document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=191368>Question 191368</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the probability of rolling a die three times, and getting outcomes of 1, 2, and 3, in this order.\r<BR>\n" );
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document.write( "I tried to work as hard as I can on this problem but I can't find a way to know the probability that will be in THE SAME ORDER as the question asks.\r<BR>\n" );
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document.write( "I did (1/6) x (1/6) x (1/6) and that got me 1/216\r<BR>\n" );
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document.write( "I don't know... Should I do (1/6)^3 x (1/6)^3 x (1/6)^3?\r<BR>\n" );
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document.write( "I hope this doesn't count as one of my daily problem limit because I really tried my hardest.\r<BR>\n" );
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document.write( "Thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #143615 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=143615\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Garamond\" size=\"+2\">\r<BR>\n" );
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document.write( "You were right the first time.  You have three independent events, each with a probability of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE {1 \over 6}\">, hence:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \left({1 \over 6}\right)\left({1 \over 6}\right)\left({1 \over 6}\right) = {1 \over 216}\">\r<BR>\n" );
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document.write( "If the order didn't matter, then you would have:\r<BR>\n" );
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document.write( "A 1 in 2 chance (i.e. 3 out of 6) ways to get a 1, a 2, or a 3, then\r<BR>\n" );
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document.write( "A 1 in 3 chance (i.e. 2 out of 6) ways to get one of the two remaining numbers, then\r<BR>\n" );
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document.write( "A 1 in 6 chance of getting the last remaining number, for:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \left({1 \over 2}\right)\left({1 \over 3}\right)\left({1 \over 6}\right) = {1 \over 36}\">\r<BR>\n" );
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document.write( "Now this all makes sense if you consider that the probability of getting 1, 2, and 3 in any other specific order is also <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE {1 \over 216}\">, and there are 6 different orders that you can have for three numbers (3! = 6), and then:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 6 \times {1 \over 216} = {6 \over 216} = {1 \over 36}\">\r<BR>\n" );
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document.write( "The exact same probability as if order didn't matter.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
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