Grebnet's Two Cents

[Websman]

were you watching it at $9.60 ?

I was at work, but I had a feeling that it was going to bounce off of the 50 day MA... I should have followed my instincts and set a limit order to buy it in that price range.

[grebnet]

Hom

I should have bought HOM today... Dangit.

were you watching it at $9.60 ?

[Websman]

I should have bought HOM today... Dangit.

[grebnet]

What a ride

Well its been a wild ride on the HOM rollercoaster. I was thinking of adding to HOM this am at 9.60 but decided My position was large enough.(woulda coulda shoulda. )

AERTA- anyone playing along on this. been creeping up nicely.

GV- maybe warming up.

RGMI- has given back some, Im waiting for the hurricaine season. I still like it.

[billyjoe]

Biggest upgrades week ending 5/26/06 under $10

ITI , NYNY , IFUL , AMY , BORL , PDF , MRB

billyjoe

[grebnet]

Hom

All that discussion about HOM options yesterday and I missed what would have been a nice drop if I had sold calls. So instead I bought more HOM at $11.10 with a trade in mind to add to profits from this great stock.

Just Gambling that it will turn : )

I think that there is a bill about to be voted on to make online gaming illegal.

That would be a drag. I have been playing texas holding, started out with 500$ now up to 1700$. I love the game but it does take alot of time in which I should be learning how to play option. I seem to only average 1/43 hands played. I fold alot. I keep telling myself, "self, your ahead, cash out and stop this gambling, do something that is more useful and $$ making".

I also think even if they pass a law to make on-line gambling illegal, there will be too many ways around it. I thought it was already illegal. I do think that playing texas holding has taught me patience in trading.

[mrmarket]

I found gambler's ruin last night in my online poker account. My balance is now 0. Went all in with a KQ top pair, but lost to QJ which both paired... oh well, no more online poker for me.

I think that there is a bill about to be voted on to make online gaming illegal.

[jiesen]

I found gambler's ruin last night in my online poker account. My balance is now 0. Went all in with a KQ top pair, but lost to QJ which both paired... oh well, no more online poker for me.

[mrmarket]

Another explanation

Casino games
A typical casino game has a slight house advantage. The advantage is the long-run expectation, most often expressed as a percentage of the amount wagered. It remains constant from one play to the next. If the long-run expectation is expressed as a percentage of the amount that the player starts with, however, then the house advantage increases the longer the player continues.

For example, the official house advantage for a casino game might be 1%, and thus the expected value of return for the gambler is 99%. However, this math would only be true if the gambler never used the results of a winning bet again. Thus after gambling 100 dollars the idealized average gambler would be left with 99 dollars, but, if he continued to bet using his 99 dollars in winnings, he would again lose 1% on average and his expected value would go down to 98.01 dollars. This downward spiral continues until the gambler's expected value approaches zero: gambler's ruin.

The long-run expectation will not necessarily be the result experienced by any particular gambler. The gambler who plays for a finite period of time may finish with a net win, despite the house advantage, or may go broke much more quickly than the mathematical prediction.

[mrmarket]

Mathematical explanation of "Gambler's Ruin"

The Gambler's Ruin
Consider a game that gives a probability q of winning 1 dollar and
a probability (1-q) of losing 1 dollar. If a player begins with
10 dollars, and intends to play the game repeatedly until he either
goes broke or increases his holdings to 20 dollars, what is his
probability of going broke?

This is commonly known as the Gambler's Ruin problem. For any given
amount h of current holdings, the conditional probability of going
broke before reaching 20 dollars is independent of how we acquired
the h dollars, so there is a unique probability p_h of going broke
on the condition that we currently hold h dollars. Of course, we
can immediately set p_0 = 1.0 and p_20 = 0.0. The problem is to
determine the values of p_h for h between 0 and 20.

The key point to realize is that in order to arrive at holdings equal
to h dollars after playing a round of the game, we must have held
either h+1 or h-1 dollars just prior to that round. When we were in
one of those states we had (by definition) a probability of p_{h+1}
or p_{h-1} respectively of going broke. Also, the conditional
probability that we just came from the state "h-1" is q (which is
the probability that we won the round), and the probability that
we just came from state "h+1" is (1-q). Now, the probabililiy of
going broke from the state p_h is just the linear combination of
these two

p_h = q * p_{h-1} + (1-q) * p_{h+1} (1)

This gives us a second-order linear recurrence relation that must
be satisfied by the values of p_h. If q and 1-q are distinct (meaning
that q is not equal to exactly 1/2), the general form of such a
recurrence is a linear combination of successive powers of any two
independent particular solutions. One particular solution is
obviously p_h = 1 for all h. Also, it's not hard to verify that
p_h = [(1-q)/q]^h is also a particular solution. Therefore, the
general solution of the recurrence is of the form

p_h = A [1]^h + B [(1-q)/q]^h

where A and B are constants to be determined by our two boundary
conditions, p_0 = 1.0 and p_20 = 0.0. Inserting these values
gives the conditions

1 = A + B

0 = A + B [(1-q)/q]^20

Setting r = (1-q)/q, this implies

r^20 1
A = - -------- B = ---------
1 - r^20 1 - r^20

Therefore, if a player is currently holding h dollars, his probability
of going broke before reaching 20 dollars is

r^h - r^20
p_h = ----------------
1 - r^20

This was based on the assumption that q does not exactly equal 1-q,
so that r is not equal to 1. If, on the other hand, q=1/2, we see
that our two particular solutions 1^h and r^h are not independent.
In this case the characteristic polynomial has duplicate roots, but
another independent solution of the recurrence (1) is given by
p_h = h. Therefore, the general form of the solution is A + Bh,
and our boundary conditions require A = 1 and B = -1/20, so the
total solution in this special symmetrical case is

h
p_h = 1 - ---
20

Hence, if we begin with 10 dollars, we have a 50% chance of going
broke before reaching 20 dollars.

