<PRE><B>Question 806336</B>
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Refer to the below table of dice results for 2 six-sided dice:

&lt;table border=&quot;1&quot; cellpadding=&quot;2&quot; cellspacing=&quot;2&quot; width=&quot;60%&quot;&gt;
&lt;tr&gt;
&lt;th width=&quot;20%&quot;&gt;Sum&lt;/th&gt;
&lt;th width=&quot;20%&quot;&gt;Ways&lt;/th&gt;
&lt;th width=&quot;20%&quot;&gt;# of Ways&lt;/th&gt;
&lt;/tr&gt;
&lt;tr&gt;
&lt;td width=&quot;20%&quot;&gt;2&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;1,1&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;1&lt;/td&gt;
&lt;/tr&gt;
&lt;tr&gt;
&lt;td width=&quot;20%&quot;&gt;3&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;1,2; 2,1&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;2&lt;/td&gt;
&lt;/tr&gt;
&lt;tr&gt;
&lt;td width=&quot;20%&quot;&gt;4&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;1,3; 2,2; 3,1&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;3&lt;/td&gt;
&lt;/tr&gt;
&lt;tr&gt;
&lt;td width=&quot;20%&quot;&gt;5&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;1,4; 2,3; 3,2; 4,1&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;4&lt;/td&gt;
&lt;/tr&gt;
&lt;tr&gt;
&lt;td width=&quot;20%&quot;&gt;6&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;1,5; 2,4; 3,3; 4,2; 5,1&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;5&lt;/td&gt;
&lt;/tr&gt;
&lt;tr&gt;
&lt;td width=&quot;20%&quot;&gt;7&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;1,6; 2,5; 3,4; 4,3; 5,2; 6,1&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;6&lt;/td&gt;
&lt;/tr&gt;
&lt;tr&gt;
&lt;td width=&quot;20%&quot;&gt;8&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;2,6; 3,5; 4,4; 5,3; 6,2&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;5&lt;/td&gt;
&lt;/tr&gt;
&lt;tr&gt;
&lt;td width=&quot;20%&quot;&gt;9&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;3,6; 4,5; 5,4; 6,3&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;4&lt;/td&gt;
&lt;/tr&gt;
&lt;tr&gt;
&lt;td width=&quot;20%&quot;&gt;10&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;4,6; 5,5; 6,4&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;3&lt;/td&gt;
&lt;/tr&gt;
&lt;tr&gt;
&lt;td width=&quot;20%&quot;&gt;11&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;5,6; 6,5&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;2&lt;/td&gt;
&lt;/tr&gt;
&lt;tr&gt;
&lt;td width=&quot;20%&quot;&gt;12&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;6,6&lt;/td&gt;&lt;td width=&quot;20%&quot;&gt;1&lt;/td&gt;
&lt;/tr&gt;
&lt;/table&gt;


Note that there are 36 different possible results, 6 of which are doubles, 2 of which are 3, and one of which is 12.  However, the 12 result is also doubles.  Because you don&#39;t specify what happens when a 12 is rolled (score 15 for the 12 which overrides the doubles or score 5 for the doubles which overrides the 12, or score 20 for meeting both criteria), I&#39;m going to go with a literal interpretation of the given rules: The player gets 5 for rolling doubles AND 15 for rolling 12.


So the probability of rolling doubles OTHER than 2 sixes = 12 is *[tex \LARGE \frac{5}{6}], the probability of rolling 12 is *[tex \LARGE \frac{1}{36}], the probability of rolling 3 is *[tex \LARGE \frac{1}{18}], and the probability of rolling anything else, which we must presume represents a loss of the initial $3 wager, is *[tex \LARGE \frac{7}{9}] (28 ways to lose out of 36 outcomes).


And then the expected payout is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ 5*\frac{5}{36}\ +\ 20*\frac{1}{36}\ +\ 15*\frac{1}{18}\ -\ 3*\frac{7}{9}]


You can do your own arithmetic, but in the long run you lose a quarter every time you play.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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