understanding stats and probability - discussion and tutorials

[pellicle]

Hi

as (in the main, lukers and the as yet unoperated) we are all medical patients and likely are interested in understanding the stuff we read. Accordingly a grounding in statistics (Stats) and its most basic maths is important. When we read papers and the analysis of studies all of it is reduced in some ways to statistical terms.

A very close friend of Stats is probability, which also is swung around the head by the same authors of papers.

People get confused (even the authors of papers) with this subject. Since the topic came up I thought that I would start the topic off here and people can add to this as they see fit.

I thought I'd start the ball rolling with a most basic probability issue: the coin toss.

Now we all know that a coin has 2 sides, and that each toss is not influnced by the previous toss or has any influence on the next. So then when we say its a 50% probability of getting a head, this implies that its a 25% chance of getting 2 heads in a row.

This does NOT mean that if you did 4 sets of coin tosses that one must be two heads in a row. It simply means its likely. Because each event has no relationship with the other.

 

[newmitral]

Hi Everyone,

Well, I got myself into this by offering to explain why my probability math is correct in response to Fundy's interpretation of annual bleed risk over on the "Valve Selection" forum area in the thread titled "Anyone wish they had chose the other option?" :

http://www.valvereplacement.org/forums/showthread.php?41059-Anyone-wish-they-had-chose-the-other-option/page3

First of all, I feel I owe a bit of introduction to support my claim that I do actually know at least a bit of what I'm talking about. I am NOT a statistician or math major, but I am an engineer with both a MS and BS degree from MIT. I did help teach some courses there during my senior and graduate years, but that was many moons ago. I do still work as a research scientist, and interact with math majors and other scientists (yes, I even know some NASA rocket scientists) on a frequent basis. So, while I think I still remember how to explain things reasonably clearly, if I come across as pedantic I assure you it is not intentional, and I'm only trying to help.

Because so many of the medical papers you will read involve statistical analysis of data and present probabilities of risk in the conclusions, it is easy to misinterpret these papers. Sometimes even the authors are wrong, although this is not usually the case if you stick to peer-reviewed journal articles.

Because this is a topic so near and dear to our hearts (pun intented), I am willing to risk public humiliation and ridicule trying to explain some of the key points here.

To begin, I want to follow up first on the example that Fundy gave in the other thread, where he claimed that if a woman were planning to have 4 children, and each child had a 25% risk of inheriting a disease, then he concluded there was a 100% probability that 1 child would have the disease if there were 4 children total with 25% chance each.

"In the 25% chance of a baby with disease. If the woman said she expected to have 4 children and another woman says 8 children in the 25% chance per child scenario. Each child is a 25% chance no matter how many children are born.

But in the first case 4 x 25% would indicate that after four children there is a 100% statistical probability of having one child with the disease. Meaning she should expect to have 3 disease free and one with the disease. "

My response to that was:

"If each child has a 25% chance (0.25 probability) of having a disease. Then the probability of each child NOT having the disease is 0.75
If you expect to have 4 children then the probability of having 4 children ALL without the disease is 0.75*0.75*0.75*0.75 = 0.316 or about 31.6% probability that all 4 children will be disease free, or about 68.4% probability that AT LEAST one of the children will have the disease, never 100%."

By way of proof for this case, I have prepared an Excel spread sheet that shows each and every possibility in this scenario, which you can download from this link:

http://www.wstco.com/ValveReplacementInfo/4KidsWith25PercentChanceEach.xls

To represent the 25% chance of disease for each child, I assigned 4 possible states for each child; A, B, C, and D, where "A","B" & "C" represent a disease-free outcome for that child, and "D" represents the child having the disease. Thus, there is a 25% probability for each individual child that he/she will have the disease.

The spreadsheet presents all possible permutations/outcomes of 4 children having 4 possible states each. There are 256 permutations, or possible outcomes for this proposed situation (4 to the power of 4). I color coded the spreadsheet so each permutation will be easier to track.
Each of these 256 outcomes has a 1 in 256 chance of happening if no child is yet conceived.

If you go through all 256 possible permutations and count how many there are with NO child having state "D", you will get 81 such possible outcomes. I have made this a bit easier by adding the columns to the right of the permutation matrix, one titled "Are ALL children disease free".
If there are 81 cases out of 256 possible outcomes where ALL children (4 out of 4) are disease free, this is a probability of 81/256 = .3164 or 31.64%. Please notice that this number agrees with the number I gave in the other thread.

