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Messages - sqzbox

#16
Street / Re: Streets question
March 09, 2016, 07:39:47 AM
I'd like to know what other people think because I don't think this question makes any sense.  ???
#17
Street / Re: Streets question
March 09, 2016, 04:57:39 AM
Matt - I'm not sure if this answers your question or not, because TBH your question is not clearly stated. You might be surprised by the answer in which case I would surmise that you haven't made your point clearly.

Approx. 12.33

We use the Geometric distribution for finding the number of trials until the first success. The formula for the Geometric distribution is just the reciprocal of the probability of the event. The probability of the event, in this case, is 3/37 and so the reciprocal is 37/3 and so the answer to "For a street bet, what is the expected number of spins until the first hit?" is 12.33. In this case the phrase "For a street bet" means that it has been specified before the trials begin exactly WHICH street we are waiting for.

Since you state "in cycles of 12 spins" I am guessing that the question I state above is not the question you are asking. The thing is, you say "average spin number that a street will hit on" but which street? One nominated ahead? Then the above applies. If not nominated ahead then which street? Any street? OK - then the answer is 1 because whatever street hits on the first spin qualifies as "any street" - yeah?
#18
Street / Re: Streets question
March 08, 2016, 11:32:43 PM
roughly 8.
#19
Roulette Forum / Re: bet selection for columns (ND)
January 22, 2016, 11:27:37 PM
Remember plolp, that the bet selection here is based on following the colour. If you have suggested that it might be more economical to bet just one chip based on a repeat, are you referring to a repeat column? or a repeat colour?  If a repeat column then that is not within the confines of ND's concept so perhaps you should consider a separate thread which does not cite ND in the title.

In my research I have found that the results are identical for a repeat of the last column with a showing of the one before last. The sleeper, interestingly, seems lower. So, is a one column bet more economical? Perhaps - but is that more efficient? I would argue that "efficiency" is way more important.

So what exactly does "efficient" mean? Well, it's a little hard to explain because there hasn't been a definition of the term as it relates to gambling so we have to go with a little common sense here. However, if you look up the term in an engineering sense then you'll get the idea of what I am trying to espouse here.

Let's say that your sims of the columns or dozens, ignoring zero (can't avoid zero but what we are interested in is the appearance of the columns/dozens in terms of the last, the second last, and the sleeper) are, over 1000 spins which included 29 zeros and so there were 971 non-zero spins, as follows -

Last = 346 (35.63%), Before last = 343 (35.32%), Sleeper = 282 (29.04%)

So there is an almost equal chance for the L or the B, while the S is a reduced chance. So what is the efficient bet here? As already stated the most economical bet is a 1 column bet, but which column would you go for? I'd suggest that, in this scenario, the best bet, i.e. the most efficient bet, would be to go with both L and B.

Now, I am not suggesting the above as a bet - this is just an example to try and explain why the efficient bet is not always the economical bet, but is always the best bet. This thread is about selecting based on the colour anyway so now that we have an understanding of efficiency (by my definition anyhow) how would this be applied? Well, perhaps if you want to be economical as well as efficient then you could just bet column 2 if following black and column 3 if following red. Perhaps just following the last, or even the last plus the before last, is a good bet but that is not the discussion here - here it is about following the colour.

Let's say that you follow black and it is black that appears. If you bet on columns 1 and 2 then you have a 14 out of 18 chance that you get the hit. 2 chips out, 1 chip profit. If you played just column 2 then your chances are 8 out of 18 for a hit. 1 chip out, 2 chips profit. Which is the most efficient?

Discuss - and start up the fan!
#20
Straight-up / Re: shalaska method
December 26, 2015, 11:28:47 PM
Looked at that thread - it seems it was still under discussion and no final conclusion had been reached - there were several variations still being researched. Did it continue after 12-07-05?
#21
I think I would be inclined to use only the spins invested on.  That is, as a measure of efficiency the 1.19 is not really all that good because it is units per spin over ALL the spins - which includes the spins where no bet was placed. I'd be more inclined to use "units profit / number of spins bet on" as a measure of return per spin, or "earning rate". So in this case 459 (profit) / 212 (spins where bets were placed) = 2.165. This would be a measure of earning rate per spin "played" - or, IMO, just "earning rate". I'm personally of the view that spins where no bet is placed is irrelevant to mathematical calculations of earning rates and efficiencies and so on. Only where the answer is specifically relevant to these would I include them - such as, the ratio of spins measured to spins played, or perhaps earning rate OVER ALL SPINS CONSIDERED, which would be the 1.19 of course but is explicitly stated as such. If you were just quoting rates without qualification then I would assume (as most would I suspect) that you were including only those spins where bets were placed.

The house edge is not really all that relevant (IMO) when measuring success. If you wanted to measure how well you did in terms of beating the house edge then of course it would be relevant. But generally I think most would only be interested in how well did you do, what was the earning rate, what was the apparent advantage, and so on, and none of these would include reference to the house edge.

All this is just my opinion of course and I would be interested to hear from others.
#22
Oh, OK. So that is the return per spin then - your 1.19 as mentioned, but multiplying that by 100 to get a percentage (actually 119%) would seem to be a bit of a stretch - not sure what that tells us.  In any case, to calculate percentages your units of measure of the operands have to be the same to give a meaningful result. So you can calculate a percentage of units-in vs. units-out, or spins played vs. spins measured, and so on, but not when the units are, for example, units for one operand and spins for the other.
#23
120%? Where did that come from?

I'd calculate like this -

459 (profit) / 1908 (investment) x 100 (to make it a percent) = 24.0566%

As for return per spin, I suppose you could look at it as 1.19 units earned per spin but I'm not sure that that can be converted into a player's edge.

Anybody else?
#24
Good question - hopefully there will be some good responses and we can all learn lots.

