Thank you, Chris. I appreciate your post, and your point of view.

I don't think I'm "clearly lazy" on this. I may be dumb [img]/images/graemlins/smile.gif[/img]. I'll admit to that. I have spent many hours working through some of these problems, and I'm making progress. If you read through my posts, you can see I was putting considerable time and thought into these posts. I wasn't just sitting back and asking others to do it all.

My abrasive reply to you was only a little abrasive, I would say. Surely you understood the attempted humor in my light a candle remark. I'll be less abrasive, I think perhaps you were a little thin-skinned on that one.

I do have a chip on my shoulder at times, when people make posts that don't seem to advance the thought, or contribute in any way to solving the problem. I'll try to keep that in check.

You obviously have a lot to contribute. And I welcome any contribution you wish to make.

My original question was what are the odds of getting 3 of a kind on the flop when you have any pocket pair.

I was trying to use the "per-card" odds method to determine this (2/50 etc.), and was getting an answer of 11.x%, when the web site said the answer was 10.8%. Then when someone mentioned a coin toss, I tried to use the method of determining all of the relevant combinations of events with the flop cards that would get the 3 of a kind. My answer was still off.

I have since been introduced here to the use of combinations to do this sort of analysis, and I've been working in Excel to get up to speed on that.

Mike's post explained this original issue very well, as you pointed out. He used the stated odds on getting a full house to get his answer, which led me into my current investigation.

So I've evolved to a little more complex example. Namely, how to figure the odds on the chances of say getting a full-house in a five card "draw", or deal.

I am working through this on my own now. Here are some of the problems I'm having.

Using combinations, and I'm a little new to these, so I may not state this correctly, a full house consists of:

We have Combination(52,5) for the total number of 5 card draws
One Combination(4,2)
One Combination(4,3)

Now, we have 13 sets of 4 in the deck.
And, once we make our pair (4,2), or trips (4,3) one of these 13 sets of 4 is removed from the equation.

I'm wondering if this problem can be represented elegantly with a fairly simple conbination solution. I think it probably can.

I think if I can see 2 or 3 problems like this explained in detail, explaining why and how each representation of a variable is derived, that I'll be ok to do any other calculation like this.

If you could do that for me, Chris, that would be wonderful. I will continue to work on my own to solve it also.

If you could run me through this full house odds determination, I think I should be able to do most any odds calculation.

I hope you can forgive me for my previous posts. I'm actually quite good with people skills [img]/images/graemlins/smile.gif[/img], with a few obvious "errors" [img]/images/graemlins/smile.gif[/img].

Thanks, Chris.

Eric