<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
		>
<channel>
	<title>Comments on: Game theory in Mario Party</title>
	<atom:link href="http://mindyourdecisions.com/blog/2010/01/19/game-theory-in-mario-party/feed/" rel="self" type="application/rss+xml" />
	<link>http://mindyourdecisions.com/blog/2010/01/19/game-theory-in-mario-party/</link>
	<description>Articles on game theory and personal finance</description>
	<lastBuildDate>Tue, 22 May 2012 15:59:00 +0000</lastBuildDate>
	<sy:updatePeriod>hourly</sy:updatePeriod>
	<sy:updateFrequency>1</sy:updateFrequency>
	<generator>http://wordpress.org/?v=3.3.2</generator>
	<item>
		<title>By: Stephen</title>
		<link>http://mindyourdecisions.com/blog/2010/01/19/game-theory-in-mario-party/comment-page-1/#comment-13777</link>
		<dc:creator>Stephen</dc:creator>
		<pubDate>Sat, 14 Apr 2012 04:43:00 +0000</pubDate>
		<guid isPermaLink="false">http://mindyourdecisions.com/blog/?p=1922#comment-13777</guid>
		<description>I&#039;m currently studying Honeycomb Havoc as part of a math project. I have devised the obvious strategy for two players remaining, as well as strategies when three and four players remain. The strategy is circumstantial, however, and depends if one&#039;s opponents are playing with optimal strategy as well. Even in that case, there is no set solution; as was described by another post, there comes a point when a player gets to choose who is eliminated (say when there are 2 fruits left, and you can choose 1 or 2). Several of these cases arise, which is why there is no exact solution. I will add, however, that in the first round with four players, if a player has 9 fruits left on his turn, he can guarantee that he progresses onto the next round by taking 2. Give it some thought and it will come to you.</description>
		<content:encoded><![CDATA[<p>I&#8217;m currently studying Honeycomb Havoc as part of a math project. I have devised the obvious strategy for two players remaining, as well as strategies when three and four players remain. The strategy is circumstantial, however, and depends if one&#8217;s opponents are playing with optimal strategy as well. Even in that case, there is no set solution; as was described by another post, there comes a point when a player gets to choose who is eliminated (say when there are 2 fruits left, and you can choose 1 or 2). Several of these cases arise, which is why there is no exact solution. I will add, however, that in the first round with four players, if a player has 9 fruits left on his turn, he can guarantee that he progresses onto the next round by taking 2. Give it some thought and it will come to you.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: John Jr.</title>
		<link>http://mindyourdecisions.com/blog/2010/01/19/game-theory-in-mario-party/comment-page-1/#comment-7276</link>
		<dc:creator>John Jr.</dc:creator>
		<pubDate>Sun, 05 Sep 2010 16:10:13 +0000</pubDate>
		<guid isPermaLink="false">http://mindyourdecisions.com/blog/?p=1922#comment-7276</guid>
		<description>1- For more items, the losing positions would be the multiples of (number of items plus 1).

2- same question, or I didn&#039;t got it...

With more than 2 people it is not as deterministic, but you can be sure to win if, in the end of your turn, there are (number of opponents * 2 + 1) items left for the next player.</description>
		<content:encoded><![CDATA[<p>1- For more items, the losing positions would be the multiples of (number of items plus 1).</p>
<p>2- same question, or I didn&#8217;t got it&#8230;</p>
<p>With more than 2 people it is not as deterministic, but you can be sure to win if, in the end of your turn, there are (number of opponents * 2 + 1) items left for the next player.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: V Paul Smith Jr</title>
		<link>http://mindyourdecisions.com/blog/2010/01/19/game-theory-in-mario-party/comment-page-1/#comment-6257</link>
		<dc:creator>V Paul Smith Jr</dc:creator>
		<pubDate>Thu, 21 Jan 2010 07:49:20 +0000</pubDate>
		<guid isPermaLink="false">http://mindyourdecisions.com/blog/?p=1922#comment-6257</guid>
		<description>I remember that Survivor episode very clearly. I was mad that I wasn&#039;t on one of those teams. Well, mostly the team that got to go first, but even if not, it&#039;s likely that you can quickly pull it around to your favor if your opponent doesn&#039;t know what&#039;s up and doesn&#039;t happen go guess correctly every time by mistake.

Out of curiosity, how would one go about solving the best optimal play with more than 2 people?</description>
		<content:encoded><![CDATA[<p>I remember that Survivor episode very clearly. I was mad that I wasn&#8217;t on one of those teams. Well, mostly the team that got to go first, but even if not, it&#8217;s likely that you can quickly pull it around to your favor if your opponent doesn&#8217;t know what&#8217;s up and doesn&#8217;t happen go guess correctly every time by mistake.</p>
<p>Out of curiosity, how would one go about solving the best optimal play with more than 2 people?</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: RohoMech</title>
		<link>http://mindyourdecisions.com/blog/2010/01/19/game-theory-in-mario-party/comment-page-1/#comment-6254</link>
		<dc:creator>RohoMech</dc:creator>
		<pubDate>Tue, 19 Jan 2010 20:09:50 +0000</pubDate>
		<guid isPermaLink="false">http://mindyourdecisions.com/blog/?p=1922#comment-6254</guid>
		<description>Given that the two-player version of this game is pretty simple to solve, I&#039;m surprised the game doesn&#039;t also introduce an element of skill or chance into the mix:  Either making the dice spin much faster, or making the numbers change randomly.

Granted, its also luck that your competitor doesn&#039;t know the optimal strategy...</description>
		<content:encoded><![CDATA[<p>Given that the two-player version of this game is pretty simple to solve, I&#8217;m surprised the game doesn&#8217;t also introduce an element of skill or chance into the mix:  Either making the dice spin much faster, or making the numbers change randomly.</p>
<p>Granted, its also luck that your competitor doesn&#8217;t know the optimal strategy&#8230;</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: William Spaniel</title>
		<link>http://mindyourdecisions.com/blog/2010/01/19/game-theory-in-mario-party/comment-page-1/#comment-6252</link>
		<dc:creator>William Spaniel</dc:creator>
		<pubDate>Tue, 19 Jan 2010 10:11:04 +0000</pubDate>
		<guid isPermaLink="false">http://mindyourdecisions.com/blog/?p=1922#comment-6252</guid>
		<description>Nim was a homework problem in the first game theory class I took. I immediately thought back to Mario Party and this game. I now make people I&#039;m teaching game theory to solve both the two player game seen in Honeycomb Havoc and a general game where you can pick up to k items. I won&#039;t spoil the answer.

One interesting thing to note about the game is that at various points you get to choose who you want to lose. With three players left and two items remaining before the honeycomb, I can grab two and force the next player to lose or I can grab one, which will induce the next player to grab one, which in turn makes the third player lose. This is important for deciding which player goes first in the final round, which determines who ultimately wins the game.</description>
		<content:encoded><![CDATA[<p>Nim was a homework problem in the first game theory class I took. I immediately thought back to Mario Party and this game. I now make people I&#8217;m teaching game theory to solve both the two player game seen in Honeycomb Havoc and a general game where you can pick up to k items. I won&#8217;t spoil the answer.</p>
<p>One interesting thing to note about the game is that at various points you get to choose who you want to lose. With three players left and two items remaining before the honeycomb, I can grab two and force the next player to lose or I can grab one, which will induce the next player to grab one, which in turn makes the third player lose. This is important for deciding which player goes first in the final round, which determines who ultimately wins the game.</p>
]]></content:encoded>
	</item>
</channel>
</rss>

