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Ask the Wizard correction
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| June 16th, 2011 at 12:44:24 PM permalink | |
| Wizard Administrator Member since: Oct 14, 2009 Threads: 313 Posts: 6784 |
You're right, thanks for the correction. The column should be correct now.
I could count the number of girls on this forum with one hand, so don't expect one of them to answer. Personally, I think that is the mathematician from the show Numbers. It's not whether you win or lose; it's whether or not you had a good bet. |
| June 16th, 2011 at 1:47:28 PM permalink | |
| mustangsally Member since: Mar 29, 2011 Threads: 5 Posts: 170 |
Thanks for the info. He also has mentioned the same over at 2+2. Hard to know what is the truth on the net. At least you show your pics along with wife and kids. That is cool. So then how about a girls night out with Mr. S maybe once a year? 5 or less would make for a fun time. Unless you require all your girls to look like your choice of models in your photos. Not counting the Bodog Girls that is. I Heart Vi Hart |
| June 16th, 2011 at 2:18:24 PM permalink | |
| Wizard Administrator Member since: Oct 14, 2009 Threads: 313 Posts: 6784 |
Well, thanks for the nice words. Any picture of me with a pretty girl I assure you that the girl was being paid. Thanks for the nice idea, but it would never happen, trust me. It's not whether you win or lose; it's whether or not you had a good bet. |
| June 16th, 2011 at 8:06:46 PM permalink | |
| pacomartin Member since: Jan 14, 2010 Threads: 547 Posts: 6211 |
David Krumholtz plays the role of Charlie Eppes in the CBS series Numb3rs. Charlie Eppes is a young mathematical genius and professor of applied mathematics at the fictional California Institute of Science, CalSci. Since BruceZ give his location as the fiction school of Cal Sci I think that is a dead giveaway.   I know this film is over 30 years, old, but It's My Turn had Jill Clayburgh playing a serious mathematician in her mid 30's who had a sexy role as a newly divorced woman. It was probably the first time I had seen an actor playing a mathematician in a movie in anything other than a throwaway character role. Jill sadly died of cancer a few months ago. The following lines of dialogue had never appeared in a movie up to that time, Cooperman was played by a young Daniel Stern who would later act in "Home Alone".  The Snake Lemma scene in a rare display in movies, is actually mathematically correct proof. Kate Gunzinger: Let me just show you how to *construct* the map S, which is the fun of the lemma anyhow, okay? So you assume you have an element in the kernel of gamma, that is, an element in C, such that gamma takes you to 0 in C-prime. You pull it back to B, via map g, which is surjective... Cooperman: Hold it, hold it, hold it. That's -- that's not unique. Kate Gunzinger: Yes, it is unique, Mr. Cooperman. Up to an element of the image of f, all right? So we've pulled it back to a fixed B here. Then you take beta of B, which takes you to 0 in C-prime, by the commutivity of the diagram. It's therefore in the kernel of the map g-prime, hence is in the image of the map f-prime, by the exactness of the lower sequence... Cooperman: No. Kate Gunzinger: ...so we can pull it back... Cooperman: No. Kate Gunzinger: ...to an element in A-prime... Cooperman: It's not well defined! Kate Gunzinger: ...which it turns out is *well* defined *modulo* the image of alpha. And thus defines the element in the co-kernel of alpha... [draws arrow on diagram] Kate Gunzinger: and that's the "snake"! And on Monday, we'll address ourselves to [Cooperman raises hand] Kate Gunzinger: the co-homology of groups... and Mr. Cooperman's next objections. Wine loved I deeply, dice dearly -Edgar, betrayed son of Gloucester in King Lear |
| June 20th, 2011 at 3:31:40 PM permalink | |
| mustangsally Member since: Mar 29, 2011 Threads: 5 Posts: 170 |
I do not see any "flaw" in what pacomartin attempted to do. From reading some of his posts he knows how to run transition matrices and set up recursive formulas to solve many problems. He was looking into a simple way to solve a run of exact 7 in 100 trials and I have yet to see a solution other than a Markov chain, I know that other more difficult solutions do exist. My professor showed a very large formula rich in combinatorics and it just seemed to be easier using a Markov chain. Why torture oneself just to say " I finally did it"? pacomartin knows he is dealing with a probability (n/N) and it just does not work out going from "at least" to "exact" by subtraction. My class professor explained why, but his explanation went over all of our heads. And class is out. I graduated! Biology was my major, I enjoyed the math but am glad it is over.
Yes, exactly correct. The subtraction method dealing with runs works for averages but not probabilities. 0.371094 average number of streaks of length 7 or more 0.183594 average number of streaks of length 8 or more average number of streaks of length 7: (0.371094 - 0.183594) = 0.18750
I never got around to that final approach since I was only doing this example for an extra credit assignment. Very close to my computer simulation. and if I may ask... What does your matrix look like? I Heart Vi Hart |
| June 20th, 2011 at 3:45:35 PM permalink | |
| mustangsally Member since: Mar 29, 2011 Threads: 5 Posts: 170 |
This is out of my understanding. This is my matrix and seems to be close to pacomartins as in his 1st post. It is a 9x9 (Yes?) and it works in my software program, where an 8x8 does not as the Wizard has now on his ask the wizard page. My BF says there are different ways of setting up the same matrix depending on the software that is also used. I accept his answer but I do not understand it. Guess one has to know more about Markov chains. Summer vacation! 0.00000 1.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 results 0.00000 0.34262 0.17200 0.08635 0.04335 0.02176 0.01092 0.00548 0.31752 I Heart Vi Hart |
| June 21st, 2011 at 7:57:51 AM permalink | |
| CrystalMath Member since: May 10, 2011 Threads: 3 Posts: 476 | Mustangsally, I see what is going on now. The difference in the two matrices (yours and the Wizards) is that you include one more state, the top row and left column which represents the state of not playing. The first move will transition the player from a state of not playing to a state of a 0-streak. After that, you have 100 more transitions to go. Therefore, you will calculate T^101 (1 to get into the game and 100 for the dice rolls). The Wizard's transition matrix assumes that you have already decided to play the game and the player starts with a state of a 0-streak. In this case, we calculate T^100. I heart Crystal Math. |
| November 8th, 2011 at 9:56:11 PM permalink | |
| 7craps Member since: Jan 23, 2010 Threads: 10 Posts: 332 |
CrystalMath, you have an excellent working knowledge of Markov Chains. Would you share your matrix solution for the above answer? I am assuming that this is an absorbing transition matrix. Thanks kindly. Life in the Key of F#...a.crap=(gambling) - (math) b.math=(crap) / (gambling) c.gambling=(crap) / (math) |
| November 9th, 2011 at 6:38:12 AM permalink | |
| boymimbo Member since: Nov 12, 2009 Threads: 12 Posts: 2533 | Another reason why I love this site; it rekindles my love for math. I remember all of those matrices from first year algebra in University. It was my first final exam at University and of course I crammed for the exam. The exam started at 9:00am; I remember sleeping in until 9:15, and running to the exam, and being able to write it. The prof. even gave me an extra 15 minutes to finish the exam. -----
You want the truth! You can't handle the truth! |
| November 9th, 2011 at 7:47:18 AM permalink | |
| CrystalMath Member since: May 10, 2011 Threads: 3 Posts: 476 |
I actually don't have it any more and it would be pretty time consuming to recreate. I used a Markov Chain instead of a matrix though. They essentially do the samem thing but the chain will use less space in Excel. I heart Crystal Math. |
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