Standard Deviation
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if SD for 1 hour is $2,000, then SD for 15 hours is SQRT(15) multiplied by $2,000.
SQRT(15) = 3.872
Thus one SD for 15 hours equals $7,745 (3.872 x $2,000).
in order to work out the probability of losing it all you also need to have an expected return per hour. You did not state what the EV was.
assuming your EV was exactly 0/hour, then your EV for 15 hours is 0 (15x0=0).
the easiest way to calculate the probability of losing $15k is to use MS Excel.
=normdist(-15000,0,7745,TRUE) = 2.64% - meaning there is a 2.64% chance of losing $15,000 OR MORE in 15 hours. Likewise there would be a 2.64% chance of winning $15,000 OR MORE.
A result of -$15,000 falls just within the 95% confidence interval (which relates to a distance of 1.96 standard deviations from the mean).
SQRT(15) = 3.872
Thus one SD for 15 hours equals $7,745 (3.872 x $2,000).
in order to work out the probability of losing it all you also need to have an expected return per hour. You did not state what the EV was.
assuming your EV was exactly 0/hour, then your EV for 15 hours is 0 (15x0=0).
the easiest way to calculate the probability of losing $15k is to use MS Excel.
=normdist(-15000,0,7745,TRUE) = 2.64% - meaning there is a 2.64% chance of losing $15,000 OR MORE in 15 hours. Likewise there would be a 2.64% chance of winning $15,000 OR MORE.
A result of -$15,000 falls just within the 95% confidence interval (which relates to a distance of 1.96 standard deviations from the mean).
Thanks matt21. The win rate is $128 per hour and the actual SD is $1808 per hour. Not sure I follow your formula. What do you get as the probability of losing $15K or more in 15 hours?
Flash, wouldn't you agree it's a result that should happen pretty infrequently? It has happened to me once in about 50 trips, each with about 15-20 hours of play. It was a loss equal to about 37 max bets. On the up side, I have seen wins of 25 max bets ($10k) on 2 trips. Lots of trips with wins of about $3K-$5K. Of course, this sample size is pretty meaninglessly small, but I'm guessing the big wins/loss are about what should be expected.
Flash, wouldn't you agree it's a result that should happen pretty infrequently? It has happened to me once in about 50 trips, each with about 15-20 hours of play. It was a loss equal to about 37 max bets. On the up side, I have seen wins of 25 max bets ($10k) on 2 trips. Lots of trips with wins of about $3K-$5K. Of course, this sample size is pretty meaninglessly small, but I'm guessing the big wins/loss are about what should be expected.
Ok, so EV(15 hrs) = 15x$128 = $1,920bj42 said:
SD(1hr) = $1,808
SD(15hr) = $1,808x SQRT(15) = $7,002
to work out the probability of losing $15,000 or more we use the normal distribution.
where mean,x = $1,920 and SD=$7,002
=normdist(-15000,1920,7002,TRUE) [the formula that you enter into MS Excel]
=0.78%
there is a 0.78% chance of losing $15000 or more - i.e. this should happen about once in 128 [1/0.78] trips
=normdist(10000,1920,7002,TRUE) [the formula that you enter into MS Excel]
=87.57% - this figure represents the probability of winning $10k or less, or losing - in other words there is a 1-0.8757 i.e. 12.43% chance of winning $10k or more - so this should happen once in every 8 trips.
and now more interestingly.... EV for 50 trips - averaging 17 hours/tripEV = 50 x 17 x $128 = $108,800
SD = SQRT(40x17) x $1,808 = SQRT (680 hrs) x $1,808 = $47,146
immediately this indicates that you have reached N0, i.e. your EV ($108k) is now bigger than one standard deviation ($47k). In fact your EV is now equal to more than TWO standard deviations meaning that if you did a sample of 50 trips over and over, then only once in 100 samples of 50 trips should you make a zero profit or a loss. In other words your sample size is not really that small (680 hours!).
=normdist(0,108800,47146,TRUE) = 1.05%
you have clocked up sufficiently enough hours to really be able to expect to be well in the black.
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Well, you bring up yet another of many areas I don't fully undersatnd lol.matt21 said:
But since bj42 brought up that the $15K may be a trip roll, although, if so, I don't why he asked about losing $15K or more becasue that would be all he has to lose.
In other words, maybe the formula you are using is fine if one assumes one has an adequate roll to FINISH losing $15K or more but, if one doesn't have enough to roll to recover from losses of $15K or more, then maybe tbat formula may not take into account the chances of losing the $15K AT SOME POINT on the way to playing 15 hrs max.
In other words, the chances of losing that $15K on a trip may be twice, or more, the chances of finishing that way.
I don't really know but what do you/anyone think about that?
FLASH - SD is not linear like EV is. All that pesky square root stuff. You just can't do what you seemed to suggest. I hope you understand that.
hi kasi! that's a good question!
Yes the probabiltity calculated above indicates the likelihood of FINISHING $15k or more down. Logically, the likelihood of being down $15k or more AT ANY TIME should be higher than the likelihood of FINISHING at $15k or more down. So clearly this assumes that a player has sufficent bankroll to be more than $15k down.
i dont know how to calculate the likelihood of a certain result being 'touched'.
The reason I expressed the result as '$15k or more' is because of the continuous probability nature of the normal distribution. Generally with the normal distribution, i think we never calculate the likelihood of an exact result, but always the likelihood of achieving more or less than a specific outcome.
