What would be the best limit to start at?

[tinkerman]

[ QUOTE ]
Also, you need to figure out how many unknown cards there are. This is 52 - however many cards you have seen exposed so far. Don't go overboard trying to count these... on 6th st, if the hand was heads up from the beginning, there are 36 cards unknown. You know your 6, your opponents up 4, and 6 door cards from the people who folded 3rd. Obviously adjust downwards if lots of people saw 4th.

So, let's say you counted your outs, and there are 15 of them, and you've been heads up the whole way, so there are 36 cards left. That means you'll hit your draw 15/36 times. I'm going to cheat and call this 15/35 because that's almost the same anyway, and 15/35 is 3/7, which in odds speak is 3:4 (3 times you hit, 4 times you don't)


[/ QUOTE ]

I'd say as a quick rule of thumb that if you have remembered two other door cards then there are 33 cards left in the pack. This means that each out is around 3%. So in the above case 15 x 3 = 45% chance of hitting. If there are 40 unknown cards then each out is 2.5% so 15 x 2.5 = 37.5%

Of course this situation is usually an obvious play. Now assuming you had a draw to the wheel only with 4 outs. SO your chances of hitting are between 10 and 12%, eg. a 9/1

[RustyBrooks]

If you can beat him with a 7, 6, or a 2 and you've already seen a 2, 6 and an 8, then you actually have 10 outs (3 twos, 3 sixes, 4 sevens) 10/36 is basically 2/7 or 2.5:1 against. Basically not quite as often as 1/3.

In actuality you will win less than this. You are drawing to beat your opponent's 76 but he's drawing to a 6 as well. If you'll make a 7 about 1/3 of the time, he'll make a 6 less than 1/3 of the time (more like 20% of the time). So if you get a 7 you'll still be outdrawin 1 time in 5, so you don't count 7s as full outs. That doesn't change things much in this case because you still have 6 full outs for the 2s and 6s and then 3.2 "adjusted" outs for the 7s. I'd call this 9 outs. 9/33 isn't that different than 10/33 so in this case, no big whoop.

However, when your opponent's current hand might be an 85, and you have the same draw as before, now only your 2 is totally clean, because he might outdraw you and make a wheel, or a 6. This means you need to discount your 7s and 6s. The actual amount to discount takes some thought but you'll never be toooo far off to discount them by 1/4 in a case like this. If you have a much worse hand like a T and you're drawing to an 8, and your opponent current has like a 96 (9, drawing to a 6) then you need to discount more heavily, more like 1/3. This is like how in holdem if you're drawing to an open ended straight, and you think your opponent *might* be drawing to a flush, then you have to consider the outs of the flush suit as only partial outs, because when you get them you'll make your straight, but might lose to a flush.

We inflate the pot on 3rd sometimes as kind of insurance (there are other reasons to raise, and certainly reasons not to). We're trying to make sure we can see 5th without making a bad call on 4th, because we think that with more cards dealt our chances of outplaying our opponent are larger, and we're removing an obvious spot where we could get outplayed on 4th.

Personally I don't share your fatalism regarding poker, but even if you are destined to lose the next 4 times that you are drawing to a wheel, it doesn't matter, unless you think that you will lose more often, in the long run, than probability says you will. If you do, then I swear I'll never mention probability to you again [img]/images/graemlins/wink.gif[/img] If you don't, then you'll have to accept that when you are in a situation where you will win 1/3 times, but don't know which times, then you just need to relax and understand that as long as you are acting in a way where you will profit if you win 1/3 times, then, in the long run, you will profit.

Put another way if you offered me a chance to win $2 for every time that a coin came up heads, and I'd pay $1 for each toss, I'd take it. I'd take it for $20/$10 also. Or $200/$100. Or $2000/$1000. At some point I might decline because losing would wipe me out - this is why we don't play in games above our bankroll even if we are likely to win at them.

Anyway, I *might* hit 10 tails in a row. Does this mean I made a bad bet? Of course not, and I'd make the same bet again. We play the same situations over and over and over and over again, the probability levels out and as long as we make the right decision, we make money.

[Praxising]

[ QUOTE ]
Personally I don't share your fatalism regarding poker, but even if you are destined to lose the next 4 times that you are drawing to a wheel, it doesn't matter, unless you think that you will lose more often, in the long run, than probability says you will.

[/ QUOTE ]I think of it as realism rather than fatalism, I've yet to see the research that proves the hypothesis that theory of large numbers applies to poker in this way. Ultimately, you'd have to find all situations where the pot odds are the same, played out to showdown, and see what the profit/loss is over a large number of hands for a large number of players. The specific hands would be irrelevant, the limit would be irrelevant, (tho' it must be limit games of the same type in the data set) the odds are the control and the money is the irreducible result.

