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Your original thinking is very cool, and it's a very very good approximation.#ed_op#BR#ed_cl##ed_op#BR#ed_cl#I'm sorry for skiping some steps/explanations at first. #ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#
QC 寫到:To get ONLY ONE successive roll of a set of (+)(+)(+) at the END by n rolls, we have to get only one successive roll of a set of (-)(+)(+)(+)at the END by n rolls when n&gt;3.#ed_op#br#ed_cl#So, p(n+4)=(1-(p(0)+P(1)+...+p(n)))*qppp#ed_op#br#ed_cl##ed_op#br#ed_cl#That is the reason of different recurrent relationship form (+)(-)(-) or (alpha, theta, mu). OK?
#ed_op#br#ed_cl#Well, thank you very very much for correcting my annoying wrong interpretation of the question and your correct answer.#ed_op#br#ed_cl#The main point was that my assumption that "All events are independent" is wrong.#ed_op#br#ed_cl#Thank you for your correct solution and detail explanation and sorry for my misunderstanding.#ed_op#br#ed_cl#
To get ONLY ONE successive roll of a set of (+)(+)(+) at the END by n rolls, we have to get only one successive roll of a set of (-)(+)(+)(+)at the END by n rolls when n>3.
So, p(n+4)=(1-(p(0)+P(1)+...+p(n)))*qppp

That is the reason of different recurrent relationship form (+)(-)(-) or (alpha, theta, mu). OK?
#ed_op#P#ed_cl#What is the difference to get a set of "altha,theta, and mu" and to get a "(+),(+),(+)"?#ed_op#BR#ed_cl#They are really different. I've shown you the difference in the recurrent relationship. Then the result changed according to the recurrent relationship. #ed_op#BR#ed_cl##ed_op#BR#ed_cl#If you flip an unfair coin/die in order to get the pattern "(+)(+)(+)", #ed_op#BR#ed_cl#and the chance to get (+) in one roll is p, the GF is different from "(+)(-)(-)" because the recurrent relationship is different:#ed_op#BR#ed_cl#let q=1-p #ed_op#BR#ed_cl#p(n+4)=(1-(p(0)+P(1)+...+p(n)))*qppp#ed_op#BR#ed_cl#(F-pppxxx)/xxxx=qppp(1-F)/(1-x)#ed_op#BR#ed_cl#F=pppxxx/(1-qx-qpxx-qppxxx)#ed_op#BR#ed_cl##ed_op#BR#ed_cl#when you want to take p=1/2, q=1/2:#ed_op#BR#ed_cl#F=1/8*xxx/(1-x/2-xx/4-xxx/8)#ed_op#BR#ed_cl##ed_op#BR#ed_cl#we get another recurrent relationship: #ed_op#BR#ed_cl#p(n+3)=p(n+2)/2+p(n+1)/4+p(n)/8#ed_op#BR#ed_cl##ed_op#BR#ed_cl#then#ed_op#BR#ed_cl#p(0)=p(1)=p(2)=0#ed_op#BR#ed_cl#p(3)=1/8#ed_op#BR#ed_cl#p(4)=1/16#ed_op#BR#ed_cl#p(5)=p(4)/2+p(3)/4+p(2)/8=1/16#ed_op#BR#ed_cl#p(6)=p(5)/2+p(4)/4+p(3)/8=1/16#ed_op#BR#ed_cl#p(7)=p(6)/2+p(5)/4+p(4)/8=7/128#ed_op#BR#ed_cl#No problem at all, I think.#ed_op#BR#ed_cl##ed_op#BR#ed_cl#Notice that p(3)=1/8, not 1/2. #ed_op#/P#ed_cl##ed_op#P#ed_cl#If&nbsp;you flip an unfair coin to get a pattern of #ed_op#STRONG#ed_cl#(+)(+)(+)#ed_op#/STRONG#ed_cl# with the probability of (+) is #ed_op#STRONG#ed_cl#4/5#ed_op#/STRONG#ed_cl##ed_op#/P#ed_cl##ed_op#P#ed_cl#p(n+4)=(1-(p(0)+P(1)+...