> No, win rate per hour is exactly the same in CVCX and
> CVData. There are numerous stats in CVData that can be
> derived in different manners from other stats. But,
> the solution is always the same.
You're right, WR per hour is the same in either case. I did not realize IBA x Inital Avg Bet = TBA x Total Avg Bet. Either approach will yield the same solution for WR.
But any answer on the question below regarding how to figure out SD of the backcounting frequency?
> But any answer on the question below regarding how to
> figure out SD of the backcounting frequency?
I don't understand the question.
> I don't understand the question.
Suppose CVCX indicates a counter will play 25% of rounds using a Backcounting strategy. Surely, that does not mean he will play 25 rounds for every 100 rounds observed. Some hours he might play 0 rounds, other hours he might play as many as 35 rounds.
If a counter wanted to figure out the probability of only playing a total of 50 rounds during the course of 25 hours of observing shoes, he would need to know the SD for the backcounting frequency. Is it normally distributed? If so, how do you determine it?
MJ
I don't calculate that number and really don't see the purpose.
> Suppose CVCX indicates a counter will play 25% of
> rounds using a Backcounting strategy. Surely, that
> does not mean he will play 25 rounds for every 100
> rounds observed. Some hours he might play 0 rounds,
> other hours he might play as many as 35 rounds.
> If a counter wanted to figure out the probability of
> only playing a total of 50 rounds during the course of
> 25 hours of observing shoes, he would need to know the
> SD for the backcounting frequency. Is it normally
> distributed? If so, how do you determine it?
> MJ
> Suppose CVCX indicates a counter will play 25% of
> rounds using a Backcounting strategy. Surely, that
> does not mean he will play 25 rounds for every 100
> rounds observed. Some hours he might play 0 rounds,
> other hours he might play as many as 35 rounds.
> If a counter wanted to figure out the probability of
> only playing a total of 50 rounds during the course of
> 25 hours of observing shoes, he would need to know the
> SD for the backcounting frequency. Is it normally
> distributed? If so, how do you determine it?
I'm not home now, so I don't have access to all my bookmarked sites, but just go to something that calculates binomial probabilities, and fill in the numbers.
For example, the prob. of playing a hand is 25%, and then you designate the number of trials (rounds dealt) and the number of "successes" that you're looking for, for example playing only 200 rounds out of 2,500.
Don
> I'm not home now, so I don't have access to all my
> bookmarked sites, but just go to something that
> calculates binomial probabilities, and fill in the
> numbers.
> For example, the prob. of playing a hand is 25%, and
> then you designate the number of trials (rounds dealt)
> and the number of "successes" that you're
> looking for, for example playing only 200 rounds out
> of 2,500.
Thanks Don. Good idea. The highest value for N I could use for an online calculator was 1000. The standard deviation was very small, and there is positive skew rather than a normal distribution. Of course as N increases, the skew becomes less and less apparent and resembles a normal distribution.
The reason I brought this topic up is because it would be a good way for a team manager to track how productive his spotters are in so far as calling in the BP into hot shoes.
If records indicate a spotter counts down 200 shoes but signals in a BP for only 25 shoes, then the manager needs to figure out the likelyhood of this event. It could be the spotter is lazy, scared, incompetent, or just unlucky.
MJ
> Thanks Don. Good idea. The highest value for N I could
> use for an online calculator was 1000. The standard
> deviation was very small, and there is positive skew
> rather than a normal distribution. Of course as N
> increases, the skew becomes less and less apparent and
> resembles a normal distribution.
You couldn't have looked too hard. I googled "binomial calculator," and the first entry that came up was:
www.stat.sc.edu/~west/applets/binomialdemo.html
That allowed me to put in any number for n that I wanted.
Don
> You couldn't have looked too hard. I googled
> "binomial calculator," and the first entry
> that came up was:
> www.stat.sc.edu/~west/applets/binomialdemo.html
> That allowed me to put in any number for n that I
> wanted.
Different search engines. :-)
What do you think of using this idea to track the productivity of spotters? If managers use SD to determine if their wins/losses are within reason, why not use this same concept to check whether their productivity is within reason?
If the TC frequency for favorable counts is 25%, then that means on average, 1 out of 4 shoes should result in calling in the BP, correct? So, if a spotter observes 200 shoes, then the expected avg is 50 hot shoes. Any variance should be within reason.
MJ
> Different search engines. :-)
Google is google, no?
> What do you think of using this idea to track the
> productivity of spotters? If managers use SD to
> determine if their wins/losses are within reason, why
> not use this same concept to check whether their
> productivity is within reason?
It's a valid idea.
> If the TC frequency for favorable counts is 25%, then
> that means on average, 1 out of 4 shoes should result
> in calling in the BP, correct?
No, I don't think that's correct. It means that, of all the rounds available to be played in an hour, you get to play about 25 of them, if you see 100 rounds/hour. Do you understand why what you're proposing isn't exactly the same thing?
It could be -- but I'm not saying that it is -- that you get called in on every single shoe but that, during that shoe, you get to play only 25% of the hands. Now, we know from experience that it doesn't work like that, either, but it certainly doesn't work your way. You will surely get called into more than one shoe out of four, on average, but you won't get to play the entirety of the shoes that you're called in for.
>So, if a spotter
> observes 200 shoes, then the expected avg is 50 hot
> shoes. Any variance should be within reason.
See above.
Don
> No, I don't think that's correct. It means that, of
> all the rounds available to be played in an hour, you
> get to play about 25 of them, if you see 100
> rounds/hour. Do you understand why what you're
> proposing isn't exactly the same thing?
Yes, your explanation below pretty much clarifies the difference. My assumption was flawed because it assumes that all 25% of the rounds played by the BP would be played in exactly 25% of the shoes observed. For example, if 100 rounds are dealt per hour, and during the course of the hour 4 shoes are dealt at 25 rounds/shoe, then my initial premise would lead us to conclude that 1 out of 4 shoes would be played by the BP with ALL 25 rounds in a single shoe! Absurd. Those 25 rounds of BP action would more likely be divided up by some fraction of the total # of shoes observed by the spotter. But that is just a rehash of what you wrote.
It should be noted that this methodology (although flawed) should at least give a team manager a starting framework in which to assess the efficiency of the spotter. It provides a bare minimum for the number of call-ins for each spotter.
Back to the matter at hand. How do I go about determining the average # of shoes that will be played for a given backcounting frequency? CVData does provide the total # of shoes that were played for a simulation if that helps. The number of rounds the BP plays can be calculated by multiplying the backcounting frequency x # rounds simulated.
Unfortunately, I doubt this info can be used to solve the problem.
MJ
> Back to the matter at hand. How do I go about
> determining the average # of shoes that will be played
> for a given backcounting frequency? CVData does
> provide the total # of shoes that were played for a
> simulation if that helps. The number of rounds the BP
> plays can be calculated by multiplying the
> backcounting frequency x # rounds simulated.
> Unfortunately, I doubt this info can be used to solve
> the problem.
There is a ton of information in the optimal departure study of Chapter 13 of BJA3. Try there for a start.
But, here's an educated guess: The two extremes of playing only 25% of the shoes, or, alternatively, playing all the shoes but only 25% of the hands are both obviously wrong. So, why not consider something about halfway in between, that is, we may play in about 50% of the shoes, and in each of them about half the time. Just a suggestion.
Don
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