For myself, I did a little bit of research and discovered the following that is based on winnings vs. investment so is focused on the money aspect rather than spins.

It has been defined that "The player's edge is the expected return divided by the initial bet". So, for example, on a 9 chance bet the calculated theoretical players edge, based on the odds of the game, would be –

Expected return = outlay x odds where outlay is 9 and odds are -2.73% = 9 x -0.0273 = -0.2457
Initial bet = 9 of course.

So the theoretical players edge for this bet works out to be 9 x -0.0273 / 9 = -0.0273 = -2.73%. No surprises there.

However, if your question is about the players edge based on your actual results (and I suspect that it is) then probably I would use the definition for EV. Quoting blackjackinfo.com -
"I think you are wanting to calculate w/l% ie: %Win/Loss – Also called EV or IBA (Initial Bet Advantage.) This is the amount won or lost divided by the initial bet. this is how QFIT's simulator help describes expected value, errhh your edge over the house."

So if you invested 100 units and got 120 back then your return is 120 – 100 = 20 and I would calculate the edge (for that particular data set) as –

20 / 100 = 0.2 = 20%

#25
Thanks Bayes - I feel vindicated!  :)

There are too many undefined variables in the original question - as you say. What is the spread provided by the house in which we are playing? 1 to 100 is a different game from 1 to 25 for example. What is the value of the limits? 100 units versus 25 units? again, different games. Do I have to bet every spin? How many events are in the set? Can I choose when to stop? and so on and on.

I've always maintained that the 2 most powerful tools in our arsenal are - 1) we can choose whether to place a bet or not, whereas the casino must always play, and 2) we can choose how much to bet, and the casino must always match it.  The OP says nothing about either of these powerful weapons.

Therefore we have no choice but to generalise the answer.  Given that the mathematical expectation under general casino conditions for this scenario is 20% ROI (didn't know that), and the assumption that we are talking "in the long run", then minimal profit is achieved by minimal investment which can only be achieved by flat betting.  But again, it should be stated that this assumes certain restrictions such as "must bet every spin".

#26
Perhaps we need the OP to define "minimal profit" then.  Does -1 represent a minimal profit? In which case -10 is a better result? And -100 better still? Because it is more minimal? Weird. Doesn't make a lot of sense to me. I would have thought that +1 was the lowest possible value that a minimal profit could be - it is a profit and it is the minimum that a profit can be. Anything less is not a profit it seems to me.

And surely a flat bet at minimum is the most "small %" of your bank that can be bet, and since, as your simulation suggests, the marty makes far more than a flat bet then playing the marty or reverse marty is NOT what we would want to play if we wanted to achieve minimal profit.

Betting your entire bank every time will guarantee you lose it all when the loss occurs, but if we are in a 60% win situation then 60% of the time we are in a big win situation. If the simulation ends at a win then that is an entirely different situation and may well represent a larger bankroll than flat betting.  However, if the simulation is long then probably the flat bet would end with a bigger bankroll than betting the whole thing each bet. This is just another reason why the question itself makes no sense - too undefined.

Are we in a circular argument? [smiley]aes/dont_know.png[/smiley]
#27
Um - I still stand by flat bet. You have to consider ALL possible outcomes. If you use a parlay, reverse marty or normal marty, or whatever, all these imply increasing individual bet amounts and so your average bet MUST be higher than your average flat bet (assuming you bet the minimum of 1 unit). Depending on exactly WHEN the betting sequence ends you could be left with a LARGE profit, or perhaps nothing. But, ON AVERAGE, somewhere between 0 and a high value which will be, I contend, larger than simply flat betting.

When flat betting it is true that you will end up with a profit - but the question was "minimal profit" not "zero profit" (the latter of course is an oxymoron anyway and in my view would not be part of the set of possible valid outcomes).
#28
Math & Statistics / Re: a solution to roulette?!
April 20, 2015, 11:25:38 PM
Nah - doesn't help. The ROBE situation has been discussed to death and, by itself, gives no advantage. The reason is that although the probabilities for the 3 possible outcomes (Win, Lose Break even) are different between betting RO versus RE, or BO versus BE, the important one is the difference between losing and winning and, on a Euro wheel, is 0.027 in both cases.

Unfortunately the author doesn't go far enough in his explanation to give any useful guidance for developing a successful method, although he hints that it may be possible. But the point of his article was, I think, to demonstrate that the layout may be such that the traditional way of viewing the game, as random and hence may be analysed using the random walk or Markov chain models, is not valid. And the point of that, is that it invalidates Gambler's Ruin. If that is true then it is good news indeed.
#29
I suspect that the reason you don't get the sort of response you are hoping for is because the question actually makes no sense. Here's why.

You say "winnings/profits" and provide an example W/L string - this implies that your bet selection creates a profitable W/L series, or, more wins than losses. This would be the holy grail which we all know does not exist. Therefore most people's immediate reaction would be "oh, here's another kook" and not take it seriously. (No offence - just saying ..).

Secondly, the lowest will simply be, as Xander has stated, simply a flat bet because - in order to do worse than that (which is what you are asking for discussion on) you would have to somehow manage to have a progression where your average winning bet is lower than your average losing bet, in which case you will always, without fail, progress towards and eventually reach a loss.  And since that would not be a winning series it does not fail within the definition you refer to in your original question of "winnings/profits".

So your question is inherently logically contradictory and therefore you are unlikely to get meaningful answers.

#30
Nope - not getting this at all. "Streams"? What are these? Splits? What splits? Nbr 4, for example, can be in 3 different splits. What makes you decide on a particular bet? Are you looking at gaps here? If so what constitutes a bet choice? Sorry Horus, maybe I am being particularly thick here, but none of this is making any sense to me at all.