+p(n)))*qppp#ed_op#/P#ed_cl##ed_op#P#ed_cl#F=pppxxx/(1-qx-qpxx-qppxxx)#ed_op#BR#ed_cl#=64/125*xxx/(1-x/5-4xx/25-16xx/125)#ed_op#/P#ed_cl##ed_op#P#ed_cl#we get another recurrent relationship: #ed_op#BR#ed_cl#p(n+3)=p(n+2)/5+p(n+1)*4/25+p(n)*16/125#ed_op#/P#ed_cl##ed_op#P#ed_cl##ed_op#TABLE style="WIDTH: 79pt; BORDER-COLLAPSE: collapse" cellSpacing=0 cellPadding=0 width=105 border=0 x:str#ed_cl##ed_op#COLGROUP#ed_cl##ed_op#STRONG#ed_cl##ed_op#COL style="WIDTH: 79pt; mso-width-source: userset; mso-width-alt: 3726" width=105#ed_cl##ed_op#/STRONG#ed_cl##ed_op#TBODY#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; WIDTH: 79pt; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" width=105 height=22#ed_cl##ed_op#STRONG#ed_cl#p(0)=0 #ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(1)=0 #ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(2)=0 #ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(3)=0.512#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(4)=0.1024#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(5)=0.1024#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(6)=0.1024#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(7)=0.0499712#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(8)=0.03948544#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(9)=0.02899968#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(10)=0.01851392#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(11)=0.01339686912#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(12)=0.009353560064#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(13)=0.006383992832#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(14)=0.004488167424#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#TR style="HEIGHT: 16.2pt" height=22#ed_cl##ed_op#TD style="BORDER-RIGHT: #ffffff; BORDER-TOP: #ffffff; BORDER-LEFT: #ffffff; BORDER-BOTTOM: #ffffff; HEIGHT: 16.2pt; BACKGROUND-COLOR: transparent" height=22#ed_cl##ed_op#STRONG#ed_cl#p(15)=0.003116328026112#ed_op#/STRONG#ed_cl##ed_op#/TD#ed_cl##ed_op#/TR#ed_cl##ed_op#/TBODY#ed_cl##ed_op#/TABLE#ed_cl##ed_op#/P#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#
QC 寫到:p(n) is the probability that you get only one successive
roll of a set of (alpha, theta and mu) at the end by n rolls.#ed_op#br#ed_cl#
#ed_op#br#ed_cl#
For p(n) &amp; F, I was considering "ONLY one successive roll of a set at the end" but not at least one.#ed_op#br#ed_cl#
When I was considering "at least one", I use s(n), and "S" as the generating function of s(n). #ed_op#br#ed_cl#
s(n)=p(0)+p(1)+p(2)+...+p(n)#ed_op#br#ed_cl#
S=F/(1-x)

#ed_op#br#ed_cl#
Well, I think I mixed up the question 2 and question 3 somewhere, but it's not the main point...#ed_op#br#ed_cl#
#ed_op#br#ed_cl##ed_op#br#ed_cl#
QC 寫到:In my calculation of r&lt;=1/27, I did not limit the die as fair or unfair one.#ed_op#br#ed_cl##ed_op#br#ed_cl#If you flip an unfair coin in order to get the pattern "(+)(-)(-)", and the chance to get (+) in one roll is p:#ed_op#br#ed_cl#r=p*(1-p)*(1-p)#ed_op#br#ed_cl#you will find r&lt;=4/27.#ed_op#br#ed_cl#we still don't have to consider r=1/2#ed_op#br#ed_cl#----------------------------------------#ed_op#br#ed_cl##ed_op#br#ed_cl#If
you flip an unfair coin/die in order to get the pattern "(+)(+)(+)",
and the chance to get (+) in one roll is p, the GF is different because
