< Earlier Kibitzing · PAGE 42 OF 42 ·
Later Kibitzing> 
May1818
  Marmot PFL: I have heard of viruses being used to treat patients with bacterial infections that resist antibiotics, but it doesn't seem to be approved in the USA. 

May2118
  al wazir: <johnlspouge>: I can't help continuing to think about the expression you derived for the probability that the plane is still aloft after 2N  1 time units, which is the same as the probability that the gambler in the cointossing game I described has not been wiped out: P( 2N  1 ) = 2^(2N + 1) * c( 2N  1, N  1 ). I will rewrite this as P(2N1) = 2^(12N)(2N1)!/(N!(N1)!, using the notation I prefer.
If you check this formula for the case of N = 2, you find that of the eight possible sequences of three heads and tails HHH
HHT
HTH
HTT
THH
TTH
THT
TTT
three are consistent with staying alive. Three out of eight is 0.375, and this agrees with P(3) = 2^(3) x 3!/1!2! = 0.375. However, it cannot have escaped your notice that 2^(12N)(2N1)!/(N!(N1)! is identical with the probability of getting N heads and N1 tails in 2N1 tosses. For N = 2, three of the eight possible sequences of heads and tails do indeed have two heads and one tail (namely, HHT, HTH, and THH). But there is something strange about this: they are *not* the same as the three sequences that correspond to survival. Similarly, for N = 3, where the survival probability is P(5) = 10/32 = 0.3125, ten of the 32 possible sequences have three H's and two T's, but they are not the same as the ten survival sequences. Very well. P(2N1) is the probability of getting N occurrences and N1 nonoccurrences of *something*, both with probability 0.5, out of 2N  1 tries. What is this "something"? If I could find the answer, that would allow me to derive the formula in one line. I had another thought. There must be other ways to derive your formula, e.g., by induction. P(1) = 0.5, which agrees trivially with the result of a single toss. For any N > 1, P(2N+1) = 2^(12N)(2N+1)!/(N!(N+1)!
= 2^(12N)(2N1)!/(N!(N1)! x 2N(2N+1)/4N(N+1) = P(2N1) x (2N+1)/4N(N+1) = P(2N1)  (1/4) x [2/(N+1)] x P(2N1). This has a simple interpretation. The probability of staying in the game is the same after 2N+1 tosses as after 2N1, *except* when the gambler holds precisely $2K after 2N1 tosses. In that case there is a 1/4 chance of getting TT in the next two tosses and losing. The last term describes that possibility, where the probability of ending up with $2K after 2N1 tosses is [2/(N+1)] x P(2N1). Now all I need is an independent way of showing that. 

May2118
  johnlspouge: @<al wazir>: You are restlessly inquisitive. I tremble at the thought of what you could have accomplished if your inquisitiveness had been focused ;>) In general, the kind of simple interpretation you seek indicates a 11 mapping between configurations ("decompositions" in the jargon of combinatorics). Let me use the equivalent (and possibly simpler) expression P( 2N1 ) = 2^(2N) * (2N)! / N! / N!
Represent a climbing step by "+" and a drop by "". (The following omits the apostrophes.) Given a configuration of + and  that does not cause the plane to crash in 2N  1 (or equivalently 2N) time units, it is an exercise for the reader to show that the following is a 11 mapping. Look back in time from the earliest (leftmost)  in the configuration and move successively to each later , finding for each  the last (compensating) +, and then padding the set of compensating +s with the rightmost +s remaining. Under this selection of +s, each configuration of + and  not causing the plane to crash corresponds to exactly one choice of N +s, and vice versa. Here is the inverse. Let * being one of the N selected +s; 0, one of the other N time units. The inverse mapping starts at the left * and puts a corresponding  at the first 0, and iterates, moving to the right, until running out of positions to place a . It then puts + everywhere remaining. As a worked example, consider N = 2.
There are
6 = 4!/2!/2!
allowed configurations where every  is preceded by a +, namely 00**
++++
0*0*
+++
0**0
+++
*00*
+++
*0*0
++
**00
++
Sometimes, the 11 mapping indicates interesting ways to generalize the problem by showing which generalizations are "insightful", i.e., which generalizations maintain some version of the mapping. 

May2118
  john barleycorn: let's take a simple view.
1) The plane is at 1000ft. The probability going up to 2000 ft is .5 and so is the probabilty for crashing. 2) being at 2000ft the probability of going up to 4000 ft in the next two minutes is .25, the probability of being back to 2000ft after 2 minutes is .5, and the probability for crashing is .25.
Total probability to crash after 3 minutes: .5 + .5*.25 = .625 3) being at 4000 ft after 3 minutes means one crash happens in the next four minutes with a probability of 1/16.
And being at 2000 ft after 3 minutes means probability of a crash within the next four minutes is .25 + .75*.25. Which gives the probabilitity of a crash after 7 minutes of .5 + .5*(.25 +.75*.25) + .25*.25*.5 = .75
I cannot get this number number out of the formulae given so far. Please advise. 

May2118
  al wazir: <john barleycorn>: Using the notation of my analogous gambling problem, there is one way of crashing after one minute: T, probability = 0.5;
there is one new way of crashing after three minutes: HTT, probability = 0.125;
there are two new ways of crashing after five minutes: HHTTT, probability = 1/32 = 0.03125
HTHTT, probability = 1/32 = 0.03125;
there are five new ways of crashing after seven minutes: HHHTTTT, probability = 1/128 = 0.0078125
HHTHTTT, probability = 1/128 = 0.0078125
HHTTHTT, probability = 1/128 = 0.0078125
HTHHTTT, probability = 1/128 = 0.0078125
HTHTHTT, probability = 1/128 = 0.0078125.
Total: 1/2 + 1/8 + 1/16 + 5/128 = 0.5 + 0.125 + 0.0625 + 0.0390625 = 0.7265625. P(7) = (7!/4!3!)/2^7 = 35/128 = 0.2734375 = 1.0  0.7265625. 

