ODDs math question

[ohnonotthat]

I understand.

If I "have something substantive to add" and am not here simply to "curse the darkness", I have your permission to contribute ?

Sadly I am unable to light any candles since they won't let me handle matches [img]/images/graemlins/confused.gif[/img] but I'll see if I can help with your math issues. (You'll have to get someone else to help you with your people skills).

If you have "X" chances and wish to know the chance that AT LEAST one will result in success you do not multiply your chance of succeeding by "X". In general, this rule applies whether or not the events are dependent, or in any other way related.

The odds against choosing a heart from a shuffled deck are 3-1 (or 1 in 4, or 25%)

If you are allowed to try twice your chances do not double (though they obviously do increase) and being allowed three attempts certainly doesn't triple your chances.

This should be easy to understand due to the next statement.

Being allowed four attempts doesn't quadruple your chances; if it did you would have a 100% chance of succeeding, and you obviously don't - regardless of whether you replace each card after you select it or set it to the side and make your subsequent attempts from a [slightly] depleted deck.

In the spirit of the season what say we overlook my sarcasm and your obnoxiousness and start from scratch. (It's a good offer since I don't see people here leaping through hoops to help you - most of them share my distaste for posters who are either lazy or abrasive, and you are clearly both).


What is your original question ?

It's sort of a good idea to respond to sub-posts directly beneath the sub-poster; when this policy is not adhered to it's often difficult to follow the bouncing ball.

Repeat what it is you'd like to have explained and I will explain it to you; many have extended me this courtesy and I am happy to do so for you.

- However, I reserve the right to dislike you if I see fit especially if your future remarks make this choice seem like a wise one. [img]/images/graemlins/mad.gif[/img]

Happy holidays,

- Chris

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

[ohnonotthat]

All is well. [img]/images/graemlins/smile.gif[/img]

If you want a good laugh have someone tell you the story about the girl, her mother, her grandmother, and the oversized ham; it's a very cute way of saying "I do it this way because I've always seen people do it this way.

I'll PM you with what you need.

However, I do NOT do group hugs [img]/images/graemlins/shocked.gif[/img] You'll have to accept a handshake. [img]/images/graemlins/grin.gif[/img]

- Bets wishes for the holidays,

- Chris

Thank you, Chris.

All is much appreciated.

What did the buddhist monk say to the hotdog vendor?

Make me one .... with everything ....

Have a great holiday [img]/images/graemlins/smile.gif[/img].

Eric

[AaronBrown]

You got an excellent reply to your specific question. I'll try to address your general one.

Expected values add. The expected number of matching cards to your pocket pair on the first flop card is 2/50, or 0.04. Before any flop cards are dealt, the expected value is the same for your second and third flop card as well. The 2/49 figure for the second flop card assumes that you didn't hit on the first one.

So on an average hand you will get 0.04 + 0.04 + 0.04 = 0.12 matches to your pocket pair in the flop. You can't get three matches (if you do, run for the door before you get shot), so the only possibilities are zero matches, one match and two matches.

It's pretty easy to compute the probability of zero matches, (48/50)*(47/49)*(46/48) = 0.8824. That means the chance of any matches is 0.1176. 0.0024 of the time we must get two matches, so that our average number of matches is 0.1200. Therefore, 0.1152 of the time we get exactly one match, trips but not quads (actually, it's 0.1151 due to rounding).

The 10.9% figure you cite also subtracts off full houses.

Thanks for the excellent post, Aaron.

Eric