the recurrent relationship is different:#ed_op#br#ed_cl#let q=1-p #ed_op#br#ed_cl#p(n+4)=(1-(p(0)+P(1)+...+p(n)))*qppp#ed_op#br#ed_cl#(F-pppxxx)/xxxx=pppq(1-F)/(1-x)#ed_op#br#ed_cl#F=pppxxx/(1-qx-qpxx-qppxxx)#ed_op#br#ed_cl##ed_op#br#ed_cl#when you want "a set of 3 rolls of any kinds of conbinations only", take p=1, q=0#ed_op#br#ed_cl#F=pppxxx/(1-qx-qpxx-qppxxx)#ed_op#br#ed_cl#=xxx#ed_op#br#ed_cl#that is, p(3)=1, p(0)=p(1)=p(2)=p(4)=p(5)=...=0
#ed_op#br#ed_cl#then it's so easy to see as you accept that there should be no difference for the solution for a fair or unfair situation...#ed_op#br#ed_cl#how about if I flip an unfair coin to get a pattern of (+)(+)(+) with the probability of (+) is 4/5 ?? The final r = (4/5)#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl# = 64/125 &gt; 1/2#ed_op#br#ed_cl#The main point is, r must be able to be any number between 0 and 1...#ed_op#br#ed_cl##ed_op#br#ed_cl#If you put r=1/3 in your previous equation, you will still get p(10)&lt;0#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl#See, you've changed your equation...#ed_op#br#ed_cl#What is the difference to get a set of "altha,#ed_op#span class="postbody"#ed_cl#theta, and mu#ed_op#/span#ed_cl#"&nbsp; and to get a "(+),(+),(+)"?#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl#But the new equation is still incorrect:#ed_op#br#ed_cl#
#ed_op#br#ed_cl#p(n+4)=(1-(p(0)+P(1)+...+p(n)))*qppp#ed_op#br#ed_cl#
#ed_op#br#ed_cl#how about if p = 1/2? q=1/2 too...#ed_op#br#ed_cl#I assume that p(3)=1/2#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl# =1/8 as there is no definition from your equation...#ed_op#br#ed_cl#p(0)=p(1)=p(2)=0, so p(4)=p(5)=p(6)=1/16 #ed_op#br#ed_cl#do you think p(3)=1/2 and p(4)=p(5)=p(6)=1/16???#ed_op#br#ed_cl#and also p(7)= 7/128????#ed_op#br#ed_cl#Do you still think the equation is valid?#ed_op#br#ed_cl#
Notice the word "ONLY" and "one set at the end".

In my calculation of r&lt;=1/27, I did not limit the die as fair or unfair one.#ed_op#BR#ed_cl##ed_op#BR#ed_cl#If you flip an unfair coin in order to get the pattern "(+)(-)(-)", and the chance to get (+) in one roll is p:#ed_op#BR#ed_cl#r=p*(1-p)*(1-p)#ed_op#BR#ed_cl#you will find r&lt;=4/27.#ed_op#BR#ed_cl#we still don't have to consider r=1/2#ed_op#BR#ed_cl#----------------------------------------#ed_op#BR#ed_cl##ed_op#BR#ed_cl#If you flip an unfair coin/die in order to get the pattern "(+)(+)(+)", and the chance to get (+) in one roll is p, the GF is different because the recurrent relationship is different:#ed_op#BR#ed_cl#let q=1-p #ed_op#BR#ed_cl#p(n+4)=(1-(p(0)+P(1)+...+p(n)))*qppp#ed_op#BR#ed_cl#(F-pppxxx)/xxxx=pppq(1-F)/(1-x)#ed_op#BR#ed_cl#F=pppxxx/(1-qx-qpxx-qppxxx)#ed_op#BR#ed_cl##ed_op#BR#ed_cl#when you want "a set of 3 rolls of any kinds of conbinations only", take p=1, q=0#ed_op#BR#ed_cl#F=pppxxx/(1-qx-qpxx-qppxxx)#ed_op#BR#ed_cl#=xxx#ed_op#BR#ed_cl#that is, p(3)=1, p(0)=p(1)=p(2)=p(4)=p(5)=...=0 #ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#
p(n) is the probability that you get only one successive roll of a set of (alpha, theta and mu) at the end by n rolls.

For p(n) & F, I was considering "ONLY one successive roll of a set at the end" but not at least one.
When I was considering "at least one", I use s(n), and "S" as the generating function of s(n).
