[_GUN_]

I wrote down my numbers from a tournament today.

58 hands, saw 7% of the flops, won 0 pots, saw 1 turn, 0 pre-flop raises

in 58 hands, I was dealt 0 pairs and one ace (A8o). A8o came on hand four and the pot had already been opened, so I dumped it.

what are the odds on zero pairs and one ace in 58 hands?

here's my glorious burnout.

hand #59

blinds are 50/100, I have 485...I raise all-in from about seat 6, seat 7 has 3900 and flat-calls, seat 8 reraises all-in for 1900...idiot in seat 7 takes no time at all and calls.

K-8-3-6-T is your final board. the guy called off over 10% stack from the tightest player at the table and then 1/2 of his stack from seat 8. what did he have?

seat 7 has K-10 offsuit and the reraiser has AKo and busts out to the two pair. seat 7 takes down 4500 chips with that horrible play.

[davejs1671]

I will admit that my math skills are not the best, but I do know that your session sssssuuuucccckkkkeeeedddd !!! I wish you better luck on the next 1 .

[OutOfCrown]

Originally Posted by _GUN_

I wrote down my numbers from a tournament today.

58 hands, saw 7% of the flops, won 0 pots, saw 1 turn, 0 pre-flop raises

in 58 hands, I was dealt 0 pairs and one ace (A8o). A8o came on hand four and the pot had already been opened, so I dumped it.

what are the odds on zero pairs and one ace in 58 hands?

There are 1326 possible starting hands. (52 * 51) / 2 ; we divide by 2 because (for example) AhKd is the same hand as KdAh

Of those 1326 starting hands, 78 of them are pocket pairs. There are 6 ways to make any given pocket pair (e.g., AsAh AsAd AsAc AhAd AhAc AdAc) and there are 13 types of pairs. 6 * 13 = 78.

Besides the 6 pocket aces, there are 4 * 48 = 192 other hands that include an Ace.

Given all that, we can compute what you ask, except first we need to figure out exactly what you are asking.

To make it simple, let's just ask the question: what is the probability of going N hands without seeing either a pocket pair or any type of Ace?

The odds of getting a not-pair-not-Ace (I'll call this NPNA) hand is simply the number of such hands divided by 1326 (total number of possible hands).

There are 1056 NPNA hands ( = 1326 - (78 + 192) )

Probability of 1 NPNA hand = 1056/1326 or about 79.6%

The probability of getting N NPNA hands is simply (1056/1326) ** N

So if you want to know (for rough estimation purposes) what the odds of getting 50 NPNA hands in a row are, they are ( (1056/1326) ** 50 ) which works out to about 0.00001138 or about 88,000:1 against.

Figuring out the odds of getting exactly 1 not-NPNA hand and 58 NPNA hands, with the one "good" hand coming anywhere (not necessarily 4th hand) is left as an exercise for the reader. Suffice to say, it's not likely, as the already-low "get 50 NPNA hands in a row" probability suggests.

That said, I think it's important to point out that highly unlikely things happen every single time you play poker. Take a deck of cards, shuffle them, and then look at them. The odds that those cards will be in THAT order (whatever order they happen to be in) are 1 in 52 factorial, which is astronomically unlikely.

And yet, THERE THEY ARE ... in spite of how unlikely it is that they will be in that particular order.

In other words, once it happened to you, it's no longer unlikely. And given how many poker sessions are run every day in the world, it's not surprising that someone runs into that "one in a few hundred thousand" rarity.

Well, that's enough math for one night, yes?

[davejs1671]

Originally Posted by OutOfCrown

There are 1326 possible starting hands. (52 * 51) / 2 ; we divide by 2 because (for example) AhKd is the same hand as KdAh

Of those 1326 starting hands, 78 of them are pocket pairs. There are 6 ways to make any given pocket pair (e.g., AsAh AsAd AsAc AhAd AhAc AdAc) and there are 13 types of pairs. 6 * 13 = 78.

Besides the 6 pocket aces, there are 4 * 48 = 192 other hands that include an Ace.

