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December 15[edit]

Clear definition of a tensor[edit]

I can't seem to find an understandable definition of what a tensor is, that does not make reference to any specific basis (or "frame of reference", as they seem to call it). Do you have pointers to such a definition? (Or if it's small enough to fit in the margin, perhaps you could provide it here directly.) I have seen the following, which seems simple enough, but I'm not sure it tells the whole truth:

Let $V$ be an finite-dimensional vector space over a field $K$ . A tensor of rank $m$ is then a function from $V^{m}$ to $K$ that is linear in all its $m$ arguments.

—Bromskloss (talk) 10:42, 15 December 2007 (UTC)[reply]

Does this respond to what you're looking for? A tensor is a function mapping the bases of some vector space to "blocks" of scalars (where a block generalizes the notion of matrix but takes an n-tuple of indices) satisfying a certain "linear transformation rule". Let t be a tensor, b a basis and M a matrix for relating a new basis to b by expressing it as the matrix-vector product Mb. Then the transformation rule can be expressed in the form t(Mb) = M § t(b). Considering two changes of basis in a row, we have on the one hand t(MNb) = M § t(Nb) = M § (N § t(b)), while on the other hand t(MNb) = (MN) § t(b). So we see that not just any L will do; such an operator § has to satisfy M § (N § x) = (MN) § x. --Lambiam 12:31, 15 December 2007 (UTC)[reply]

I don't want references to specific bases! —Bromskloss (talk) 21:58, 15 December 2007 (UTC)[reply]

There are no "specific bases" here; this property is meant to be universally quantified over all b, M and N. It's like saying that a function is periodic if f(x) = f(x+p) for some p ≠ 0; that does not involve a reference to a specific x. In any case, this approach would become messy when you get to the details; the approach of the "intrinsic definition" in the article on Component-free treatment of tensors referred to below is much more elegant. —Preceding unsigned comment added by Lambiam (talk • contribs) 23:14, 15 December 2007 (UTC)[reply]

Ok, mabye I should say "component free". —Bromskloss (talk) 14:59, 16 December 2007 (UTC)[reply]

One example is that if V is a vector space, then V

\otimes

V^* is isomorphic to End(V), the space of linear transformations from V to itself. End(V) makes sense without any reference to a basis: it is simply the collection of set functions of the form f: V → V, satisfying the following rules: f(v + w) = f(v) + f(w), and f(c·v) = c·f(v) [for v,w in V and c in the base field]. You don't even need to know that a vector space always has a basis for this to make sense.

The fact that I always keep in mind is that a bilinear map from V × W (to some other vector space Z) is the same as a linear map from V

\otimes

W (again to Z). If you're willing to believe that tensors exist, then this one property is enough to define them completely. (See our article on the intrinsic definition.) Tesseran (talk) 14:17, 15 December 2007 (UTC)[reply]

Odds in a hypothetical casino game[edit]

Suppose two people play a game which I will call "face-up draw poker". The game begins with each player anteing up an equal amount. There are no bets in the game other than this ante. Each player then receives five cards. The first player decides which of his five cards to hold (he can hold any number of cards from zero to five). The cards he does not hold are set aside (not returned to the deck). He then draws new cards to replace his discards, and these new cards are placed face up. The second player then follows the same procedure with his own cards. Whoever has the higher hand wins the total amount bet. If both hands rank equally, the game is a tie. It seems to me that in this game, the second player has the advantage because he has more information than the first player. Does he indeed have the advantage, and if so, how great is his advantage? —Preceding unsigned comment added by 207.210.129.15 (talk) 16:07, 15 December 2007 (UTC)[reply]

Seems pretty obvious to me that the second player has a great advantage, though I won't try to prove that ;). In any case, consider first player's attempts to make a straight: For many of the failed attempts, a pair will suffice to beat the first gamblerer. That's too much of an advantage.

To have a precise idea of such advantage, I strongly suggest that you play it yourself! Or, much better, run a simulation. That won't be easy, for you need a couple of bots to play instead of you :-| Pallida Mors 19:21, 15 December 2007 (UTC)[reply]

I would say that the advantage gained will only be slight - even if the terms are much better than you describe. Suppose the second player sees the entire final hand of the first. While having little to no hands-on experience myself, I think that from a typical initial hand there will only be one combination with a non-negligible probability to achieve after the replacement. If you happen to know that this combination doesn't beat your opponent's, you're screwed either way. Of course, you can then improve your odds by shooting for a significantly less likely combination that does win, but being significantly less likely, this strategy won't help by much.

Of course, playing second in this setting has the additional advantage that you know some cards which are no longer in your deck, optimizing your probability calculations. If you're really smart, you can determine which cards are more likely to have been replaced based on the opponent's final hand. However, the differences in probability are too small to have any significant impact.

Returning to what I think the original suggestion was (the second player can see the new cards but none of the original ones), the second player has even less information. Theoretically, of course the second player will have an advantage if he plays optimally. Practically, I doubt any human would be able to harness that advantage, and even if he does, it is too small to be readily observable. -- Meni Rosenfeld (talk) 12:49, 16 December 2007 (UTC)[reply]

Meni, I think that we have understood two different things from the OP's description. He/she talks about face-up poker, explicitly speaking also of visible redraws. With this in mind, I took it was clear that all ten initial cards were common konwledge. Were that the case, I doubt very much that the advantage could be small. As an amateur poker player, let me tell you that tries at straights and flushes represent an important good amount of hands (in this variant), and for these, the advantage the second player holds are important. Just for a comparison, a bias of 0.027 for a dollar's bet is enough to make casino owners rich.

The other interpretation of the OP's statement is that only the redraws are visible, in which case I concede that the advantage would be minor [albeit I ignore if it means less than 0.027 a dollar ;)] Pallida Mors 13:42, 16 December 2007 (UTC)[reply]

The OP didn't specify what version he is based on. He just said he called his own version "face-up draw poker" and stated that new cards are visible. If all cards are indeed visible, I suppose the advantage could be of the order you say is significant. -- Meni Rosenfeld (talk) 13:55, 16 December 2007 (UTC)[reply]

I am the original poster, and all cards are to be visible. —Preceding unsigned comment added by 207.210.129.9 (talk) 21:50, 17 December 2007 (UTC)[reply]