ODDs math question

I'm trying to get my mind around poker odds. Let's take an example.

What are the chances of getting 3 of a kind on the flop, if we have any pocket pair?

Now, on the first flop card we have this chance:

2/50 = .04

Since we have 2 "outs" that would complete our trips, and there are 50 cards left in the deck after our hole cards are dealt to us.

And for the 2nd flop card we have: 2/49
And for the 3rd flop card we have: 2/48

Now, here's where I'm having a problem. As I understand it, if any of 3 probabilities will get the desired result we ADD these 3 probablilties to get the probability that the needed event will occur within the 3 events.

So, this would mean we'd add:

2/50 + 2/49 + 2/48
or .04+.041+.042 (last 2 rounded)

or .123 or 12.3 %

According to this web site:
http://www.homepokergames.com/odds.php

The odds are 10.9% to flop 3 of a kind when we have any pocket pair.

Where am I going wrong in my calculations?

Thanks all.

Eric

But to make a set from the second and third cards a condition for the first (and second) cards must be met, so its not simply 2/49 and 2/48.

I just took a look at the link and the site writes 'Flopping a set or better (with a pocket pair) - 7.5/1 (11.8%)' There are 3 ways to flop a set or better,
event A - set from the first card = 2/50
event B - set from the second card = 48/50*2/49
event C - set from the third card = 48/50*47/49*2/48
Depending on the other cards that fall you may end up with better hand than a set
therefore the probability of hitting a set of better is event A + event B + event C = 11.8%

If you just want the probability of flopping a set, you must now subract the probability of making a boat or quads (0.74% and 0.25%), which can be calculated in a similar way. This gives the 10.8% for hitting a set.

Thanks for you answer. Here are my comments.

I'm really asking for the answer to the general probability question:

If any one of 3 events a,b, or c, will cause event y to occur, and the probabilities of a, b, and c occurring are:
a=d
b=e
c=f
Is the probability that y will occur equal to?:
d + e + f = Probability of y occuring

And if that is the correct formula for the odds of y occuring, where did I go wrong in my calculations.

With a pocket pair we have the following probabilites for a,b, and c.
a = (2 outs)/(52-2 or 50 cards) = .04
b = (2 outs)/(52-3 or 49 cards) = .041
c = (2 outs)/(52-4 or 48 cards) = .042

So why aren't the odds of y (getting trips) equal to:
.04 + .041 + .042 = 12.3%

The web site says it's 10.8%

Remember, we're ONLY concerned with the case of having any pocket pair, and getting 3 of a kind on the flop. NOT any other case. Let's keep it simple.

If you have any further insight, please comment.

Thank you for your help.

Eric

No, this is wrong. Consider flipping a fair coin twice. The probability of getting heads the first flip is .5. The probability of getting heads the second flip is .5. Let's say you "win" if you get heads either time. According to your way of looking at this you win every time since .5+.5=1.

Thanks for the reply. You make a good point /images/graemlins/smile.gif.

Now can you enlighten us any on the correct answer to my problem /images/graemlins/smile.gif?

Are you here to light a candle or just curse the darkness /images/graemlins/smile.gif.

Eric

Here's the way the coin flip works.

If we flip a coin twice we can get 4 outcomes:
2 heads probability .5*.5 = .25
2 tails .5*.5 = .25
1 heads 1 tails = .5*.5 = .25
1 tails 1 heads = .5*.5 = .25

3 of these outcomes gives us our required 1 head result. So the odds of getting at least 1 head out of 2 coin tosses is:

.25+.25+.25 = .75 = 75%

In our pocket pair/trips example we have the following possible results for the 3 flop cards, excluding the possibilities with 2 3rd cards because they would give us 4 of a kind (3rd card = the card that completes the 3 of a kind):

1=other 2=other 3=other Prob=.96*.96*.96 = .885
1=3rd 2=other 3=other Prob= .04*.96*.96= .037
1=other 2=3rd 3=other Prob = .037
1=other 2=other 3=3rd Prob = .037

Total = 3.7%+3.7%+3.7% = 11.1%

This gets me closer to the 10.8% from the web site.

Thanks for the example of the coin toss to stimulate my thinking. Perhaps one of the experts here can fill in the last few blanks to correct my errors.

Thanks to all for your help.

Eric

The probabilty of 'other' occuring is not 0.96 every time. You are assuming the events are independent. But the probabilty of the events depend upon what happened before, so you need to account for this. In your coin flip example the events are independent, the coin has no 'memory' so you were correct in using 0.5 each time.

For exmaple if 1=3rd then P(2=other) = 48/49 = 0.98 (not 48/50)

The 3 answers I get are all 0.038367
So 3*0.038 = 11.5%

This is still wrong because the 'other' cards can still produce a full-house. So 11.5% - P(full-house) = 11.5 - 0.74 = 10.76% to hit only a set.
Please correct me if you think this method is wrong.
Thanks,
Mike

That method looks correct to me, Mike.

Now I want to understand how to get the full-house value /images/graemlins/smile.gif.

I'll work on that for now.

Thanks, Mike.

Eric

[MagicFlea]

if you look more closely at mike's answer it has everything you need...

Mike's answer is very helpful.

The full-house calculations add considerably more complexity in my opinion.

If you have anything to contribute other than making such comments, please feel free to make substantive contributions to the discussion.

Thank you.

Eric

[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 /images/graemlins/confused.gif 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. /images/graemlins/mad.gif

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 /images/graemlins/smile.gif. 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 /images/graemlins/smile.gif, with a few obvious "errors" /images/graemlins/smile.gif.

Thanks, Chris.

Eric

[ohnonotthat]

All is well. /images/graemlins/smile.gif

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 /images/graemlins/shocked.gif You'll have to accept a handshake. /images/graemlins/grin.gif

- 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 /images/graemlins/smile.gif.

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