s(n)=p(0)+p(1)+p(2)+...+p(n)
S=F/(1-x)
QC 寫到:What is the range of probablity "r" to have three successive rolls of "1", "2", and "3" with a k-sided die labled with number "1" to k?#ed_op#br#ed_cl#(k&gt;=3)#ed_op#br#ed_cl##ed_op#br#ed_cl#let the probablity to get "1" in one roll be P1#ed_op#br#ed_cl#let the probablity to get "2" in one roll be P2#ed_op#br#ed_cl#...#ed_op#br#ed_cl#let the probablity to get munber k in one roll be Pk#ed_op#br#ed_cl##ed_op#br#ed_cl#P1+P2+P3+...+Pk=1#ed_op#br#ed_cl#1&gt;= Pi &gt;=0 for very i=1 to k#ed_op#br#ed_cl##ed_op#br#ed_cl#r=P1*P2*P3#ed_op#br#ed_cl#&lt;=((P1+P2+P3)/3)^3#ed_op#br#ed_cl#&lt;=(1/3)^3#ed_op#br#ed_cl#=1/27#ed_op#br#ed_cl##ed_op#br#ed_cl#we don't have to consider r=1/2 because r&lt;=1/27#ed_op#br#ed_cl#
#ed_op#br#ed_cl##ed_op#br#ed_cl#How about if I flip a coin or roll an unfair die?#ed_op#br#ed_cl#Or if I just need a set of 3 rolls of any kinds of conbinations only, which obviously r = 1.#ed_op#br#ed_cl#p(0)=0, p(1)=p(2)=1, but p(3) = (1-p(0))*r = 0&nbsp; , is it possible ???#ed_op#br#ed_cl#I don't think that there should be any limit on the probability on a single event probability r.#ed_op#br#ed_cl##ed_op#br#ed_cl#If you still do not see the problem:#ed_op#br#ed_cl#
#ed_op#br#ed_cl#p(3)=p(4)=p(5) = r#ed_op#br#ed_cl#
#ed_op#br#ed_cl#Should it be possible?#ed_op#br#ed_cl#
#ed_op#br#ed_cl#(1) p(n) is the probability that you get only one successive roll of a set of (alpha, theta and mu) at the end by n rolls.#ed_op#br#ed_cl#
#ed_op#br#ed_cl#Actually, we are considering at least one successive roll of a set but not ONLY one...#ed_op#br#ed_cl##ed_op#br#ed_cl#Anyway, p(4) = p(3) means that p(4)-p(3)=0#ed_op#br#ed_cl#Should the probability of 3 rolls same with the probability of 4 rolls to get a successive roll set?#ed_op#br#ed_cl#Shouldn't we gain a larger chance of having one more roll?#ed_op#br#ed_cl##ed_op#br#ed_cl#
What is the range of probablity "r" to have three successive rolls of "1", "2", and "3" with a k-sided die labled with number "1" to k?#ed_op#BR#ed_cl#(k&gt;=3)#ed_op#BR#ed_cl##ed_op#BR#ed_cl#let the probablity to get "1" in one roll be P1#ed_op#BR#ed_cl#let the probablity to get "2" in one roll be P2#ed_op#BR#ed_cl#...#ed_op#BR#ed_cl#let the probablity to get munber k in one roll be Pk#ed_op#BR#ed_cl##ed_op#BR#ed_cl#P1+P2+P3+...+Pk=1#ed_op#BR#ed_cl#1&gt;= Pi &gt;=0 for very i=1 to k#ed_op#BR#ed_cl##ed_op#BR#ed_cl#r=P1*P2*P3#ed_op#BR#ed_cl#&lt;=((P1+P2+P3)/3)^3#ed_op#BR#ed_cl#&lt;=(1/3)^3#ed_op#BR#ed_cl#=1/27#ed_op#BR#ed_cl##ed_op#BR#ed_cl#we don't have to consider r=1/2 because r&lt;=1/27 #ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#
QC 寫到:#ed_op#p#ed_cl#
#ed_op#span style="text-decoration: underline;"#ed_cl##ed_op#/span#ed_cl#G@ry 寫到:#ed_op#/p#ed_cl##ed_op#p#ed_cl#Do you really think that p(3)=p(4)=p(5) = r, whereas p(6) = r(1-r)&nbsp;≠ r ?#ed_op#br#ed_cl##ed_op#br#ed_cl#
#ed_op#/p#ed_cl##ed_op#div#ed_cl#yes.#ed_op#/div#ed_cl##ed_op#div#ed_cl#p(3)=p(4)=p(5) = r#ed_op#/div#ed_cl##ed_op#div#ed_cl#p(6) = r(1-r)&nbsp;#ed_op#/div#ed_cl##ed_op#div#ed_cl#p(7)=r*(1-2r)#ed_op#/div#ed_cl##ed_op#div#ed_cl#p(8)=r*(1-3r)#ed_op#/div#ed_cl##ed_op#div#ed_cl#p(9)=r*(1-4r+rr)#ed_op#/div#ed_cl##ed_op#div#ed_cl#&nbsp;#ed_op#/div#ed_cl##ed_op#div#ed_cl#I don't think there is&nbsp;any reason to&nbsp;consider "p(6)=r" or not.#ed_op#/div#ed_cl##ed_op#div#ed_cl#p(6) = r(1-r)&nbsp;, no problem.#ed_op#/div#ed_cl#
