<tr class="even">
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<td style="text-align: left;">? \ ?</td>
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<td style="text-align: left;">992</td>
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<td style="text-align: left;">A spare (31 options) and anything (32)</td>
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</tr>
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</tbody>
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</table>
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<p>Which gives 1088 possible ways of bowling the 10th frame.</p>
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<p>So this gives a total count of possible 5 pin bowling games of <span class="math inline">\(1088 \times 64^9\)</span> which is about <span class="math inline">\(2 \times 10^{19}\)</span></p>
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</div>
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</div>
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</section>
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<section id="nothing-fancy" class="level2 page-columns page-full">
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<h2 class="anchored" data-anchor-id="nothing-fancy">Nothing fancy</h2>

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