You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
Copy file name to clipboardExpand all lines: index.xml
+17-40Lines changed: 17 additions & 40 deletions
Original file line number
Diff line number
Diff line change
@@ -23319,38 +23319,15 @@ Note
23319
23319
</div>
23320
23320
</div>
23321
23321
<div class="callout-body-container callout-body">
23322
-
<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>
<p>For a single frame there are three possibilities:</p>
23325
-
<table class="caption-top table">
23326
-
<thead>
23327
-
<tr class="header">
23328
-
<th style="text-align: left;">Frame</th>
23329
-
<th style="text-align: left;">Number of Possibilities</th>
23330
-
<th style="text-align: left;">description</th>
23331
-
</tr>
23332
-
</thead>
23333
-
<tbody>
23334
-
<tr class="odd">
23335
-
<td style="text-align: left;">? ? ?</td>
23336
-
<td style="text-align: left;">32</td>
23337
-
<td style="text-align: left;">All three balls are thrown</td>
23338
-
</tr>
23339
-
<tr class="even">
23340
-
<td style="text-align: left;">X</td>
23341
-
<td style="text-align: left;">1</td>
23342
-
<td style="text-align: left;">A single ball is thrown, a strike</td>
23343
-
</tr>
23344
-
<tr class="odd">
23345
-
<td style="text-align: left;">? \</td>
23346
-
<td style="text-align: left;">31</td>
23347
-
<td style="text-align: left;">Two balls are thrown, a spare</td>
23348
-
</tr>
23349
-
</tbody>
23350
-
</table>
23351
-
<p>So there are 64 possible ways of bowling a single frame of bowling <em>that is not the tenth frame</em>.</p>
23322
+
<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.</p>
23323
+
<p>First let’s consider a single frame (within the first 9 frames). The order that pins are knocked down – the score per ball – matters because of the way strikes and spares are counted. So we are trying to answer the question “how many ways can 5 pins be divided into 4 categories (hit by ball 1, 2, or 3, or left standing)?” This is the sum of a multinomial and is well known, for <em>n</em> objects divided into <em>m</em> categories, the number of ways is <img src="https://latex.codecogs.com/png.latex?m%5En">: <img src="https://latex.codecogs.com/png.latex?4%5E5%20=%201024"></p>
23352
23324
<p>The tenth frame has some extra rules, so lets go through it:</p>
23353
23325
<table class="caption-top table">
23326
+
<colgroup>
23327
+
<col style="width: 25%">
23328
+
<col style="width: 45%">
23329
+
<col style="width: 29%">
23330
+
</colgroup>
23354
23331
<thead>
23355
23332
<tr class="header">
23356
23333
<th style="text-align: left;">Frame</th>
@@ -23361,28 +23338,28 @@ Note
23361
23338
<tbody>
23362
23339
<tr class="odd">
23363
23340
<td style="text-align: left;">? ? ?</td>
23364
-
<td style="text-align: left;">32</td>
23365
-
<td style="text-align: left;">Anything but a strike or spare in the first two balls</td>
23341
+
<td style="text-align: left;">1024</td>
23342
+
<td style="text-align: left;">Any combination of the first set of 5 pins</td>
23366
23343
</tr>
23367
23344
<tr class="even">
23368
23345
<td style="text-align: left;">X ? ?</td>
23369
-
<td style="text-align: left;">32</td>
23370
-
<td style="text-align: left;">First ball a strike, two non strikes</td>
<td style="text-align: left;">Every follow up to every spare except ?\- which were counted in ???</td>
23381
23358
</tr>
23382
23359
</tbody>
23383
23360
</table>
23384
-
<p>Which gives 1088 possible ways of bowling the 10th frame.</p>
23385
-
<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>
23361
+
<p>Which gives 2289 possible ways of bowling the 10th frame.</p>
23362
+
<p>So this gives a total count of possible 5 pin bowling games of <img src="https://latex.codecogs.com/png.latex?2289%20%5Ctimes%201024%5E9"> which is about <img src="https://latex.codecogs.com/png.latex?2.8%20%5Ctimes%2010%5E%7B30%7D"></p>
<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>
<p>For a single frame there are three possibilities:</p>
497
-
<tableclass="caption-top table">
498
-
<thead>
499
-
<trclass="header">
500
-
<thstyle="text-align: left;">Frame</th>
501
-
<thstyle="text-align: left;">Number of Possibilities</th>
502
-
<thstyle="text-align: left;">description</th>
503
-
</tr>
504
-
</thead>
505
-
<tbody>
506
-
<trclass="odd">
507
-
<tdstyle="text-align: left;">? ? ?</td>
508
-
<tdstyle="text-align: left;">32</td>
509
-
<tdstyle="text-align: left;">All three balls are thrown</td>
510
-
</tr>
511
-
<trclass="even">
512
-
<tdstyle="text-align: left;">X</td>
513
-
<tdstyle="text-align: left;">1</td>
514
-
<tdstyle="text-align: left;">A single ball is thrown, a strike</td>
515
-
</tr>
516
-
<trclass="odd">
517
-
<tdstyle="text-align: left;">? \</td>
518
-
<tdstyle="text-align: left;">31</td>
519
-
<tdstyle="text-align: left;">Two balls are thrown, a spare</td>
520
-
</tr>
521
-
</tbody>
522
-
</table>
523
-
<p>So there are 64 possible ways of bowling a single frame of bowling <em>that is not the tenth frame</em>.</p>
492
+
<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.</p>