#ed_op#br#ed_cl#Well, if then, do you think that p(n) can be negative for some n?#ed_op#br#ed_cl#Let me show you:#ed_op#br#ed_cl#how about if r=1/2?#ed_op#br#ed_cl#p(8) = 1/2*(1-3/2) = -1/4 &lt; 0#ed_op#br#ed_cl##ed_op#br#ed_cl#The definition of p(n) is inaccurate.#ed_op#br#ed_cl#
#ed_op#P#ed_cl#
Do you really think that p(3)=p(4)=p(5) = r, whereas p(6) = r(1-r)&nbsp;≠ r ?#ed_op#BR#ed_cl#
#ed_op#/P#ed_cl##ed_op#DIV#ed_cl#yes.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#p(3)=p(4)=p(5) = r#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#p(6) = r(1-r)&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#p(7)=r*(1-2r)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#p(8)=r*(1-3r)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#p(9)=r*(1-4r+rr)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#I don't think there is&nbsp;any reason to&nbsp;consider "p(6)=r" or not.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#p(6) = r(1-r)&nbsp;, no problem.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#
Well, I'm sorry that I misunderstood some of your points.#ed_op#br#ed_cl#But I finally find out what's the problem:#ed_op#br#ed_cl##ed_op#p#ed_cl#
#ed_op#/p#ed_cl##ed_op#p#ed_cl##ed_op#span class="postbody"#ed_cl#p(n+3)=(1- (p(0)+p(1)+p(2)+...+p(n)) )*r#ed_op#/span#ed_cl##ed_op#/p#ed_cl##ed_op#p#ed_cl#
#ed_op#br#ed_cl##ed_op#/p#ed_cl#The above formula is not appropriate.#ed_op#br#ed_cl#Assume that p(0)=p(1)=p(2)=0 [ It's really obvious...]#ed_op#br#ed_cl#Please consider the case that n=0, 1 and 2#ed_op#br#ed_cl#Do you really think that p(3)=p(4)=p(5) = r, whereas p(6) = r(1-r)&nbsp;≠ r ?#ed_op#br#ed_cl##ed_op#br#ed_cl#
generating function 我玩過20幾年了, 再加上用excel驗算, 應該是不會錯的. Sorry for eliminating some steps.#ed_op#BR#ed_cl#^___^ #ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#
#ed_op#P#ed_cl#
#ed_op#/P#ed_cl##ed_op#P#ed_cl#if you set x=0, the divided by zero situation make the whole thing undefined.#ed_op#BR#ed_cl#
#ed_op#/P#ed_cl##ed_op#DIV#ed_cl#r=1/24^3=1/13824#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#F=rx^3+rx^4+rx^5+(1-r)rx^6+....#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#F/xxx=r+rx+rx^2+(1-r)rx^3+...#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#I can't find anything undefined even when x=0.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#For H(x)=1/x, even H(0) is undefined, we can still deal with H(x), right?#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#Do I use F(0) or (F(0)/0^3)&nbsp;anywhere in my calculation? No.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#在某些點沒定義的函數仍然可以做式子的運算, 沒關係的.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#
#ed_op#P#ed_cl#(1) p(n) is the probability that you&nbsp;get only one successive roll of a set of (alpha, theta and mu) at the end by n rolls.#ed_op#BR#ed_cl##ed_op#BR#ed_cl#(2) what do you think "F" if I say F=1+x+2x^2+3x^4? Of caurse I meant F=F(x). #ed_op#BR#ed_cl#Do you think "F" or "G" really matter the result?#ed_op#BR#ed_cl#I've defined that F is the generating function ! Even I can eliminate the statement that "F is the generating function", because I've define that F=p(0)+p(1)x+p(2)x^2+...#ed_op#/P#ed_cl##ed_op#P#ed_cl#(3) what is the coefficient of x^n of (F/xxx)? It's p(n+3), right?#ed_op#BR#ed_cl#p(n+3)=(1-(p(0)+p(1)+p(2)+...+p(n))*r#ed_op#BR#ed_cl#F/xxx#ed_op#BR#ed_cl#=sigma{p(n+3)*x^n|n=0 to infinite}#ed_op#BR#ed_cl#=r* sigma{(1-(p(0)+p(1)+p(2)+...+p(n))*x^n|n=0 to infinite}#ed_op#BR#ed_cl#=r*(sigma{x^n|n=0 to infinite}-sigma{(p(0)+p(1)+p(2)+...+p(n))*x^n|n=0 to infinite})#ed_op#BR#ed_cl#=r*(1/(1-x)-F/(1-x))#ed_op#BR#ed_cl#=r*(1-F)/(1-x)#ed_op#BR#ed_cl##ed_op#BR#ed_cl#(5) In geometric distribution, p(n)=p*(1-p)^(n-1).#ed_op#BR#ed_cl#In this case, you see that p(1)=0, p(2)=0, p(3)=r, p(4)=r, p(5)=r.#ed_op#BR#ed_cl#You can tell the difference, right?#ed_op#BR#ed_cl#That's why I said it's close to geometric distribution, but not a geometric distribution exactly. #ed_op#/P#ed_cl##ed_op#P#ed_cl#(6) I've verified the result with Excel. I think my result in this problem is correct.#ed_op#/P#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#