Matt
Yes the probabiltity calculated above indicates the likelihood of FINISHING $15k or more down. Logically, the likelihood of being down $15k or more AT ANY TIME should be higher than the likelihood of FINISHING at $15k or more down. So clearly this assumes that a player has sufficent bankroll to be more than $15k down.
i dont know how to calculate the likelihood of a certain result being 'touched'.
The reason I expressed the result as '$15k or more' is because of the continuous probability nature of the normal distribution. Generally with the normal distribution, i think we never calculate the likelihood of an exact result, but always the likelihood of achieving more or less than a specific outcome.
Matt
Thanks rukus. Much appreciated your confirmation that I may have not yet gone totally insane lol.rukus said:
My main point is always to try and understand as best as one can the assumptions behind stuff.
I only mentioned it because, perhaps, it makes bj42's results of losing the $15K once in 50 sessions of 15 hours each even "more to be expected" compared to the more traditional analysis of 1 in 128 like Matt did that may have assumed enough to overcome losing $15K at some point along the way.
The simplest answer, maybe only me lol, is always take, or have access to, your entire roll on a trip and one doesn't have to any longer worry about Trip ROR, how much of total roll to take on a trip lasting x long etc no matter how long one plays on a trip becasue Trip ROR and "lifetime ROR" are then equal.
Would that be true?
Yes, you have it lol.matt21 said:
In simplest terms, with a coin-flip of a max of 100 flips, wherein I am 10 paces to the right of the Grand Canyon to start and take one-step-left with every "head" result, you would say the chances of me falling over the edge =
Sqrt(100)=10 and since mean is where I start from, I'd have a 16% chance of finishing 10 or more units down and dying as I walked over the canyon edge.
But, with adequate roll, I'd bounce back 1/2 the time from when I may have lost all after 70 or 80 tosses or whatever.
So one has to double that 16%, to 32%, of losing all no matter how much roll one started with.
It's the difference between looking at things from an "end-point" perspective versus "at some point" while playing.
In other words, even with a coin-flip, one's chances of being behind by AT LEAST some amount during a trip is about double the chances of being behind by that amount at the END of the trip.
When, playing BJ, as opposed to coin-flipping, the chances, from the end-point perspective can be triple or more the chances of the "at some point" perspective.
I don't know how to calculate those chances of "touching" at some point, as you eloquently put it, either.
I'd just recognize my limitations and trust what the software told me about the "double-barrier" syndrome.
PS - this Grand Canyon experiment I probably massacred above in trying to explain it is from pp 123 at the bottom to pp 124 in BJAIII.
To me, at the time I read it, I still remember it as an "epiphany" in awaking me to the difference from chances of finishing down by X from "end-point" vs "some-point while trying to play that much" lol.
agree here in spirit. your RoRs wouldnt be the same since you play many more hours in the lifetime than over one trip :grin:. with the same bankroll for trip and lifetime, your ROR is determined based on EV, SD, (both the same on trip and lifetime) and hours played (obviously different for trip and lifetime).Kasi said:
in general, i will have access to 60% of my BR for any sizeable trip beyond the 1-2 day excursion. for those 1-2 day trips ill usually have about 40% ready, for anything like a week or longer 75% or more.
As you note, Don S does provide some formulas in the same ROR section you paraphrase to gauge how much BR you need available based on your game's EV and SD and hours planned to be played. they do in fact account for the premature hitting of the barrier problem. it is where my estimates of trip stakes above come from. i personally couldnt solve the equations with a closed-end solution, just through a good approximation using VBA and excel; if someone does have a closed-end solution to these let me know!
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Not that it matters, or am surprised in the slightest, but I like the large %'s of roll you take on various trip lengths.rukus said:
For trip ROR I use the formula on p 132 which I interpret to mean "chances of going bust at any point during a trip of some length". The formula assumes no Goal combined with a time constraint. It's easy to duplicate in Excel.
The formulae on pp 136 & 137 assume that "achieving goal before going bust in an amount of time." stuff. Those are the ones I don't know how to apply lol.
But, then again, on a trip, I guess I have an infinite goal anyway so maybe who really cares anyway that much how much roll to take to achieve a goal in so long without going bust, like I take a 50 unit roll with a goal to win 80 units in a 20 hour trip, etc lol.
I only replied on the slim chance you needed the p 132 formula lol.
You know, I just thought, in case you maybe spread on trips, if you can find it somewhere, there's an article somewhere called "Risk, Ruin and Trip-Stake Wipeout" by Bryce Carlson. Not saying I understood it all but I remember reading it and thought it was pretty cool lol. I think it sort of debunked a little the traditional way of how much to spread in multiple hands to keep risk the same on trips. In effect saying the traditional way of keeping risk constant is only valid assuming a constant target win but not valid for a constant number of rounds like on a trip. It used expected number of max bets to be made on a trip to calc the Trip ROR assuming it's the max bets wherein one loses or wins alot anyway.
On the other hand, who gives a rat's ass lol.
Sorry I misunderstood when you said "$15,000 in 15 hours is just $1,000 per hour This is a completely unremarkably "normal" result - as youFLASH1296 said:
have already stated a priori that the SD is $2,000 per hour ! " knowing that the SD was $2K in one hour.
So, now I ask, since you now "know", if you believe, the chances of losing that $15K in 15 hours is somewhere around 1-3% , just let's say, why do you now characterize this probability as " rather high" whereas before you characterized it as "unremarkably normal"?
Was "unremarkably normal" a typo or something?
I don't know - does a square root function fit the definition of "exponential"?