Poker Math and the Monty Hall Problem

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.
http://en.wikipedia.org/wiki/Monty_Hall_problem
:banghead::banghead::banghead::banghead:

I think the guys that have played 100s of thousands of hands would be able to give you a more accurate answer. But I don't think there is a specific number of hands that would give you an exact answer. Which is why most people have swings in there results because sometimes the odds are with you and sometimes there not.

Monty Hall is not hard, it's super easy. If you picked the right one to start, you win if you stay lose if you switch. If you picked one of the two wrong ones to start, you win if you switch, lose if you stay. It's the easiest switch ever.

As for the poker question it's just binomial. Odds of not flopping a set is (48/50)(47/49)(46/48) ~= 88.24%

Odds of this 37 times in a row = .8824^37 = just under 1%. So every 100 times you play 37 pocket pairs, you will flop 0 sets in that sample an expected 1 out of those 100.

aRoseIsaRose said:

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.

Click to expand...

This is not highly debated in the sense that serious mathematicians are divided about the answer, neither is it just Marilyn vos Savant's opinion. It is provable mathematically.
It's true that it does provoke a lot of discussion (like this one)

The probability is not transferred to the other door as you suggest, it was always there

I think the easiest way to look at it is
Suppose you were given the chance to choose between
a. 1 door
or
b. the other 2 doors (at least one of which, will definitely have a goat behind it)
which would you choose?

33% or 66%

when Monty offers you the chance to change your mind you will effectively be choosing the 2 doors at 66% except that now Monty has revealed which one has the goat.

Easy choice. It is 2:1 on that changing is correct

I thought that on the flop there is 11.76% chance that you will get a set with a pair in your hand. .1176 * 37 = 4.3512 I could not see it if I took a pair of 2's out of the deck, shuffle the cards, and then deal a flop 37 times and not get at least one set. Some of these hands, probably around 1/3 went to the river. Maybe 3 or 4 went to the turn.

If you flipped a coin twice and it came down heads both times, you wouldn't think it that strange.
If you flipped a coin 10 times and it came down heads every time you would think that unusual but possible
If you flipped a coin 100 times and it came down heads 70 times, that is off from the probability of 50/50 but it's easy to see how that is still possible
but if you flip a coin 100,000 times it will come down much closer to the expected 50/50, maybe 55/45 but it would be very unlikely to be 70/30 or higher.
The probability is just that, only a probability not a certainty, and it is only when you have a large enough sample size that you will see the true effect.
As for what that sample size should be for your example of pairs making a set on the flop, I don't know the answer to that, — but you (along with me and every other poker player) can safely continue to use the probability of 11.76% as that is the best information that any of us have.
There are people and organisations who monitor the pokersites to ensure that these probabilities run true within a limit of deviation from the expected value over large samples. They will let us know if anything is genuinely wrong

The standard deviation, over your sample size, is about 2. You would expect to flop about 4 sets. So you're 2 standard deviations below expectation. So, there's about a 2.5% chance of this happening. For every group of 37 pairs you see a flop with about 1 in 40 will have no flopped sets.

It's unlikely. But, over such a small amount of hands, it's not that unlikely. It's more likely than hitting a one-outer on the river, and people manager to do that all the time.

There is a mathematical solution to your question, but it's not simple. It involves using statistics.

Basically, you're asking how large of a sample size do we need, given a certain probability and standard deviation, before we can be reasonably certain that our results will be reasonably close to the expectation?

The answer to that question is complicated and relies on what you mean by being reasonably close. When we're dealing with something that has high amounts of variance (flopping a set, for example) we need an even larger sample size than when dealing with something with low variance (flipping a coin).

In my own databases, I have thousands of hands. Take one site, I had 4,116 pairs where I saw a flop. I would expect to flop a set or better (quads / full house) 484 times. I happened to make that 487 times. So I was slightly above expectation. The standard deviation of this sample is about 20. So I am only off by 6% of the expectation. That's not bad. Over this sample, 33% of the time I can expect to be more than 20 hands above or below the 484 flopped sets.

Anyway, the short answer to your question is that you need LOTS of hands. I would be comfortable with any sample size greater than 100 hands. At that point, you're well into the outer edges of probability.

Pairs seeing a flop in a row with no sets.
25: Probability 3.39% (339 times in 10,000 opportunities)
50: Probability 0.5% (50 times in 10,000 opportunities)
75: Probability 0.08% (8 times in 10,000 opportunities)
100: Probability 0.03% (3 times in 10,000 opportunities)

As you can see, increasing the sample size can have dramatic impacts on the probability. Of course, 100 hands in a row without hitting a set would be pretty impressive.

Besides the not hitting sets with pairs, there is one other thing that I mentioned in my original post. I got 37 pairs in one day and didn't hit a set. Before I go to bed I email PurePlay and say hey doesnt your site ever flop a set to a pair. Well the next day, I go all my 107 hands without getting a pair. I was thinking that this might be a coincidence or it might be PurPlay's way of telling me do not complain about never getting sets with pairs. No Pairs- No Complaints. That was all I could play for that day because I had something else to do. The reason I posted this is because when most people start talking about how rigged up a site must be is because they do not understand odds and probabilities at all plus could generally be not a good player. I am not the greatest at math but I can figure odds and outs and apply it to a game and I am a fishy beginner.

I responded to that in your original post. The std. dev of not getting a pair over 107 hands is about 2.4339. You expect to get 6.294 pairs. The probability of getting 0 pairs, given those numbers, is 4.9%. It's low but about 1 session in 20 (of 107 hand sessions), you'll get no pairs. If these events were truly connected (and not coincidence) then the odds would be about 800-1 against.

But, we have no proof that they're truly connected. We have two samples, which we can't be sure represent reality. If you had multiple sessions where you flopped no sets, followed by calls to support that resulted in a no pair session immediately after, we could assume something is up.

As it is, this is just coincidence. It's unlikely, but meaningless.

aRoseIsaRose said:

I thought that on the flop there is 11.76% chance that you will get a set with a pair in your hand. .1176 * 37 = 4.3512 I could not see it if I took a pair of 2's out of the deck, shuffle the cards, and then deal a flop 37 times and not get at least one set. Some of these hands, probably around 1/3 went to the river. Maybe 3 or 4 went to the turn.

Click to expand...

You don't seem to understand how to correctly figure out the probability of a set of independent trials. Multiplying 11.76% (the probability of flopping at least a set) by 37 (the number of trials) is meaningless.

As hackmeplz said, you can figure out the probability of not flopping a single set in 37 trials by raising the individual probability (1 - 11.76%) to the 37th power. That gives 0.8824^37 = 0.976%. So the probability of not hitting a single set in 37 flops is just under 1%. While it isn't very likely, it's going to happen 1 out of every 100 independent set of 37 pocket pairs - there's nothing you can do about it.

As for the Monte Hall problem, think of it this way. When you are given the choice of 3 doors, one of which has a car behind it, you are 33% likely to have picked the right door (assuming you can't smell the goats behind the other doors ). That means you are 67% likely to have picked a door with a goat behind it. Obviously whether you picked the right door or not, he has at least one door with a goat behind it that he can show you.

So just because he shows you a goat doesn't mean you know anything more about the door you picked, and you're still only 33% likely to have chosen the right door. So would you rather keep your 33% likely door, or would you rather switch to the door that is 67% likely to have a car behind it?