I'm sure that f played out, the hands do win as often as probability says they will within an error range. I taught genetics, I know what a balanced polymorphism is and it depends for it's "balance" on the Law of Large Numbers. But I also know that law is indifferent to the individual, I also know there is founder effect, and while I am convinced the hand will win as expected, I am not convinced a player does. And even if the player does, I'm not at all sure that means all players will profit. Most especially in Razz.

[RustyBrooks]

Realistically the odds we're talking here do not require large numbers to work themselves out, because the probabilities are not that small. In fact it's pretty easy to figure out how many times you need to look at a given situation to see if it settles down to, say, within 1% of the expected value, and the number is something like 1500 times. I don't know about you, but I've had way more than 1500 flush draws in my life. Probably more like 50,000.

[jbrennen]

Actually, you probably need a lot more trials than that to have a strong chance of being within 1% of the expected value.

For instance, the number of heads expected in a coin flip is 0.5 with 0.5 standard deviation.

Flip 10,000 coins, and the expected value goes up by a factor of 10,000. The standard deviation goes up by the square root of that, or by a factor of 100.

So flip 10,000 coins, and you expect 5,000 heads with a standard deviation of 50. So +/- 1% from the expected value is one standard deviation. The chance to fall within 1 std. dev. of a normally distributed probability is about 2/3. So if you flip a coin 10,000 times, the chance that it will NOT be within +/- 1% is a relatively high 1/3 probability.

To get +/- 1% to represent 3 standard deviations (giving you about a 99% chance of being within that range), you'd need to flip 90,000 coins.

[RustyBrooks]

That doesn't seem right. I'll haul out the formula after lunch. Real quick, though, I wrote a program that flips coins. It does N trials, K times.

For each time, I run N flips, and record the number of heads. If the average # of heads is more than 1% from the EV then I increment a counter. Here are my results, running 100 times with different trial sizes.

1000: 47/100
5000: 86/100
10000: 93/100
15000: 98/100
20000: 100/100 (I ran this 5 times, same result every time)

So for a coin flip, the worst case looks like 20k which is worst than I thought, so I need to look at my formula again.

I ran this with 1000-sized runs also, similar results. I can post code if you like.

[RustyBrooks]

Still too lazy to look up my formula, but one program tweak later and as far as I can see it, the 95% confident mark (95% of the time, it's within 1% of EV for N trials) happens for N=~8800

50/50 should be the worst case scenario, also.

In fact for grins I just tried 75/25 and you get to 95% confidence with N=~7200

For 90/10 I get N=~3600

[jbrennen]

[ QUOTE ]
Still too lazy to look up my formula, but one program tweak later and as far as I can see it, the 95% confident mark (95% of the time, it's within 1% of EV for N trials) happens for N=~8800

[/ QUOTE ]

I think your random number generator must have some "issues" if that's true.

The actual probability that if you flip a coin 8800 times, that the number of heads will be 4400 +/ 1% is only 65.7%.

[RustyBrooks]

My RNG is fine. Possibly the program is flawed in some way.

The formula I'm going on is this

N >= p * (1 - p) * (z_alpha/2) / epsilon^2

p is the EV
z_alpha is the Z value for your confidence interval.
epsilon is your acceptable margin of error

z_alpha I get as 1.6449 for confidence interval of .95
p * (1-p) is obviously .25

When I plug-n-chug I get N >= 8224
Which matches my experimental results.

Now, I admit that I am no statistician, but I consulted with one before starting to use this formula for my simulation work. I use this simulation to figure out how many iterations I need to do, to be "close enough" to the real value for simulation.

Maybe we should post this to the probability forum?

[jbrennen]

Rusty, I think most likely it's just a semantic difference in how we're treating +/- 1%.

To me, if the EV is 4400 (in 8800 trials), +/- 1% is the range [4356...4444]. The +/- 1% being one percent of the EV.

It looks like you're using +/- 1% being the range [4312...4488]. The +/- 1% being one percent of the number of trials.


Using my interpretation of 1%, the chance of being within the range is 65.7%. Using your interpretation, the chance of being within the range is 94.1%.


I'll show you why I think my interpretation is more useful: say that the chance of winning a particular game is 1% per attempt, and you win 200 times in 10000 trials. Your win rate is 2% per attempt. Are you running 1% above expectation, or are you running 100% above expectation?