document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1158491>Question 1158491</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi Professor:\r<BR>\n" );
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document.write( "I have a question related to the probability, a stock price is currently at $10, what will be the probability the stock price will be above $10 after 5 days? Assume the stock price will be randomly moved according to the following exact percentages: -3%, -2%, -1%, 1%, 2% and 3% (so there are total 6 different possible percentages movement and the movement are purely random among each day). \r<BR>\n" );
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document.write( "My approach to this problem is the following: there are total 6 different movements, and the total trials are 5 days, so the total permutations will be 6^5=7776, but I just can't continue the rest due to my limited math knowledge, could you help me out on this?\r<BR>\n" );
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document.write( "Thank you for reading my email and I am looking forward to hearing from you soon. \r<BR>\n" );
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document.write( "Alex <BR>\n" );
document.write( "California\r<BR>\n" );
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document.write( "Hi Tutors:\r<BR>\n" );
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document.write( "Thank you for answering my questions (#1158324) above (at Answer 781251). \r<BR>\n" );
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document.write( "I think the answer (781251) just solved one of the permutation, what if the percentage change were repeatable negative percentages, these percentage changes are all independent, so it could have like: -3% every single day, or -3% for 3 days then another 2 days -2%, etc.  I wanted to know what are the probability under all the permutations, what are the chances that the stock price at the end of 5th day will be above $10?\r<BR>\n" );
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document.write( "again, thanks<BR>\n" );
document.write( "Alex<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #781450 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=781450\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "original response deleted... I'm looking at this further....<br><BR>\n" );
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document.write( "In my earlier attempt at solving this, I tried to simplify the problem by treating all the percent increases and decreases as being relative to the original price.  That way, for instance,  increases of 3% and 3% would exactly balance decreases of 1%, 2%, and 3%.<br><BR>\n" );
document.write( "However, that is probably not how the problem was intended.<br><BR>\n" );
document.write( "If the percentage increases or decreases are treated as multipliers, there is NO permutation of 5 of the 6 possible percentage changes that results in an ending price exactly equal to the starting price.<br><BR>\n" );
document.write( "So I looked at the problem again....<br><BR>\n" );
document.write( "Certainly the other tutor missed the point of the problem, finding the percent increase or decrease if each of the changes is applied once.<br><BR>\n" );
document.write( "It appears to me that a purely analytic solution would be extremely tedious, making it necessary to examine each permutation of 5 of the 6 percentage changes.<br><BR>\n" );
document.write( "So I built an excel spreadsheet with all 6^5=7776 permutations of 5 of the 6 and identified the ones that produced a product greater than 1.<br><BR>\n" );
document.write( "ANSWER: 3588 of the 7776 permutations of 5 of the 6 percentage changes produce a product greater than 1.  Therefore, the probability that the stock price will be above the original $10 after 5 days is 3588/7776, or about 46.142%.<br><BR>\n" );
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document.write( "In response to the reader's question, here is how to make the spreadsheet.<br><BR>\n" );
document.write( "The 6 numbers you are using repeatedly are 1.03, 1.02, 1.01, 0.99, 0.98, and 0.97.<br><BR>\n" );
document.write( "There will be 6^5 = 7776 rows in the table.<br><BR>\n" );
document.write( "In column A there will be 6^4=1296 entries of 1.03, followed by 1296 entries of 1.02, followed by... etc.<br><BR>\n" );
document.write( "In the first 6^4=1296 entries in column B, there will be 6^3=216 entries of 1.03, followed by 216 entries of 1.02, followed by ... etc.<br><BR>\n" );
document.write( "Then you can copy those first 1296 entries in column B 5 more times to complete column B.<br><BR>\n" );
document.write( "So you can see building the spreadsheet is not really as big a task as it seems, because you can do a huge amount of copying and pasting.<br><BR>\n" );
document.write( "Similarly, in column C, you have repeated blocks of 36 entries of 1.03, followed by 36 entries of 1.02, ... etc.<br><BR>\n" );
document.write( "And in column D the repeated pattern is blocks of 6 of each of the 6 numbers.<br><BR>\n" );
document.write( "And finally in column E the 6 numbers repeat in a cycle through the whole 7776 entries.<br><BR>\n" );
document.write( "Now the spreadsheet shows each possible permutation of 5 of the 6 possible numbers exactly once.<br><BR>\n" );
document.write( "Now in column F, make each entry the product of the entries in columns A through E of that row.  Those numbers, being the product of 5 numbers between 0.97 and 1.03, will all be close to 1.  We are of course interested in how many of them are greater than 1.<br><BR>\n" );
document.write( "To count how many of the 7776 entries in this column are greater than 1, there are 2 steps.<br><BR>\n" );
document.write( "(1) in row 1, column G, enter<br><BR>\n" );
document.write( "=IF(F1>1,1,0)<br><BR>\n" );
document.write( "Then copy that formula down through all 7776 rows of column G.<br><BR>\n" );
document.write( "That will put a 1 in column G wherever the value in column F is greater than 1 and a value of 0 wherever that value is less than 1.<br><BR>\n" );
document.write( "Then to count the number of permutations for which the product is greater than 1, simply sum the entries in column G.<br><BR>\n" );
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