493
+
<p>First let’s consider a single frame (within the first 9 frames). The order that pins are knocked down – the score per ball – matters because of the way strikes and spares are counted. So we are trying to answer the question “how many ways can 5 pins be divided into 4 categories (hit by ball 1, 2, or 3, or left standing)?” This is the sum of a multinomial and is well known, for <em>n</em> objects divided into <em>m</em> categories, the number of ways is <spanclass="math inline">\(m^n\)</span>: <spanclass="math inline">\(4^5 = 1024\)</span></p>
524
494
<p>The tenth frame has some extra rules, so lets go through it:</p>
<tdstyle="text-align: left;">Every follow up to every spare except ?\- which were counted in ???</td>
553
528
</tr>
554
529
</tbody>
555
530
</table>
556
-
<p>Which gives 1088 possible ways of bowling the 10th frame.</p>
557
-
<p>So this gives a total count of possible 5 pin bowling games of <spanclass="math inline">\(1088 \times 64^9\)</span> which is about <spanclass="math inline">\(2 \times 10^{19}\)</span></p>
531
+
<p>Which gives 2289 possible ways of bowling the 10th frame.</p>
532
+
<p>So this gives a total count of possible 5 pin bowling games of <spanclass="math inline">\(2289 \times 1024^9\)</span> which is about <spanclass="math inline">\(2.8 \times 10^{30}\)</span></p>
"text": "Trying everything\nWhile hanging out at the lanes a few obvious impossible scores got thrown out: a 1, obviously, but also a 449 – 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.\nMaybe we can brute-force this and try every conceivable bowling game? However there are a lot of possible bowling games. As a first pass, there are thirty balls thrown in a game and each ball has up to fourteen possible pinfall scores (0 through 15 excluding 1 and 14). This would give 1430 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.\nBut 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 13 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.\n\n\n\n\n\n\nNote\n\n\n\nI 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):\n\\[\n\\binom{5}{0} + \\binom{5}{1} + \\binom{5}{2} + \\binom{5}{3} + \\binom{5}{4} + \\binom{5}{5} = 32\n\\]\nFor a single frame there are three possibilities:\n\n\n\nFrame\nNumber of Possibilities\ndescription\n\n\n\n\n? ? ?\n32\nAll three balls are thrown\n\n\nX\n1\nA single ball is thrown, a strike\n\n\n? \\\n31\nTwo balls are thrown, a spare\n\n\n\nSo there are 64 possible ways of bowling a single frame of bowling that is not the tenth frame.\nThe tenth frame has some extra rules, so lets go through it:\n\n\n\nFrame\nNumber of Possibilities\ndescription\n\n\n\n\n? ? ?\n32\nAnything but a strike or spare in the first two balls\n\n\nX ? ?\n32\nFirst ball a strike, two non strikes\n\n\nX X ?\n32\nTwo strikes, one anything\n\n\n? \\ ?\n992\nA spare (31 options) and anything (32)\n\n\n\nWhich gives 1088 possible ways of bowling the 10th frame.\nSo this gives a total count of possible 5 pin bowling games of \\(1088 \\times 64^9\\) which is about \\(2 \\times 10^{19}\\)"
1106
+
"text": "Trying everything\nWhile hanging out at the lanes a few obvious impossible scores got thrown out: a 1, obviously, but also a 449 – 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.\nMaybe we can brute-force this and try every conceivable bowling game? However there are a lot of possible bowling games. As a first pass, there are thirty balls thrown in a game and each ball has up to fourteen possible pinfall scores (0 through 15 excluding 1 and 14). This would give 1430 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.\nBut 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 13 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.\n\n\n\n\n\n\nNote\n\n\n\nI 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.\nFirst let’s consider a single frame (within the first 9 frames). The order that pins are knocked down – the score per ball – matters because of the way strikes and spares are counted. So we are trying to answer the question “how many ways can 5 pins be divided into 4 categories (hit by ball 1, 2, or 3, or left standing)?” This is the sum of a multinomial and is well known, for n objects divided into m categories, the number of ways is \\(m^n\\): \\(4^5 = 1024\\)\nThe tenth frame has some extra rules, so lets go through it:\n\n\n\n\n\n\n\n\nFrame\nNumber of Possibilities\ndescription\n\n\n\n\n? ? ?\n1024\nAny combination of the first set of 5 pins\n\n\nX ? ?\n35 -1 = 242\nEvery follow up to a strike except X–, which has already been counted in ???\n\n\nX X ?\n25 -1 = 31\nEvery follow up to 2 strikes except XX-, which has already been counted in X??\n\n\n? \\ ?\n25 × (25-1) = 992\nEvery follow up to every spare except ?\\- which were counted in ???\n\n\n\nWhich gives 2289 possible ways of bowling the 10th frame.\nSo this gives a total count of possible 5 pin bowling games of \\(2289 \\times 1024^9\\) which is about \\(2.8 \\times 10^{30}\\)"
0 commit comments