QC 寫到:#ed_op#div#ed_cl#My method:#ed_op#/div#ed_cl##ed_op#div#ed_cl#&nbsp;#ed_op#/div#ed_cl##ed_op#div#ed_cl#Let F=p(0)+p(1)x+p(2)x^2+...#ed_op#/div#ed_cl##ed_op#div#ed_cl#F is the genereating function of p(n)#ed_op#/div#ed_cl##ed_op#div#ed_cl#&nbsp;#ed_op#/div#ed_cl##ed_op#div#ed_cl#let r=1/24^3=1/13824#ed_op#/div#ed_cl##ed_op#div#ed_cl#because p(n+3)=(1-(p(0)+p(1)+p(2)+...+p(n))*r#ed_op#/div#ed_cl##ed_op#div#ed_cl#F/xxx=r*(1-F)/(1-x)#ed_op#/div#ed_cl##ed_op#div#ed_cl#F*(1-x)=rxxx(1-F)#ed_op#/div#ed_cl##ed_op#div#ed_cl#F=rxxx/(1-x+rxxx)#ed_op#/div#ed_cl##ed_op#div#ed_cl#&nbsp;#ed_op#/div#ed_cl##ed_op#div#ed_cl#E(n)=p(1)*1+p(2)*2+...#ed_op#/div#ed_cl##ed_op#div#ed_cl#=&nbsp;F'(1)#ed_op#/div#ed_cl##ed_op#div#ed_cl#=(3rr-r*(3r-1))/rr#ed_op#/div#ed_cl##ed_op#div#ed_cl#=1/r=13824#ed_op#/div#ed_cl##ed_op#div#ed_cl#&nbsp;#ed_op#/div#ed_cl##ed_op#div#ed_cl#--------------------------#ed_op#/div#ed_cl##ed_op#div#ed_cl#
#ed_op#br#ed_cl##ed_op#br#ed_cl#Sorry, I think you get something wrong:#ed_op#br#ed_cl##ed_op#br#ed_cl#First, you did not define what is p(n);#ed_op#br#ed_cl#Even though you said
p(n+3)=(1-(p(0)+p(1)+p(2)+...+p(n))*r
Definition for p(0), p(1) and p(2) are definitely needed, [I assume that they are all 0 yet]#ed_op#br#ed_cl#And I guess that p(n) is the probability that you can get one successive roll of a set of (al#ed_op#span style="font-weight: bold;"#ed_cl#p#ed_op#/span#ed_cl#ha, theta and mu) by n rolls#ed_op#br#ed_cl##ed_op#br#ed_cl#Second, I assume that F = F(x) when not specified. And this notation is used for a cdf, we use G(x) for a pgf...#ed_op#br#ed_cl##ed_op#br#ed_cl#Third, p(n+3) is equal to G#ed_op#sup style="font-style: italic;"#ed_cl#(n+3)#ed_op#/sup#ed_cl#(0)/(n+3)! but definitely not G(x)/x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl##ed_op#br#ed_cl#And even if you set x=0, the divided by zero situation make the whole thing undefined.#ed_op#br#ed_cl##ed_op#br#ed_cl#Forth, can you show how come "1-(p(0)+p(1)+p(2)+...+p(n))" = "(1-F(x))/(1-x)"?#ed_op#br#ed_cl#G(x) = p(0)+p(1)x+p(2)x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+...+p(n)x#ed_op#sup#ed_cl#n#ed_op#/sup#ed_cl##ed_op#span style="font-weight: bold;"#ed_cl#+p(n+1)x#ed_op#/span#ed_cl##ed_op#sup style="font-weight: bold;"#ed_cl#(n+1)#ed_op#/sup#ed_cl##ed_op#span style="font-weight: bold;"#ed_cl#+p(n+2)x#ed_op#/span#ed_cl##ed_op#sup style="font-weight: bold;"#ed_cl#(n+2)#ed_op#/sup#ed_cl##ed_op#span style="font-weight: bold;"#ed_cl#+... infinitely~~#ed_op#/span#ed_cl##ed_op#br#ed_cl#It will #ed_op#span style="font-weight: bold;"#ed_cl#NOT#ed_op#/span#ed_cl# stop at any finite probability term.#ed_op#br#ed_cl##ed_op#br#ed_cl#Fifth, E(n)=1/p is a strong characteristic of a geometric distribution... and it is definitely not the answer as you've said that this is not a geometric distribution also, too.#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl#
