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[QUOTE="aRoseIsaRose, post: 2126660, member: 95870"]
I asked a poker math question earlier and I was told that getting 37 pairs and never flopping a set was probably not a high enough sample size and that not getting one set on the flop (different hands went beyond the flop some did not) was maybe on par because there were not enough hands to I guess make the odds start working where at least one pair turned into a set. I do not understand what makes a sample size. If  5.8 percent of pairs flop a set, then with 37 sets I think that I would at least hit one seeming that some hands went beyond the flop. Where does the "enough hands" start. I've seen people explain how after so many 1000's of hands that what they were dealt came up to what was expected percentage.

If something should happen 5.8 percent of the time, when does it start to seem strange when 0 times happen.  My question is how many hands does it take to make a good "sample size'?

Is there a mathematical solution or is it like the following:
                      The Monty Hall Problem (Best as I can tell the story)

If you are on the games show Let's Make a Deal and you have a chance to pick one of three doors and if you pick the door with the car you win it. 
     You have 3 doors to pick from and one of the doors has a car behind it and the other 2 have goats behind them, and after you pick one, Mointy will always go  to one of the other doors and open a door that has a goat behind it. Monty  now asks you if you would like to change your mind and pick the other door or keep the one you already picked.
     Ok you started out with a 33.33 percent chance of picking the right door in the first place but since Monty will always go and open the door that has a goat behind it, that door loses its 33 percent and it adds to the 33 percent to the door you didn't pick at first. 
      Supposedly the percentage cannot be added to your door because you already picked it or do you think it would split between the two remaining doors or or that each door remains at 33 percent. Well you already have one door opened so it can't have any percentage left now. or 0 percent if you would.    
           Would you change doors to the higher percentage door or do you still think that each door remains at a 33 percent chance. 
     Or now that one of the doors is open, do you have a 50/50 chance of having the right door seeming there are two doors left.  
                   This is known as the Monty Hall Problem. It is a highly debated mathematical problem that even the person with one of the highest or the highest  IQ  in the world, Marilyn Savant, has a hard time getting people to see the math that she came up with to, in her opinion, is the right answer.  
                  If you think about it too hard and read the different answers and explantions that people come up with, it can make you dizzy going around in circles on the steps you could use to solve the problem.     
     [URL]http://en.wikipedia.org/wiki/Monty_Hall_problem[/URL]
:banghead::banghead::banghead::banghead:
[/QUOTE]

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