document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=125652>Question 125652</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A urn has 9 white balls and 11 black balls.  A ball is drawn and then replaced.  If you draw a white ball you win 5 cents, if you draw a black ball you lose 5 cents.  You have a dollar to gamble with, your opponent has 50 cents.  IF you play until one of you loses his money, what is the probability you will lose your dollar? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #92167 by </B><A HREF=/tutors/aboutme.mpl?userid=kev82><B>kev82(151)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=kev82><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=92167\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is a one dimensional random walk with absorbing barriers, more commonly known as the gambler's ruin problem.\r<BR>\n" );
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document.write( "At this point note that there are 3 outcomes I win, you win, game goes on indefinitely, so P(I win)+P(you win) does not necessarily add to 1.\r<BR>\n" );
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document.write( "Let's think of this in terms of 5 cents. to make it easier. Initially I have 1 dollar(20x5cents) and you have half of that (10x5cents). The game is over when one of us has nothing and the other has 30x5 cents.\r<BR>\n" );
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document.write( "Let i be the number of 5 cents someone has, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p_i\"> be my winning probability from that position and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?q_i\"> be your winning probability. Let <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p\"> be the probability I win a game 9/20 and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?q\"> be the probability you win a game 11/20.\r<BR>\n" );
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document.write( "Now, the probability I win with i coins is the probability I win this game, and that I win with i+1 coins, or that I lose this game and win with i-1 coins. (Not these two events are mutually exclusive) So\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p_i = pp_{i+1} + qp_{i-1}\">\r<BR>\n" );
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document.write( "If we have 0 coins we can't win, sp <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p_0=0\"> and if we have 30 coins we have won so <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p_{30}=1\">. Our task is to find <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p_{20}\">.\r<BR>\n" );
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document.write( "I can write <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p+q=1\"> so I can write <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p_i = pp_i+qp_i\">. Substituting this in and rearranging gives\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?(p_{i+1}-p_i) - \frac{q}{p}(p_i - p_{i+1})\">\r<BR>\n" );
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document.write( "Now let me define <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?z_i = p_{i+1}-p_i\">. This leads to the recurrence <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?z_{i+1}=\frac{q}{p}z_i\"> This has the well known solution <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?z_n = z_0\left(\frac{q}{p}\right)^n\">. The proble is that we don't know <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?z_0\">. Let's move on though and consider\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\sum_{i=1}^{n-1} z_i\">\r<BR>\n" );
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document.write( "Notice that it is a collapsing sum and actually evaluates to <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p_n - p_1\">. Doing the actual sum gives a standard geometric series which I'm sure you can evaluate. This is great because we know <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p_{30}=1\"> so using this we can calculate <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p_1\">. Given this we can either use the p recurrence, or the sum of the z recurrence (we now know <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?z_0\">) so we can calculate <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?p_{20}\">. I get it as\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\frac{11^{20}9^{10} - 9^{30}}{11^{30}-9^{30}}\">\r<BR>\n" );
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document.write( "Which I calculate to be about 13.2% </SPAN>\n" );
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