QC 寫到:let 1/a,1/b,1/c be the 3 root of 1-x+rxxx=0#ed_op#/div#ed_cl##ed_op#div#ed_cl#套用卡丹公式 得#ed_op#/div#ed_cl##ed_op#div#ed_cl#&nbsp;#ed_op#/div#ed_cl##ed_op#div#ed_cl#a~ 0.9999276515687950#ed_op#/div#ed_cl##ed_op#div#ed_cl#b~ -0.0084693831120445#ed_op#/div#ed_cl##ed_op#div#ed_cl#c~ 0.0085417315432198#ed_op#/div#ed_cl##ed_op#div#ed_cl#&nbsp;#ed_op#/div#ed_cl##ed_op#div#ed_cl#S=F/(1-x)#ed_op#/div#ed_cl##ed_op#div#ed_cl#=rxxx/(1-x+rxxx)(1-x)#ed_op#/div#ed_cl##ed_op#div#ed_cl#~1/(1-x)-1.0001447278597800/(1-ax)-0.0041815685963448/(1-bx)+0.0043262968712862/(1-cx)#ed_op#/div#ed_cl##ed_op#div#ed_cl#&nbsp;#ed_op#/div#ed_cl##ed_op#div#ed_cl#s(n)~1-1.0001447278597800*a^n-0.0041815685963448*b^n +0.0043262968712862*c^n#ed_op#/div#ed_cl##ed_op#div#ed_cl#&nbsp;#ed_op#/div#ed_cl##ed_op#div#ed_cl#when n &gt;100, b^n and c^n ~0#ed_op#/div#ed_cl##ed_op#div#ed_cl#so#ed_op#/div#ed_cl##ed_op#div#ed_cl#s(n)~1-1.0001447278597800*a^n#ed_op#/div#ed_cl##ed_op#div#ed_cl#0.9~1-1.0001447278597800*a^n#ed_op#/div#ed_cl##ed_op#div#ed_cl#n&gt;= log((1-0.9)/1.0001447278597800)/log(0.9999276515687950)=31827.17955#ed_op#/div#ed_cl##ed_op#div#ed_cl#the smallest n=31828#ed_op#/div#ed_cl##ed_op#div#ed_cl#&nbsp;#ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl#
#ed_op#br#ed_cl#The first part is wrong, and certainly this part will get wrong too...#ed_op#br#ed_cl#
#ed_op#DIV#ed_cl#My method:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#Let F=p(0)+p(1)x+p(2)x^2+...#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#F is the genereating function of p(n)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#let r=1/24^3=1/13824#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#because p(n+3)=(1-(p(0)+p(1)+p(2)+...+p(n))*r#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#F/xxx=r*(1-F)/(1-x)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#F*(1-x)=rxxx(1-F)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#F=rxxx/(1-x+rxxx)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#E(n)=p(1)*1+p(2)*2+...#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=&nbsp;F'(1)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=(3rr-r*(3r-1))/rr#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=1/r=13824#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#--------------------------#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#let 1/a,1/b,1/c be the 3 root of 1-x+rxxx=0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#套用卡丹公式 得#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#a~ 0.9999276515687950#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#b~ -0.0084693831120445#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#c~ 0.0085417315432198#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#S=F/(1-x)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=rxxx/(1-x+rxxx)(1-x)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#~1/(1-x)-1.0001447278597800/(1-ax)-0.0041815685963448/(1-bx)+0.0043262968712862/(1-cx)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#s(n)~1-1.0001447278597800*a^n-0.0041815685963448*b^n +0.0043262968712862*c^n#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#when n &gt;100, b^n and c^n ~0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#so#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#s(n)~1-1.0001447278597800*a^n#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#0.9~1-1.0001447278597800*a^n#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#n&gt;= log((1-0.9)/1.0001447278597800)/log(0.9999276515687950)=31827.17955#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#the smallest n=31828#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