May2118
  Marmot PFL: <P(7) = (7!/4!3!)/2^7 = 35/128 = 0.2734375 = 1.0  0.7265625.> Going back to work I saved from 5/15 that's the number I also got. 

May2118
  johnlspouge: Turn the probability problem into a counting problem by pretending the plane can continue to fly underground, so all its up or downone paths are equally likely. Then count paths that occur in N time units, and paths corresponding to the subset where the plane does not crash. Division then gives the probability you seek. 

May2118
  johnlspouge: My answer is the same as <al>s, without his numbers. 

May2118
  johnlspouge: Essentially... 

May2118
  john barleycorn: < <al wazir: ... P(1) = 0.5, which agrees trivially with the result of a single toss. For any N > 1, P(2N+1) = 2^(12N)(2N+1)!/(N!(N+1)!
= 2^(12N)(2N1)!/(N!(N1)! x 2N(2N+1)/4N(N+1)
= P(2N1) x (2N+1)/4N(N+1) = P(2N1)  (1/4) x [2/(N+1)] x P(2N1). ...> Well, these are a dangerous just by looking at them. some wellplaced braces could structure them. 

May2118
  john barleycorn: < johnlspouge: Turn the probability problem into a counting problem by pretending the plane can continue to fly underground, so all its up or downone paths are equally likely....> I looked at it that the plane cannot go underground. After a crash the plane can neither crash again nor go up again. (it looks so reasonable but unsymmetric :)) that is why I broke up the case of having reached 2000ft after 3 minutes into the case of being crashing after 2 minutes (.25) and the probability of being still in the air after 2 minutes and then crahing in the remaining 2 minutes (.75*.25) In fact, as some poster mentioned you cannot evade the Catalan numbers in this problem. but I do not see them in any solution presented here. 

May2118
  johnlspouge: Catalan numbers count binary trees of various kinds. It should not be hard to find them if you are looking ;>) 

May2118
  Marmot PFL: Turn the probability problem into a counting problem by pretending the plane can continue to fly underground, so all its up or downone paths are equally likely....> Maybe Malaysia Airlines Flight 370 is still flying, somewhere beneath the Indian Ocean. 

May2118
  al wazir: <Marmot PFL: Malaysia Airlines Flight 370 is still flying, somewhere beneath the Indian Ocean> with probability 1.0. 

May2118
  john barleycorn: < johnlspouge: Catalan numbers count binary trees of various kinds. It should not be hard to find them if you are looking ;>)> haha yes they are ubiquitous in branching processes. but you forgot them here :) 

May2118
  john barleycorn: < al wazir: <john barleycorn>: Using the notation of my analogous gambling problem, there is one way of crashing after one minute: ... P(7) = (7!/4!3!)/2^7 = 35/128 = 0.2734375 = 1.0  0.7265625.> Well, not even in my dreams.
1. there are 128 outcomes possible.
2. to win the player has to start with a head which cuts it down to 64 events. 3. out of that 64 events the outcomes of having 6H or 6T, or 5H, 1T, as well as having 4H,2T or 3H, 3T are symmetrical. 4. all in all the Player loses in 1+6+15+10+64= 96 times. 5. and 96/128= .75 

May2118   nok: Those still pondering can have a look at the wiki's Fig. 1 where the ground is drawn as a red diagonal with the "sky" below it. They count the paths that crash at 2n+1 without crashing earlier. https://en.wikipedia.org/wiki/Catal... 

May2118
  john barleycorn: < nok: Those still pondering can have a look at the wiki's Fig. 1 where the ground is drawn as a red diagonal with the "sky" below it. They count the paths that crash at 2n+1 without crashing earlier.> Thank you for the link. the problem is here you have to sum up the earlier crashes as well. the best way to do that is that after the plane has reached 2000ft height after 1 minute to go on with 2 minute time intervalls. If you take D for going down 1000ft and U for going up 1000ft then it comes to handling properly (D+U)^n with n even. 

May2118
  WannaBe: Y'all lost me at < 

May2118
  john barleycorn: <WannaBe: Y'all lost me at > that is why Stumpers page was invented :) 

May2118
  WannaBe: No, I mean I am lost, I was looking for the Kenneth Rogoff page. Knew I shoudda made a left turn at Albuquerque. 

May2218
  john barleycorn: Yeah, same feelings here. I have laid out two baits for <al wazir> but he is not responding with a post of at least 5 inches long.
Where has the world come to when <al wazir> skips his routine? 

May2218
  al wazir: <john barleycorn: I have laid out two baits for <al wazir> but he is not responding>. You mean, you're waiting for me to tell you why you got 96 instead of 93? Nah. I'll let you find your mistake yourself. 

May2318
  john barleycorn: <al wazir: ...
Nah. I'll let you find your mistake yourself.> haha, no need for me to find them myself as I built them in. Most obvious, 6!/3!/3! = 20 not 10. and the claim about the symmetry is another story. Anyway, the correct number of losses in a 7 game play of Head/Tails is 64+29 and so probability is 93/128. 

May2318
  john barleycorn: Read 6!/3!/3! as (6!/3!)/3! 



< Earlier Kibitzing · PAGE 42 OF 42 ·
Later Kibitzing> 