Obviously we can replace 20 with any other threshold we choose. For
any given initial holdings, if we increase our upper target from 20
to some larger number, we see that our probability of going broke
before reaching that number also increases. If we have no "quit
while we're ahead" target, and simply intend to play the game
indefinitely, our probability of eventually going broke approaches
1.0 (which presumably is why this problem is called the Gambler's
Ruin).

In the above discussion we considered only the case when each step
changed our holdings by one unit, up or down. We can also treat the
more general problem of allowing more than two possible outcomes of
each round, and allowing the steps to be of arbitrary sizes. For
example, we might consider a game that has three possible outcomes,
with probabilities a, b, and c changing our holdings by the amounts
-1, +1, and +2 respectively. In this case the same reasoning that
led to equation (1) leads to a third-order recurrence

p_h = c * p_{h-2} + b * p_{h-1} + a * p_{h+1} (2)

If we replace 20 with some arbitrary fixed threshold T, then we have
three boundary conditions

p_0 = 1.0 p_T = 0.0 p_{T+1} = 0.0

noting that it's possible to end on either T or T+1. In this more
general case we usually must simply solve the recurrence (2) in the
traditional way, by finding the roots of the characteristic polynomial,
and then expressing p_h as a linear combination of the hth powers of
those roots, subject to the boundary conditions.

This problem is essentially an example of a one-dimensional random
walk. Of course, we can also represent this by a Markov model,
and recursively generate the probabilities of having each particular
value of holdings after the nth round of play, beginning from some
specified initial holdings. This is an example of a diffusion
process, with absorbing states at 0 and T, where all the probability
eventually accumulates.

[mrmarket]

[QUOTE=grebnet]I just sat on my porcelain throne where I get some of my more enjoyable research .

Why the House Always Wins
It is beyond the scope of this article to compare trading to each of the familiar games one finds in casinos, but a fundamental principle that all casinos and all successful traders understand is that, irrespective of the game being played, if the house has a mathematical edge over the player, the law of averages ensures that in the “long-run” the house will win. In fact, for our purposes, an excellent definition of the “long-run” is enough games so that someone making good (bad) bets will win (lose). The number of games required for the desired or dreaded outcome is contingent on how good or bad the bets are.

QUOTE]


Actually, the real reason that the house always wins is because it has more money than you. The mathematical proof is as follows;

Let two players each have a finite number of pennies (say, for player one and for player two). Now, flip one of the pennies (from either player), with each player having 50% probability of winning, and transfer a penny from the loser to the winner. Now repeat the process until one player has all the pennies.

If the process is repeated indefinitely, the probability that one of the two player will eventually lose all his pennies must be 100%. In fact, the chances and that players one and two, respectively, will be rendered penniless are

(1) P1 = n2 / (n1+n2)
(2) P2 = n1 / (n1+n2)

i.e., your chances of going bankrupt are equal to the ratio of pennies your opponent starts out to the total number of pennies.

Therefore, the player starting out with the smallest number of pennies has the greatest chance of going bankrupt. Even with equal odds, the longer you gamble, the greater the chance that the player starting out with the most pennies wins. Since casinos have more pennies than their individual patrons, this principle allows casinos to always come out ahead in the long run. And the common practice of playing games with odds skewed in favor of the house makes this outcome just that much quicker.

This holds true even if the player has a slight edge. Of course, if the player has a brain, he will "quit when he is ahead"..usually the Titans have a problem quitting when they are ahead...but 2 weekends ago, we scorched the Hilton for $16,000....jejejejeje

[grebnet]

Ezen

Here's two that I'm watching Greb...

SVL

EZEN

I've owned both and sold both for a profit, but could have made more. I may get back in one or both of them soon.

Ive owned EZEN since $1.79 have traded several times and own a decent chunk still.

SVL , i dont know but the news after market that insiders will be selling 8 million shares at $3.50 will drag the stock for a while.

[Websman]

Here's two that I'm watching Greb...

SVL

EZEN

I've owned both and sold both for a profit, but could have made more. I may get back in one or both of them soon.

[grebnet]

CNXT---Not my style

cnxt has just been upgraded............ I was in at 2.20 and should of cash out a month ago.. looking at chart I see a lot of gaps to be filled...

I am "hoping" since I didnt stick to plan. What do you think?

CNXT is not my style. Not that it wont go up , but I as a rule limit my stock to ones with positive earnings and little to no debt .

CNXT fails both of those initial screening criteria. so not for me.
It does have a nice revenue growth pattern and appears to be getting ready to turn positive on earnings,BUT has a huge amount of debt. That is a drag on any company.

Good luck