Showing cards when all in (and before decided how many times to run it) do you show?

Except you are not right no matter how much you think you are.

You are verbally lazy.
You are intellectually lazy.
You are a lazy dealer, not even caring to try and learn or get better.

The last part is the worst. You don’t even care enough about your “job” to even try to get good. This is the ultimate in laziness.

Always told my kids…find what you like to do so you will do it well and you will ultimately do well.

I am not a dealer. Never said I was. Doesn’t mean I think it is below me. I respect and value good dealers. Never intended to be a dealer but couldn’t now due to certain physical issues.

Not being a dealer doesn’t mean I can’t recognize a good one. Doesn’t mean I don’t know the rules and procedures. I have been on the other side of the table for many years.

I suspect I could ID 3 flaws in your mechanics wo even seeing you deal. Won’t as there is no way to prove it. Why do I think this? Because there are certain things all lazy dealers do because they are lazy. And yes you are lazy. It comes through on every post. You live by the ‘good enough is good enough’ philosophy. You take the easy way if you can even wh3n you know it isn’t the best.

Think you misunderstand completely... way of the mark as it were. I don't get paid for dealing and never will. I deal in the game I'm playing in or maybe have just been knocked of and I've got time to kill. I do it as a favour only and take my turn ..... you'd be surprised how many people can't even shuffle the deck properly. "thanks for dealing" is all I ask and people say that.. good dealer.

It's ungreatful people and experts in the audience like yourself that make dealing not fun. I would high card it if your miserable nit picking pus was at the table.

We are all talking about casino dealing, because this is the casino and cardroom forum.

Home games are elsewhere. https://forumserver.twoplustwo.com/24/ho... Some home game dealers (or bar league, etc) hang out here because they want to know how the casino side handles things, or they want to improve their skills, or whatever. But home games do not have to follow rules and procedures the same way that casino dealers do.

Casino players don't, as a rule, really care about how fun the dealers think their job is. (We do here, since many of us are or know and hang out with dealers as well. Or because we want good dealers to stay as dealers. But it's not a primary goal of any ruling or procedure discussion.)

ok i'm sorry

There are many instances in your short posting history where you chose the far lazier approach of sticking to your wrong beliefs instead of doing any amount of research or examination. I don't know who you're trying to fool.

Well you fell for it. But I have retracted it

I started this thread thinking you were just a bit of a jerk--at least when it comes to this one particular peeve of yours.

Then I learned you're a dealer and I see that you're an active, material detriment to any game you're involved in.

As you more forcefully you dig your heels in regardless of how many people tell you in no uncertain terms you're wrong and should rethink your behavior, my armchair psychologizing ass feels like you're an active detriment to society at large.

Anyway, hope you're trolling, and if not, stay away from fire arms and school children!

lol it's only the cards I want on their backs, not the people.

I found a situation.

Set over Set. If you flip the cards over and find that someone only has ONE out to win, then there is a decision as to whether you would want to run it more than once - and thus the weaker hand could only ever win one of the runs. If you run it twice the higher set has already locked up half the pot min.

And what did the comparative EV calculation of running it once versus twice in this scenario show you?

That "decision" is whether or not you want a 1/44 chance of winning 100% of the pot or a 2/44 chance of winning 50% the pot.

Do the math and let us know which option will give that player the higher EV.

Still irrelevant. I’ll calculate trailer’s EV as it’s simpler.

RI once - (I’m assuming only a river left to be dealt) trailer has one out with 8 known cards. EV is p/44 where p is the size of the pot.

RIT - trailer hits his out with probability 1/44 on the first run. He hits the second with probability 43/44 * 1/43 =1/44. Thus he hits an out with probability 2/44 or 1/22. He wins p/2 when he does so, so EV is p/2 * 1/22 which is p/44 - exactly the same

It's actually a fantastic example for demonstrating that the EV doesn't change. In thinking that you found a counterexample, you might have found the trivial one!

If we choose to run it 4 times but the 1st time the LOWER SET gets there, then we just give him 25% of the pot and don't have to bother with the other 3 times.

So i've cracked it

That’s irrelevant. You could agree to run it 44 times and then you wouldn’t have to deal cards at all. The EV is still the same. In the run it 44x case, the one out player wins 1/44 of the pot with probability 1, so his EV is p/44.

In your case, he hits the first deal with probability 1/44. He hits card 2 with probability 43/44*1/43, or 1/44. For card 3 it’s 43/44*42/43*1/42, again 1/44. He hits card 4 with probability 43/44*42/43*41/42*1/41, again 1/44. The total probability of hitting one of the four can be obtained simply by adding these (valid precisely because he can’t hit more than once), to get an overall probability of 4/44, or 1/11. He wins 1/4 of the pot, so EV is 1/11*p/4 or p/44, same as when running once.

In general, the pattern should be clear. Running N times gives him probability of N/44 of hitting. He wins p/N when he hits. EV is therefore N/44 * p/N = p/44 - the value of N is irrelevant. N=1 gives the same EV as N=2,3,4 etc.

Running it twice reduces variance and keeps fish in the game longer. The main reason to run it once is to prevent people from calling you light.

.

I come to these threads specifically to learn what the rules are and how some of them are still subject to interpretation depending on The specific facts. I appreciate the time and effort posters make to clarify such strange instances where the rules provide answers and guidance how a proper ruling should be made to make the game as fair as possible.

Thanks. We do try

It's still better to see what the hands are before deciding to run it more than once.

What if someone has a super combo draw compared with being dominated?

With the combo draw it might be good to run it 4 times so that some of their outs are wasted on the river (they have already hit the turn but then waste a future out by hitting again on the river).

More information is usually better

It's never good to run it 4 times.

The EV of running it once and running it 4 times is identical.

What about the opposite where they miss the turn but one of your outs is wasted on the river?

Again your intuition fails and math proves that EV is unaffected. For simplicity I only worked it out for running it twice, but the “wasted outs” still factor in. Suppose you have 15 outs and you run the turn and river twice. There are three cases for run 1: zero outs, one out or two outs hit. The probability of zero is 29/44*28/43 or 0.4292. For one out it’s 2*15/44*29/43 or 0.4598. For two it is 15/44*14/43 or 0.1110

Now we consider each in turn - for zero outs the chance of zero again on run 2 is 27/42*26/41 or 0.4077. Since this is the only case we lose both runouts we can calculate our probability directly of losing 0.4922*0.4077 =0.1750. We chop in this case with either one or two outs on run 2, so that has probability 1-0.4077=0.5913, contributing 0.5913*0.4292=0.2542 to our chop probability (this ainÂ’t the final value, since more chops are to come)

Next we consider one out on run 1. If we hit zero on run 2 we chop, otherwise we scoop. We got zero with probability 28/42*27/41 or 0.4390. This contributes 0.2019 to our chop probability. Since our chop+lose+scoop probabilities must add to 1, IÂ’ll skip the rest of this case. We calculate chop and already know lose probability so we donÂ’t have to calculate scoop.

Finally, in similar manner we consider two outs on run 1 and zero on run 2. The probability of zero on run 2 in this case is 29/42*28/41 or 0.4715. This contributes 0.4715*0.1110 or 0.0523 to our chop probability.

We get a total of 0.5084 for our chop probability when we add all three cases. Our loss probability was 0.1750 so our scoop probability is 0.3166. Our expected share of the pot is half the chop plus the scoop or 0.5708.

Suppose we ran in once. We expect to lose with probability 29/44*28/43 or 0.4292. We therefore win with probability 0.5708, giving us exactly the same expected share of the pot as with RIT.