I may experiment with this product some during my upcoming vacation (yeah, I also told myself that about Flare and Five3D). The Business and Education products look somewhat interesting. Anyone have any experience with this tool? Thoughts and opinions are welcome.
Want more insight into the killer team creating a killer product? Check out some recent interviews with the PV3D core team.
I originally discussed this problem in the old Singularity blog as a result of it having been mentioned on a Numb3rs episode. Numerous people asked me if the show made a mistake. I’ve been around enough to recognize the old “Let’s Make a Deal” problem from the famous old game show of the same name.
When the problem was first posed to me (a very long time ago), I made the correct choice but only because I took the ‘intuitive’ answer and chose the opposite. My first analysis of the problem (done in the early 90’s along with a Fortran simulator) is reposted below. Shortly after the first Singularity post, a Flash simulator was provided at the lessrain blog.
A recent conversation about the non-intuitive nature of many results from Statistics lead me to druge this up. Perhaps I’ll do a Flex simulator some time in the future.
Original Singularity post follows
Here is the modern version of “Let’s Make a Deal”. You are a contestant on a game show. The host offers you a choice of three doors. Behind one of the doors is a lifetime pass to Flash In the Can. Behind the other two doors is a rotten apple or some equally undesirable item. You receive whatever is behind the door you select.
Suppose you select door #1. The host opens door #3 to reveal one of the rotten apples. The host gives you an option to either stick with door #1 or switch to door #2.
Would you be better off to switch, stay, or does it not make any difference? Conventional wisdom states that since only two choices remain and both are equally likely, the probability that either door contains the prize is 1/2. So, switching or staying makes no difference. In reality, your chances of winning the prize double by switching. The solution to the problem is non-intuitive and sparked a great deal of debate when the solution was proposed in a column by Marilyn vos Savant. The intuitive approach treats the problem as one of absolute probability, when it is actually an exercise in conditional probability.
A simple (but non-rigorous) approach to the correct solution breaks the problem into two sets. S is the singleton consisting of your selected door. S’ is the set of the two doors you did not select. If P(S) is the probability that the prize is behind the door you selected, then P(S) = 1/3. P(S’) = 2/3. When the host reveals what is behind one of the doors in S’, neither P(S) or P(S’) changes. The host revealed nothing about S, but did reveal something about S’. By switching rom S to S’, you double the probability of selecting the door containing the prize. For a somewhat more rigorous analysis, consider the game conditions
1: Each door is equally likely to contain the prize.
2: There is only one prize. The other two doors hide something undesirable.
3: The host knows in advance which door contains the prize and *always* opens a door containing the non-desirable item before offering the choice to switch or stay.
The location of the prize door and contestant’s selection make no difference to the analysis, so for simplicity, let’s stick with the original contestant selection of door #1.
Define the events
H1: Host opens door #1
H2: Host opens door #2
H3: Host opens door #3
P1 Prize behind door #1
P2 Prize behind door #2
P3 Prize behind door #3
Recall that P(A|B) is the probability of event A given that event B occurred. Note that P(H1|any event) = 0 and P(P1) = P(P2) = P(P3) = 1/3 . For events A and B, P(A and B) = P(A)P(B|A).
From the initial conditions, P(H2|P1) = 1/2, P(H2|P2) = 0, P(H2|P3) = 1, P(H3|P1) = 1/2, P(H3|P2) = 1, P(H3|P3) = 0
The table of intersections is below. The ij-th entry in the table is the probability of Pi and Hj. For example, P(P1 and H2) = P(P1)P(H2|P1).
H1 H2 H3 P1 0 (1/3)*(1/2) (1/3)*(1/2) P2 0 (1/3)*0 (1/3)*1 P3 0 (1/3)*1 (1/3)*0
As a check, the row sums should each be 1/3. For example, P(P1) = P(H1 and P1) + P(H2 and P1) + P(H3 and P1) or all the ways the prize could be behind door #1. The column sums are the probability the host opened a particular door.
You select door #1. The host opens door #3 to reveal the rotten apple. You are given the option to stay with door #1 or switch to door #2. To determine which choice is better, compare the probability that the prize is behind #1 given that the host opens #3 to the probability that the prize is behind #2 given that the host opens #3. P(P1|H3) = P(H3 and P1)/P(H3) P(P2|H3) = P(H3 and P2)/P(H3). Using the intersection values from the table,
P(P1|H3) = ‘stay’ = (1/6)/(1/3 + 1/6) = (1/6)/(3/6) = 1/3
P(P2|H3) = ‘switch’ = (1/3)/(1/3 + 1/6) = (1/3)/(3/6) = (1/3)/(1/2) = 2/3.
This is a very old problem that has been discussed numerous times online, so google “Monty Hall” or “Let’s Make a Deal” for more information. Also, check out that lessrain simulator and be glad that Flash development does not require statistics 🙂
From Electric Rain,
“Electric Rain is opening new doors for Adobe® Flash® designers with an update to Swift 3D® v5.0 enabling export to the increasingly popular Papervision3D format. This new feature will be available in conjunction with the release of Swift 3D v5.0 for the Macintosh® platform, as well as a free update to existing Swift 3D v5.0 Windows users, adding even more export flexibility to the industry leading 3D tool for Flash.“
Sounds very interesting for those wanting a solid, entry-level modeling tool for Papervision work.
After living in the clutches of the M$ empire for most of my career, I’m seriously considering joining the rebel alliance and purchasing a Mac Book Pro (17″). I’m interested in all opinions from current owners; in particular, those currently running FlexBuilder and/or Flash CS3 with Leopard. Thanks!!
I rarely comment on products as there is plenty of that available from other blogs. Every now and then, I come across something really worth a comment. The keyboard on my Dell M60 has become sticky after several years of use (or abuse). I recently started using an external keyboard, but that setup was clumsy, especially for travel.
A couple months ago, I saw the Logitech Alto in a Dell store and ordered one for myself.
This product has delivered beyond my best expectations. I feel more like I’m working on a desktop system and the entire setup folds easily for travel. After using the Alto day-to-day and during over half a dozen business trips, I feel comfortable recommending it to others who may be interested in a better setup for their notebook.
I’ve responded to a blog post and a few private e-mails recently on this topic. Two companies seem to have solutions – OrgPlus from Human Concepts and ILOG Elixir. I don’t have any experience with Human Concepts, but I did work extensively with ILOG in my SGI days. I have high confidence in anything they develop.
CMOrgChart was an entry in the Flex Derby, but not sure if the source is available.