Given all that, we can compute what you ask, except first we need to figure out exactly what you are asking.

To make it simple, let's just ask the question: what is the probability of going N hands without seeing either a pocket pair or any type of Ace?

The odds of getting a not-pair-not-Ace (I'll call this NPNA) hand is simply the number of such hands divided by 1326 (total number of possible hands).

There are 1056 NPNA hands ( = 1326 - (78 + 192) )

Probability of 1 NPNA hand = 1056/1326 or about 79.6%

The probability of getting N NPNA hands is simply (1056/1326) ** N

So if you want to know (for rough estimation purposes) what the odds of getting 50 NPNA hands in a row are, they are ( (1056/1326) ** 50 ) which works out to about 0.00001138 or about 88,000:1 against.

Figuring out the odds of getting exactly 1 not-NPNA hand and 58 NPNA hands, with the one "good" hand coming anywhere (not necessarily 4th hand) is left as an exercise for the reader. Suffice to say, it's not likely, as the already-low "get 50 NPNA hands in a row" probability suggests.

That said, I think it's important to point out that highly unlikely things happen every single time you play poker. Take a deck of cards, shuffle them, and then look at them. The odds that those cards will be in THAT order (whatever order they happen to be in) are 1 in 52 factorial, which is astronomically unlikely.

And yet, THERE THEY ARE ... in spite of how unlikely it is that they will be in that particular order.

In other words, once it happened to you, it's no longer unlikely. And given how many poker sessions are run every day in the world, it's not surprising that someone runs into that "one in a few hundred thousand" rarity.

Well, that's enough math for one night, yes?

Yeah what he said. Uh my head hurts, I am going to go lay down now. Thank You Professor OutOfTown. Now can you sum up Quantum physics in a short paragraph please. Just joking of course, It's pretty cool to see that explained in real terms Thanks.

[ipgyst]

I don't see any flaws in the previous answer. Here's how I did it:

First card dealt not an ace = 48/52 (or 12/13)
Second card dealt not an ace and doesn't match the first card = 44/51
Odds of being dealt a non-ace, non-pair hand are 12/13 * 44/51, or 528/663, or 176/221

If you take 176/221 to the 58th power, you get the odds of getting ZERO aces or pairs in 58 hands. This is .0000018407, or 1 in 543272.

First card ace (1/13) second card non-ace (48/51) multiplied is 16/221
First card non-ace (12/13) second card ace (4/51) multiplied is also 16/221
Odds of being dealt one and only one ace are 32/221

So you get (176/221)^57 multiplied by (32/221)^1 multiplied by 58 (because the one-ace hand could occur in any of the 58 hands). This is .000019411, or 1 in 51517.

So if you sat down to play 58 hands of holdem about half a million times, on average one of those times you would get zero aces or pairs, and on average ten times you would get no pairs and one ace. Not at all likely, but possible.

[OutOfCrown]

Agree with ipgyst (just finished my own work and got the same answer he did).

Also, I neglected to point out (and was asked in PM) that I used ** to mean ^, i.e., X ** Y means X to the Yth power.

Ever gotten pocket Aces back to back? Odds of that are about 48000:1 against (slightly worse than that if you want the exact number).

Unlikely things, good and bad, happen all the time. This is one area where common sense / intuition really works against most people.

[whataboutj]

Sorry to hear baout the crappy session Gun. That just means poker karma is due for your next game.

Thanks for the math explanation - those calculations alway amaze me

Side note - After reading the probabilit of getting pocket Aces back to back it makes me wonder about the insane rush of cards I got in 4 hands about 3 weeks ago at my poker league game:
K-K
A-A
rags
A-A

so top starting pair 2 out of 4 and the 2nd best startig pair 1 out of 4 -- good times

On another side note: had the kings cracked by a full boat hit on the river and flopped trip A's both times and rivered quad A's the second Needless to say it was awesome!!!!

[smoore]

Originally Posted by whataboutj

That just means poker karma is due for your next game.

LIES! I ran bad for about 7 months. It's *possible* that I'm on an upswing now.