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#1
12-23-2005, 10:27 PM
 Guest Posts: n/a
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
#2
12-24-2005, 12:02 AM
 Guest Posts: n/a
Re: ODDs math question

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.
#3
12-24-2005, 01:39 AM
 Guest Posts: n/a
Re: ODDs math 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.

Eric
#4
12-24-2005, 01:59 AM
 Guest Posts: n/a
Re: ODDs math question

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.
#5
12-24-2005, 02:05 AM
 Guest Posts: n/a
Re: ODDs math question

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

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

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

Eric
#6
12-24-2005, 04:00 AM
 Guest Posts: n/a
Re: ODDs math question

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
#7
12-24-2005, 08:04 AM
 Guest Posts: n/a
Re: ODDs math question

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
#8
12-24-2005, 08:43 AM
 Guest Posts: n/a
Re: ODDs math question

That method looks correct to me, Mike.

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

I'll work on that for now.

Thanks, Mike.

Eric
#9
12-25-2005, 03:12 AM
 MagicFlea Senior Member Join Date: Mar 2005 Posts: 106
Re: ODDs math question

if you look more closely at mike's answer it has everything you need...
#10
12-25-2005, 07:56 AM
 Guest Posts: n/a
Re: ODDs math question

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

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