If you again go through all 256 possible permutations and count how many there are where AT LEAST one child had the disease (state "D") you will get 175 such outcomes out of the total possible 256 outcomes. So, 175/256 = .6836 or 68.36%. Again, this agrees with the number I presented for this case in the other thread.

The sum of all the possibilities DOES equal 1 since 0.3164 + 0.6836 = 1.00
Notice that the two possibilities above are mutually exclusive, and do cover all the situations. Either all children are disease-free or there are 1 or more children with the disease. There are no other logical options. 81 disease free outcomes plus 175 with some disease outcomes makes up the full 256 set of permutations.

A different question, not specifically given before, is "What is the likelihood that EXACTLY ONE child will have the disease, and the other 3 will be free of it?". Counting the number of permutations that have exactly 1 child with the disease gives 108 such outcomes. So the probability of EXACTLY ONE child having the disease out of 4 is 108/256 = 0.4219 or about 42.2%. This case is NOT mutually exclusive from the case of AT LEAST ONE child having the disease, but rather it is one component of that greater aggregate possibility. Also, note that this is the probability of ANY SINGLE ONE out of the 4 being the lone unlucky child, not the probability of specifically child #1 or specifically child #2 having the disease - those would be back to the 25% probabilities we started with.

How many cases are there where EXACTLY 2 children out of 4 have the disease? There are 54 such cases, giving a probability of 54/256 = 0.2109 or 21.9%

There are 12 cases where EXACTLY 3 children out of 4 have the disease, giving a probability of 12/256 = 0.0469 or 4.7%

There is exactly 1 case where all 4 children out of 4 have the disease, giving a probability of this outcome of 1/256 = 0.0039 or about 0.4%

Once again, the total of all the possible mutually-exclusive outcomes is:
81 none of the 4 children has disease
108 where 1 child out of 4 has disease
54 where 2 children out of 4 have the disease
12 where 3 children out of 4 have the disease
1 where all 4 children have the disease
---------------------
256 total possible outcomes

Notice that the case of AT LEAST ONE child having the disease given earlier (0.6836) DOES equal the sum of the 4 more restrictive cases where EXACTLY 1, 2, 3 or 4 children have the disease ( 0.4219 + 0.2109 + 0.0469 + 0.0039 = 0.6836). It may be confusing that I am adding these probabilities here, but this is because these are essentially "simultaneous" probabilities as given, not sequential probabilities of 4 things happening in a row.

Hopefully this helps prove that the math I used in the other thread is correct and does give the correct result. By going through each of the individual permutations, this is a brute-force way of getting the same answer, but perhaps it is easier to see that it is correct.


I'll try to monitor this thread frequently throughout the weekend to get into more explanations, or other case examples if I can. But, I have a be spending a lot of time up in the attic this weekend to get some air-sealing and insulation done before the colder Winter weather really hits us here.

 

[pellicle]

applause

laudations for effort
Qualificathions for me? No stats major, but a bit os use: BSc, Ass Dip Elec Eng, B Info Tech, M Sc

 

[Fundy]

Well I now understand where my confusion lies. After looking to my old textbook and finding the usage of the algorithm newmitral used to calculate his 64% number. The part that confused me was the multiple occurences of a bleed event happening with some people. Thus almost 64% of the people experience at least one bleed event over the 25 year period. Some of those 64 will have multiple bleed events. Those multiple bleed event occurrences explain why things didn't totally add up.

So if there are 100 people under consideration almost 64 of them will have a major bleed event. But some of those 64 have more than 1; I still maintain that the total bleed events are 100 but they only occur to 64 people. 36 of the people will have zero bleed events.

I guess it would be wrong to say that 100 bleed events amongst 100 people would be 100% chance of having a bleed occur when there's near certainty that 34 will have no bleed events. For some reason I failed to comprehend the significance of a person having multiple bleed events when newmitral explained it. But for some reason I understood the textbook more easily.

newmitral, Is that where you figure I was misunderstanding the bleed event scenario.

 

[Fundy]

If I understand the calc for the lottery example of 10 people each with a ticket for a draw. Each person has a 10% chance of winning the draw. If there is 20 draws then each person's chance of winning at least once is 87.84%.

Total number of winners is 20 still. But since some will be multiple winners you have to figure out the probability that a person won't win at all. That number is equal to 90% per draw multiplied by itself 20 times or more simply .9 to the power of 20. which is 12.6%. representing the probaility of a person not winning at least once.

The complement of the number, 87.84% represents all other outcomes. That being a person winning once, a person winning, twice, three times, even the occurence of a person winning all 20 times even.

So the probability of a person winning at least once is 87.84%. Some must however win multiple times so that total wins is equal to 20.

Look correct?

 

[newmitral]

Hi Fundy,

You are basically correct on the lottery example. You explain your logic properly, but I think there is a typo in the math. You are correct that if you have a 0.1 probability of winning each individual draw, you have a 0.9 probability of not winning. So, in 20 lottery draws, there is 0.9 to the power of 20 which = 0.1216 or 12.16% chance of not winning any of the draws, not 12.6%. So, you have a 1-0.1216 = 0.8784 or 87.84% chance of winning the draw AT LEAST once in 20 times. Since this number equals the number you quoted, I'm assuming the first was just a typo. You are correct that this calculation is NOT a 100% probability because some people (maybe you) may win more than once. The odds of you winning EXACTLY once will be different.

I think there is some confusion with your terminology when you say something, like the bleed, has a 100% probability. A 100% probability means it is absolutely positively guaranteed to happen. This is almost never the case with most any statistical event, by definition. If it is a statistical problem it is not a certainty or you wouldn't be calculating the odds.

If you ever get an answer that comes up with 100%, or more than 100%, you know you are most likely wrong somewhere. Although sports coaches like to ask for 110% from their players, this is not a possibility in mathematics as you can never have more than 100% of anything.

 

[Mom2izzy]

Wow...thanks for the details. I also do well in math courses, have a math minor and recently completed some additional stats classes (not including much on the probability side)...I still couldn't get the details right, but I was pretty certain that it wouldn't be cumulative...thanks for the clarification.

There's one other thing that perplexes me about all of this...in the lottery, coin toss and even the baby example, we are talking about "random" occurrences. Bleeds/strokes that may be attributed to ACT don't seem to fit a "random" model. Proper INR management would lower risks; whereas, certain lifestyle choices (e.g. extreme sports) may increase it. To me, it seems like there are just way too many variables to give a number that would apply to everyone on ACT.

My main point was to make sure people were not confused or scared by the 100% statistics being thrown out there. Thank you for helping with that. I think I may have encountered some people who LOVE Excel even more than I do ...amazing!!

 

[newmitral]

Hi Again,

I think I need to add some further explanation as to why the lottery example is different from the bleed example. These are really two different problems with different sets of rules.

The way the lottery problem is set up, you DO have a guaranteed winner for each of the 20 draws. You have the SAME 10 participants and there is one and only one guaranteed winner in each draw. 20 guaranteed winning events (not necessarily 20 separate individuals) So, that is how you establish your 1 in 10 chance of winning for each participant. You started with a known guaranteed outcome and calculated individual participant probalities.

The bleed problem is fundamentally different. The medical study that arrived at the 4% risk of bleed per year had to start with a much larger sample than just 100 people, and had to look over a longer period of time than just 1 year. Otherwise the study would be flawed as the sample would not have been statistically representative of the general population under study. To interpret the 4% annual risk per person you have to first understand that while there is a 0.04 probability that any given person will have a bleed, this does NOT mean that there is a guarantee of exactly 4 bleeds per 100 people each and every year. If you follow the same (lets say) 1,000 people, it is perfectly possible to have some years where there will be 40 bleeds (40/1,000 = 0.04) but there is also some probability that there will be 60, only 20 or even zero bleeds in a given year. On average, over many years you should see about 40 bleeds each year out of 1,000 people if the author's study is correct. The nature of the problem is different from the lottery example because in the bleed scenario you are not starting with a guaranteed outcome and calculating individual probabilities from there. Instead, you are assigning average risks from a larger sample population down to an individual case and calculating that individual's risk over time (number of years).

If there is a 4% probability that a given individual in a pool of 100 will have a bleed in one year, then you can calculate the probability that NOBODY in the pool of 100 people will have a bleed in a given year. That probability is the individual's probability (1.0-0.04 = 0.96) multiplied for each individual in the pool. So, there is a 0.96 to the power of 100 = 0.01687 or 1.687% probability that out of 100 people, NONE of them will have a bleed in a given year, even though each individually has a 0.04 probability of having a bleed. There is a 1-01687 = 0.98313 or 98.313% chance that AT LEAST one of those 100 people will have a bleed. Again, note that you are NOT guaranteed 100% that even 1 person out of your pool of 100 will have a bleed any given year, let alone 4.

 

[Fundy]

newmitral I'm going to throw another bleed risk discussion at you, if you don't mind looking at it quickly.

For instance one that should closely apply to myself if i were on warfarin. Many of the major bleed risk stats thrown out are usually something like 1% annually under age 70 and 2% over age 70. Which is less than the previously discussed study. I was 45 at replacement date and I'm ballparking my life to last until 90.

So I'm saying I would have 25 years of 1% annual risk of a major bleed event and 15 years of 2% risk. Does that seem about right regarding what is often cited for risk?

So here's my calculation of the probability I would have had at least one year where a major bleed event occurs to me, if I were on warfarin.

No probability of a bleed is 0.99 to the power of 25 multiplied by 0.98 to the power of 15. ( 0.99^25 * 0.98^15 )
which is .5745 or 57.45% chance I will have no bleed events at all in my approximate lifetime.
The complement of that is .4255 or 42.55% which represents having at least one major bleed event.

So if I were on warfarin and based on the stated bleed risks above of 1% and 2%; there is a 42.55% chance that I would have at least one year where there is one or more major bleed events happening.

That seems about right doesn't it? I forget what the mortality rate per bleed event usually is cited at with regard to those above numbers though, so I'll have to skip attempting that one.

 

[newmitral]

There's one other thing that perplexes me about all of this...in the lottery, coin toss and even the baby example, we are talking about "random" occurrences. Bleeds/strokes that may be attributed to ACT don't seem to fit a "random" model. Proper INR management would lower risks; whereas, certain lifestyle choices (e.g. extreme sports) may increase it. To me, it seems like there are just way too many variables to give a number that would apply to everyone on ACT.

Hi Mom2izzy,

I posted my second explanation before I saw your question, but I think it partially adresses it. To be more more specific, in the lottery, coin toss or baby birth, these examples all start with an event happening, and then calculate the probabilities of a given outcome for that event. You have a guarante that an event will occur and calculate probabilities related to it.

With the bleed/stroke situation you are starting with the probabilities of a given event happening and calculating the odds of it happening to a given person withing a given time period. The bleed/stroke problem is fundamentally different and has to be approached with a different perspective. You do NOT have a guaranteed event, you are calculating the odds of the event occurring.

Your comment about "way too many variables" is right on. That is why, when you see a paper that quotes a certain probability/risk of an event happening to a patient, you really have to read the paper carefully to see if it truly relates to the same situation you are in. With stroke/bleeds in particular, many of those papers are from studies of atrial fibrulation patients, not valve replacement patients. Even if they are for valvers, some of the studies are for mechanical valves only, some are for mixes of bio-prosthetic and mechanical, some are for older valve designs (ball & cage) and some don't separate results for aortic versus mitral valve risks.
Many of the numbers quoted for risks of bleed/stroke for mechanical valve studies don't separate those on ACT who are within their target INR ranges from those who are outside their target ranges. So, to really match the risks/probabilities presented in any of the journal articles you read, or discussions here on this forum, it takes a lot of scrutiny to determine which ones truly apply to you for whatever particular valve type you have, or situation you are in.

 

[newmitral]

So if I were on warfarin and based on the stated bleed risks above of 1% and 2%; there is a 42.55% chance that I would have at least one year where there is one or more major bleed events happening.

That seems about right doesn't it? I forget what the mortality rate per bleed event usually is cited at with regard to those above numbers though, so I'll have to skip attempting that one.

Hi Fundy,

I believe your math and the chance of bleed you arrived at is right this time, based on the annual probabilities you started with.

If I recall, the paper referred to in the other forum area quoted a risk that 20% of all the major bleed were fatal. So, with the 42.55% chance you will have a bleed at all in 40 years, there is then only an 8.51% risk of having a fatal bleed in 40 years with the assumptions you started with.

[update: see corrected calculation in later post]

As with my comment to Mom2izzy, I urge you, and others, to carefully examine the source of those probabilities to see if they are correct for your particular valve replacement situation.

Not to start another discussion topic here, but I do want to reiterate that many of the papers I read present a combined risk of thromboembolism and bleed events.
For mechanical valve patients this makes sense, since you are balancing those two risks to arrive at an optimal INR range that has the minimum overall risk.

Bio-prosthetic valves are not risk free. Although you don't have the bleed risk associated with possibly higher than target INR from a lifetime on warfarin, the bio-prosthetic valves do themselves raise the risk of thromboembolism, just not high enough to justify or overcome the added risk associated with anti-coagulation. That is why bio-prosthetic valve patients are placed on warfarin for only their first 3 months (the period of highest risk for thromboembolism) and then the warfarin is stopped for them. After the first 3 months, the added risk of ACT is higher than the residual thromboembolism risk. But the residual thromboembolism risk is NOT zero.

The papers I have read (I'll see if I can dig up the references) generally conclude that a mechanical valve patient who stays within his/her target INR taking warfarin has approximately the same overall risk of undesired adverse event (bleed/stroke) as that of a bio-prosthetic valve patient (stroke-only) not taking warfarin. The numbers that I believe for that are in the 1% annual risk range. So, you probably have about the same 8% risk of fatal stroke over 40 years with a bio-prosthetic valve as the fatal bleed/stroke risk from the mechanical valve with proper ACT.
I give that opinion with the caveat that it is only my recollection from having read lots of articles when I was looking into the state of the art after already having had my valve replaced. I could be mis-remembering the facts. So take this with a grain of salt until I can find the specific reference articles where I read that conclusion.

You never completely eliminate the risks in either valve choice. All you can do is manage the things you can control to minimize the risks for your particular situation.

 

[Freddie]

Al Lodwick posted this VR years ago. Look at the upper left corner of this link and then click on the link below it:

http://web.archive.org/web/20091231184542/http://www.warfarinfo.com/

This whole topic has been visited time and time again and proven wrong.

 

[newmitral]

Hi Freddie,

I appreciate the link, and I also question the applicability of the bleed risk numbers quoted in the other thread. But, I offered to help folks with the math associated with applying the probabilities to a given situation. I don't want to divert this thread into a discussion of the accuracy of the probabilities one is starting with. That topic would be appropriate for the regular forums.

I will quickly say that while I have great respect for Al Ludwick, in this case I think he is not presenting the whole story. He presents a very small number of deaths by bleeding (32 total per year in the US) from what was stated as the cause of death on the death certificates - counting only those listing bleeding from overdose of warfarin. But, the studies that speak of bleed risk from high INR generally include hemorrhagic stroke, which would usually be listed as stroke being the primary cause of death on the death certificate, and therefore not be counted in Al's number. The stroke would be the primary cause of death, but the warfarin might have been a secondary contributor.

That's why I stated that you always have to scrutinize the risk numbers presented by anyone, including Al Ludwidk, to see if they are applicable to the specific situation you are addressing.
I personally think the 4%/year number is way too high, but I also think the 0.00001%/year number is way too low when assessing the overall risk of adverse events.

Again, I thank you for the link, but I think this discussion of what the actual risk is should be addressed in a thread in the primary forums. I want to keep this particular thread to a discussion of how to do the math once you have a risk number you want to use.

I do realize that I'm guilty of straying from the math topic myself in the above posts. I'll try harder to stick to the topic here.

 

[pellicle]

I do realize that I'm guilty of straying from the math topic myself in the above posts. I'll try harder to stick to the topic here.

Well it is that type or forum, so I for one don't mind

 

[Duffey]

Well it is that type or forum, so I for one don't mind

Right on! It's Small Talk Forum; the rules are REAL relaxed here, and I personally am enjoying the discussion.:smile2:

 

[newmitral]

OK, if you don't mind my following up on my prior post about tissue valves having the same residual risk of stroke as a properly anti-coagulated mechanical valve, I did locate at least one of the papers that I had previously read which gives that conclusion.

The paper I have as the reference is:
"Mitral Valve Replacement and Thromboembolic Risk"
Pub: The Journal of Heart Valve Disease 2004
Authors: Stefano Benussi, Alessandro Verzini, Ottavio Alfieri
Division of Cardiac Surgery, S. Raffaele University Hospital, Milan, Italy

Here is a link to the paper on the web :
http://www.icr-heart.com/journal/maysup2004/12211_Naples_24_r1.pdf

As you can guess from my forum user name, I am personally more interested in the statistics related to mitral valve replacement than aortic, so my own reading concentrated on that.

This paper gives a nice consice summary of the problem, and makes the statement on page 2:
"Available evidence shows that thromboembolic risk
is equivalent for patients with a bioprosthesis (not
receiving oral anticoagulants) and for those with a
mechanical device under a proper anticoagulation regimen"

I would like to believe that this is true, and so I do I haven't found any scholarly articles to the contary for my particular situation so far.
So, I am proceeding under the assumption (delusion??) that my risk of stroke/bleed with my mechanical valve, as long as I stay anti-coagulated within my target range, is no worse than it would have been if I had chosen the tissue valve option. So, indeed, it is a tradeoff of the minor (in my opinion) hassle of taking the daily pill and being a bit more careful with extreme sports versus the risks and major (in my opinion) hassle of having one or more reops with a tissue valve in my lifetime.

I also hope that the On-X folks will show via their Proact study that my valve design might have a lower thromboembolism risk than some of the older models, and possibly allow me to lower my target INR, maybe even (pipe dream??) lowering the overall adverse event risk (bleed/stroke) with my particular mechanical valve below that of the tissue valve option - provided I stay within my target INR range. But, like the percutanious tissue valve reop option, this is a hope for the future, not yet a mainstream option.

 

[Fundy]

newmitral, I definitely want to thank you for your time spent presenting the probability calcs and discussion. I definitely had a hard time finding the percentage of one or more wins useful to me in reference to the roulette example you used. I definitely still needed to apply the average wins per spin calcilation to win payouts versus amount spent wagering. I found it necessary to determine if 38 should be breakeven and that 35 for a win is reason to not play, and that 40 per win is profitable to want to play.

However, even if such a game could be found and be profitable. The calculation applied to roulette would explain why I would go home broke so often from a profitable situation if the bet amounts I place are too large.

Once again, thanks for your time spent explaining to me why the average in many situations are basically useless in itself.

 

[newmitral]

Hi Fundy,

You're quite welcome. I'm happy to help, and it is acutually a good exercise for me to see how much statistics and probability I do actually remember. It has been a long time since college, and I don't do these sorts of calculations all that frequently.

But, the Las Vegas gaming situation is one area where the casinos prey on folks who make many common misinterpretations of how these probabilities apply. Many folks assume, as in the roulette example, that the odds will "even out" somehow and if they watch the wheel and "red-7" has not come up in several spins, that somehow it is "due" to hit soon. The casino folks love people like that, because the probabilities don't work that way. In roulette, each new spin starts the probabilities over, and the past results have absolutely no impact on the next and future spins, unless the wheel is rigged - in which case all bets are off :wink2:

Roulette, and pretty much all of the table games, are designed so that even if you do make the average number of expected wins in the long term, you still come away a loser. That's the other "house advantage".

The only game where the odds might actually be turned to your favor was (is??) in blackjack if you were able to count cards. The casinos have changed the rules after this was uncovered by some folks at my old alma-mater, so it's a lot harder to do now with multiple decks and more frequent reshuffling. Also, while card counting is NOT illegal, if the casinos think you are doing it, they can and will ban you from playing. I want to make it clear that I was not in any way involved with the MIT blackjack team, particularly since this did not happen untill after I gradulated. If you don't know the story of the MIT Blackjack team, google it, or better yet, see if you can find and watch the History Channel documentary about it, or one of the other similar films.

The reason that blackjack is the only game where you could win, at least back in 1980 or thereabouts, was that the hands being dealt actually DID depend on the past deals. As the cards were dealt from the deck, your odds for the current and future hands changed based on how many high or low cards had already been dealt in prior hands, which changed the odds of getting a good or bad card from the remaining ones left in the deck to be dealt. So, if you kept track, you could tell when the odds were in your favor, and bet more heavily. When the odds were not in your favor, you bet less money. This was the only game where this strategy would have a chance of working. Subsequently, the casinos increased the frequency at which they shuffled the decks and started over. They also went to multiple deck blackjack deals, so that the potential odds shift would be much smaller. They also now watch for any pattern in your betting to see if you are increasing/decreasing your bets in a manner that would indicate you are counting the cards, in which case they ban you from playing. So, I'm definitely NOT recommending that anyone try this.

For a while, there was also a video poker game where the payback table was such that if you made exactly all the right moves with regard to keeping or tossing out cards in the poker hand, you had a slightly positive payback expection in the long term. There are several sites you can google about video poker strategies that describe that game and how the payback tables effect your odds of coming away a winner. Naturally, those games with the old payback tables are no longer available in the casinos. With the most favorable payback tables used today, even if you make all the right moves, you will still eventually lose in all the video poker terminals that are in the casinos today.

