Calculating Odds

[Dovberman]

Holdem Odds Calculation Issue

Using probability methods, you should be able to calculate the odds of flopping three of a kind when you have unpaired hole cards.

Consider all the possible flops where ‘a’ and ‘b’ are your unpaired hole cards.
‘x’ is any card not matching a hole card

The probability of two or more events occurring at the same is found by multiplying the individual event probabilities.

Hole Flop Probability Calc Odds

a,b x,a,a (48/50 * 3/49 * 2/48) .00245
a,b a,a,x (3/50 * 2/49 * 48/48 ) .00245
a,b a,x,a (3/50 * 48/49 * 2/48) .00245
a,b x,b,b (48/50 * 3/49 * 2/48) .00245
a,b b,b,x (3/50 * 2/49 * 48/48 ) .00245
a,b b,x,b (3/50 * 48/49 * 2/48) .00245
'a,b x,x,x (3/50 * 2/49 * /48) .00005 Ignore

The probability of each possible flop is added.

6 times .00245 = .0147 or 1.47 percent.

The “Winners Guide to Texas Hold’em Poker” by Ken Warren on page 199 shows a percentage of 1.57. Other sources also publish odds closer to 1.57.

What am I missing ?

Thanks.

[Lost Wages]

Given unpaired hole card a and b with x <> a, b then:

Flops of aax = ((3*2)/2)*44 = 132
Flops of bbx = ((3*2)/2)*44 = 132
Total possible flops = (50*49*48)/6 = 19,600

Probability of flopping trips = (132 + 132)/19,600 = 1.35%

I don't own the Warren book so I can't comment on his figure. The probability of flopping trips, a full house or quads is 1.45%.

Lost Wages

[DeadRed]

When I calculate it and let the "other" card make either quads or a boat, I am getting the 1.47%. I can't seem to find that missing .1% as well.

2*C(3,2)*C(48,1) / C(50,3)

2 allowing for either card a or b to be the trips.
C(3,2) choosing two of the three available 3 cards to make the trips. (3)
C(48,1) choosing the last card on the flop. Changing the 48 to 44 will get you trips only. (48 or 44)
C(50,3) Everybody's favorite number, 19600.

2*3*48/19600 ~ 1.47%
2*3*45/19600 ~ 1.35%

[Dovberman]

We are getting the same 1.47 percent answer using 3 different methods. That's good enough for me. I am developing a pop-up odds calculator and was confused by my variance with the poker books. I used to teach statistics and could not understand why valid techniques were not matching Doyle or Ken Warren. However, they do not describe how they derived the figures in their tables.

[DeadRed]

I am a college math instructor with a love of combinatorics and one of my colleagues with a Ph.D. in stats has said that he has noticed many texts with incorrect numbers in places. So, with no offense to the authors, I would not be surprised if they were off a bit in some places.

[Dovberman]

You will be interested in my next post on the subject, "Factoring Unavailable Cards". This considers the probability of cards you need being dealt to other players.

[scottjack]

perhaps I'm missing something as well, but wouldn't the probability of x, the card that doesn't match either of your hole cards be 44/50, not 48/50 ?
there are 3 remaining a's and 3 remaining b's in the deck. subtracting these 6 from the 50 cards remaining leaves 44.

Also, while it's true that the probability of two or more events occurring at the same is found by multiplying the individual event probabilities, this is only so when the events are independent . i don't think this would apply here.

For instance, the probability that the 3rd 'a' falls, IS affected by the fact that the 2nd has already fallen.

As an example, (assuming neither of the hole cards is a 7)the probability that the event of "first flop card is a 7" is 4/50. if the event "second flop card is a 7" were independent of the first, then its probability would also be 4/50 and the probability of both events occurring would be (4/50)^2 however, the probability of these 2 both occurring is (4/50)*(3/49).

i think this may get messy, but there's going to have to be some conditional probabilities figured into things before it's all said and done.


I'm sure i could very well be wrong on all of the above. Just thought i might shed some light on your quandary.

Hope this helps a little.