### [數學]cool!!!

#ed_op#DIV#ed_cl#Very very cool!!!#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#It's almost a geometric distribution, but not a geometric distribution exactly, I think.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#I think the answer of Q3 is 31828, which is very close to yours.#ed_op#/DIV#ed_cl#

### Re: [數學]FAMAT fall 2007 interschool (A)

#ed_op#span style="background-color: rgb(255, 255, 0);"#ed_cl#Verified WRONG SOLUTION REPLY#ed_op#/span#ed_cl##ed_op#br#ed_cl#please ignore it for correct solution#ed_op#br#ed_cl##ed_op#br#ed_cl#
QC 寫到:#ed_op#div#ed_cl#2. Suppose you have a 24-sided fair die, with each face labeled with exactly one of the 24#ed_op#br#ed_cl#letters from the Greek alphabet. If you were to begin rolling the die, and keep rolling#ed_op#br#ed_cl#until you have three successive rolls of alpha, theta, and mu (in that order), what is the expected number of rolls until you reach your goal?#ed_op#/div#ed_cl##ed_op#div#ed_cl#
#ed_op#br#ed_cl#The probability distribution is a geometric distribution.#ed_op#br#ed_cl#Let every set of 3 rolls be a single event, the probability of a event of (alpha, theta and mu)#ed_op#br#ed_cl#is (1/24)#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl# = 1/13824#ed_op#br#ed_cl#then the expected value for the number of events occur for the distribution#ed_op#br#ed_cl#is 1 / (probatility of single event) = 13824#ed_op#br#ed_cl#The number of rolls needed for 13824 sets of 3 dies = 13824+3-1 = 13826 rolls.#ed_op#br#ed_cl#&nbsp;#ed_op#br#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#br#ed_cl#
QC 寫到:#ed_op#br#ed_cl#3.Assume the same conditions as in question 2, except now you're rolling blindfolded#ed_op#br#ed_cl#and someone else will record the letters that you roll, but they won't stop you when#ed_op#br#ed_cl#your goal is reached. How many rolls would you need to make before stopping to have#ed_op#br#ed_cl#greater than 90% probability of having three consecutive rolls of&nbsp;alpha, theta, and mu (in that order) somewhere in the sequence of your rolls?#ed_op#/div#ed_cl#
#ed_op#br#ed_cl#cdf of a geometric distribution = 1-(1-p)#ed_op#sup#ed_cl#k#ed_op#/sup#ed_cl##ed_op#br#ed_cl#cdf of the distribution of a set of 3 dies = 1-(1-1/13824)#ed_op#sup#ed_cl#k#ed_op#/sup#ed_cl# &gt; 90%#ed_op#br#ed_cl#=&gt; 0.1 &gt; (1-1/13824)#ed_op#sup#ed_cl#k#ed_op#/sup#ed_cl# =&gt; log(0.1) = -1 &gt; k log(1-1/13824)#ed_op#br#ed_cl#=&gt; -1 / log(13823/13824) &lt; k&nbsp; || 13823/13824&lt;1 =&gt; log(13823/13824)&lt;0#ed_op#br#ed_cl#=&gt; k &gt; 31829.7xxx =&gt; Number of rolls = 31830+3-1 = 31832 rolls#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl#