Quantifying Good & Bad Decisions

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Originally posted by: mrmarcus12LVA
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Originally posted by: snidely333 Each day at Casino X:
10% of patrons win an average of $10,000
90% of patrons lose an average of $1,200

You have bimodal distribution but the casino wins average of $8,000 each day.

If you plot the daily win/loss for the casino for a year it will be a normal distribution with mean of $8000.
Um, getting back to the IP's question, are there other possible sets of results (# of winners and # of losers) with the exact same mean and the exact same SD?

Is it possible there are MORE winners in the 99% game than in the 101% game, even though they both have the same SD?



Normal, Mean and sd define the distribution.
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Originally posted by: FrankKneeland
At least a few of you got the correct answer, which is approximately 15.9%. There was no need to over complicate the answer since we only specified win/lose as the criteria. The percentage of people that will fall within + or - 1 standard deviation of mean is 68.2% of the remainder, half will do worse and half will do better. Therefore 100% - 68.2% / 2 = 15.9%. As I said it was easy math. Now what does it mean and what are the in-obvious implications.

That is to say:

Group A: Out of the 100 people Playing 1 million hands of 99% return machines, about 16 of them would be ahead at the end of a full year of play...even though they were playing with a negative expectancy.

Group B: Out of the 100 people Playing 1 million hands of 101% return machines, about 16 of them would be stuck at the end of a full year of play...even though they were playing with a positive expectancy.

Obviously, Group A would have 84 losers and Group B would have 84 winners.

Now imagine this:

Before running the experiment you told each group what their expectancy was, and whether or not they were supposed to win or lose. You even told them how much they were supposed to win or lose, which was $50,000. During the one year experiment none of the subjects was allowed to discuss their results or talk about the experiment publicly (for instance on-line forums).

Your job is to reason out which folks in each group would be the most vocal at the end of the experiment and what they might have to say about things.

Group A The 84% losers: They were told they should lose playing these games and they lost.
1. What do you think would be on their mind, and what if anything would they be saying if they posted on on-line forums.
2. And how likely do you think it would be that they would be talking on on-line forums.

Group A The 16% Winners: They were told they should lose, but they won. Same questions as above.

Group B The 84% Winners: They were told they should win and they won. Same questions as above with careful attention to whether you'd think they would even bother posting on forums. Keep in mind everything went exactly as planned for these folk and they likely don't feel they have anything to learn or share on a forum.

Group B The 16% Losers: They were told they should win, but they lost instead. Same questions as above, with extra attention to what they would likely now think and be saying about positive expectation gambling.

When answering these questions consider this seemingly unrelated question: When was the last time you got up in the morning, brushed your teeth, drove to work, had an uneventful day, drove home, had dinner, watched a TV show, and went to sleep...and then felt the need to talk to everyone about your completely unremarkable and normal day???.???


If anyone answers these question I have a couple of follow up questions and conclusions.
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Originally posted by: snidely333 Normal, Mean and sd define the distribution.
Um, getting back to the IP's question, there was no mention of any normal distributions. Nor any mention that either game was being played for optimal EV by anyone.

In fact, we can assume that the 100 people playing the 99% game are NOT playing for optimal EV.

ASSUMING all the players play for optimal EV (a terrible, grossly irrational assumption), after 100,000,000 hands of each game, the one group will have $101,000,000 and the other will have $99,000,000, and the casino will have what it started with (???). But the number of winners and losers in each group is an entirely different question from the total return to each group.

Could the group that made $1 million actually have FEWER winners than the group that lost $1 million, merely by chance?
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Originally posted by: mrmarcus12LVA
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Originally posted by: snidely333 Normal, Mean and sd define the distribution.
Um, getting back to the IP's question, there was no mention of any normal distributions. Nor any mention that either game was being played for optimal EV by anyone.

In fact, we can assume that the 100 people playing the 99% game are NOT playing for optimal EV.


You can assume that if you'd like to get bogged down with pointless details. It was a hypothetical situation, not a real one. There was no reason to complicate it. I kept it simple.

If I'm telling you a joke that begins, "Two guys walk into a bar" are you the type that asks, "what's the name of the bar?"?

Your points are valid if this was a real life study. It is not, so the things you are bringing up are not relevant to the discussion. It would be nearly impossible to train 200 people to play perfectly, so let's just assume the games are 99% & 101% return with errors. Does that solve your quandary?

The problem would be simplified if both groups played the same game, but one played with 99% EV, on average, and the other group played with 101% EV, on average.
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Originally posted by: snidely333 Luckily, the math police will not kick you in the head 5 times. But if they did, I would understand.
They only do that if you don't correct your answer within the split second allowed.

I find it odd that even though there has been considerable commentary, no one has directly answered or attempted to answer any of the direct questions in the second part of the exercise.
The 84% groups don't really have a good story to tell unless it involves getting the number of the cocktail waitress. The 16% group has more of a story, but the losing side won't really want to tell theirs unless it's of the type of "oh, you think you had a bad week, let me tell you about mine..." (or to bash all AP players as losing addicts). Those 16% who beat the odds and won always want to brag and tell their story.
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Originally posted by: FrankKneeland
When was the last time you got up in the morning, brushed your teeth, drove to work, had an uneventful day, drove home, had dinner, watched a TV show, and went to sleep...and then felt the need to talk to everyone about your completely unremarkable and normal day???.???
Umm, . . . if responding to lettuceman's "What did you have for dinner tonight?"-thread counts, it was two days ago.

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Originally posted by: KayPea
The 84% groups don't really have a good story to tell unless it involves getting the number of the cocktail waitress. The 16% group has more of a story, but the losing side won't really want to tell theirs unless it's of the type of "oh, you think you had a bad week, let me tell you about mine..." (or to bash all AP players as losing addicts). Those 16% who beat the odds and won always want to brag and tell their story.


Yes I like your answers and I think you have predicted where I'm going with this thread.

I'll wrap this up in a day. Busy tonight.
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