Video poker can still make your losses happen a lot slower than many of the other slot machine games if you play all the right moves. You still eventually lose, it just takes longer. So, on a dollar per hour of entertainment value, it might be OK. Personally, I'm not a gambler. I only know about the video poker thing because I went to Las Vegas for a friend's wedding several years ago and looked into which game would be the most likely to make whatever money I was willing to toss away for entertainment last as long as possible.

 

[Fundy]

I've heard of that MIT Team. Seen intereviews or discussions where Andy Bloch was talking about the MIT BlackJack team, I believe he either started it or was a part of the few that started it if I remember correctly. I've seen Chris Ferguson do some very impressive simple demos of card counting, and its definitely useful for seven card stud especially large player games. I doubt you could beat a good counter if you couldn't count. no matter how well you knew the card math. Basically he took a deck, turned each card over, one on top of the next (fast) and knew exactly what the last card was each time. Simple demo but damn impressive to me.


That's another area where my understanding often became confused, during hold em hand discussions. Most people are always referring to odds using the ratios such as 3:1 but I have a tendency to almost always reduce to my own figurings substituting 1 in 4 instead, and adjusting by including the bet size in pot count. Quite often I'd be mistaking someone's mention of 3:1 and discussing it as if he said 1 in 3. For some reason I found it easier dealing in percentages.

Sometimes I play the micros a lot(20-40)hrs a week, and sometimes like now I rarely play(maybe two sessions in six months). Been playing with the same $100 since Oct 2007. But still bouncing around that $ level.

What is the name of the documentary you mentioned. I see references to him featured in a book "Bringing Down the House", would that be the name? Maybe I can find a torrent for it somewhere?

Are you sure about the mortality rate as a factor of the percentage rather than the average. I'd have thought it would be 13% rather than your above calculation of 8.51%. As it is expressed as a percentage of the total. Or am I confused about this somehow?

 

[newmitral]

Are you sure about the mortality rate as a factor of the percentage rather than the average. I'd have thought it would be 13% rather than your above calculation of 8.51%.

Hi Fundy,

No, I'm not sure, and thinking about it a bit more I was too quick and cavalier with my first answer.
Looking at that problem a bit more carefully, I think neither the 8.51% nor the 13% answer is correct.

I am sure that the probability of a combined event is the product of their individual probabilities. So, if you have a 1% risk of an event in a given year, and 20% of the events are fatal, then you have a 0.01*0.20 = .002 probability of a fatal event in any given year.
Where I messed up is not taking into account the fact that if you have a fatal event in, let's say, year 9, then you are out of the picture, and you don't get to have any events, fatal or otherwise, in subsequent years.

So, to do the calculation again, after considering the rules of the problem a bit more accurately, here's what I get:

With your 1% assumed event risk for 25 years, you have to NOT have a FATAL event each of those 25 years.
So, if the probability of a fatal event is 0.002/year, then the probability of not having a fatal event is
1.0 - 0.002 = 0.998/year
To go 25 years without the fatal event is 0.998^25 = 0.95118

For the next 15 years, you wanted to use a 2% or 0.02 probability of having an event, again assuming 20% of those are fatal, that gives a probability of 0.004 of a fatal event per year for those next 15 years. So, you have a probability of NOT having a fatal event of 0.996 for those years.
To go 15 years without a fatal event is then 0.996^15 = 0.94165

So the probability of getting through BOTH the 25 and 15 year periods without a fatal event is the product of those two numbers:
40 year combined probability = 0.95118 * 0.94165 = 0.89568

The odds of having a fatal event in that 40 year period, with your starting assumptions, is then 1.0-0.89568 = 0.10432
or about 10.43%

Again, I'm not sure I trust the 1% and 2% risk numbers, or the 20% fatality number, but that's how I would do the math.

I could still be mistaken, and there could be an additional wrinkle to this sort of problem where an early fatality takes you out of the picture for subsequent years. So if there are any actual statisticians who want to chime in, please do so.

What is the name of the documentary you mentioned.

I saw it several years ago on a cable channel. I didn't remember the exact title, so I googled it, and it seems the title was "Breaking Vegas". The video, surprisingly, is available on YouTube.