What are the odds?

We all know (or all should know) that the odds of hitting a gutshot are 11:1 against.

But what if there are two missing cards.
This is based on a hand I had today: https://forums.pokerstrategy.com/forum/thread.php?postid=3256695#post3256695

It is a single-draw game.
I have seen 5 cards, so the deck contains 47 unknowns.
If we take them in groups of 2, there are 1081 two card groups.
It hit a straight, we need to draw 54o or 54s
Any offsuit 2-card hand represents 12 combinations, and any suited such hand has 4 combinations.

So of the 1081 combinations only 16 will give us the straight.

So the odds are 1065:16 or 67:1

Stated as a percentage the chance of drawing two cards an making a straight are 1.5%

"Rolled up" means your two hole cards and your up-card are all the same rank

Each rank has 4 combinations of 3 cards
There are 13 ranks, so there are 52 combinations of cards in the deck.

In a deck of 52 cards, there are 22,100 3-card combos total.
So 22048 combos aren't rolled-up, and 52 are.
So the odds are 22048:52 against such an event,
So 424:1

Maybe this was asked odds of

Aa vs Kk vs QQ vs AK preflop?

Originally posted by mkBISCUIT
Maybe this was asked odds of

Aa vs Kk vs QQ vs AK preflop?

If you mean "If I hold KK, what are the odds that another player holds AA?"
let me know, and I'll see what I can do.

Note that I'm not a mathematician -- so I may not be able to find an answer.

Naa .. I am saying 4 way hand

That might be above my skill level.
I'll have a whack at it after supper tonight.

You guys heard about Equilab, right?

Once when I was watching Skodljivec's PLO video and I was playing 2 tables of PLO zoom I got an interesting hand - KJJK all different suits, lets say K♦:J♣:J♥:K♠:
I raised, all folded to ~80bb Big Blind who 3bet. His stats were 9/3/2! So naturally I put him always on AAxx. I paused the video because I had to focus.
Flop was AKXr and I just folded my set to his pot size bet which is quite interesting hand on it's own...

When I continued the video, Skodljivec had the exact same 4 cards in the exact same order - K♦:J♣:J♥:K♠:
So what are the chances of that happening? 1/52 * 1/51 * 1/50 * 1/49 = 1/6'497'400

But I want to ask a deeper question. Provided that the timeframe of playing poker in my lifetime is 40 years, what are the chances of watching that exact video, that exact minute mark of the video, that exact hand.

Originally posted by sandraardnas
You guys heard about Equilab, right?

Yes, absolutely.
AFAIK Equilab will give you the equity of one hand vs another, but won't give you the drawing odds.
If you know how, please let us know...

Let's start with two hands and work our way up
I don't think it matters whether we are 6-max or full ring.

There are 6 combinations of AA
There are 1326 two-card hands possible. 1320 of them aren't AA, so the odds of gettting AA are 1320:6 or 220:1 against.
Stated as a percentage the chances of getting AA are 6/1326 or 0.4525%
Once that hand is dealt, there are 1325 combinations left, and 6 of those are KK so the chance of the second player getting KK is 6/1325 or 0.4528%
For QQ it is 6/1324 or 0.4532%
Since there are 2 aces and 2 kings gone, there are only 2 of each left. That would yield 6 combos, but two of them are the other pairs, so there are 4 combinations of AK, and only 1323 possible combos left.
So for AK it is 4/1323, or 0.3023%
For all of these to happen on the same hand, just multiply the percentage probabilities together:
0.4525% x 0.4528% x 0.4532% x 0.3023% = 0.00000002807414320%

Stated as odds, the odds of this happening are 3,561,996,505:1
I am a bit surprised.

I looked this up later, and the site I found says that the odds of completing a flush when your first three cards are all suited is 5:1

I'll leave my calculations in place while I try to figure out what's wrong.
[edit: OK I figured out where I went wrong. I'll leave the error up until tomorrow. See if you can figure it out]

Here it matters whether there are other cards of that suit showing in other players' hands.
To start, we will assume that we have 10 outs to our flush all live.

At first this looked quite difficult, then it looked even more difficult.
In an 8-handed game we know 10 cards, and 3 of them are in our hand, and all the same suit.
So of the 42 cards left, 10 of them match our suit.
We will get 5 more cards we need only 2 of them to match our suit.
Those 42 cards represent 850668 5-card combinations.
If we take out the 10 cards we need, there are 32 left, and there are 4,960 3-card combinations of those, and 45 two card combinations of our outs.
If those are multiplied together, we see the number of 5-card outcomes with two of our suit: 4960 x 45 = 223,200
Since there are 850668 combinations in total, the chances are 223,200/850,668 = 26.24% or 2.8:1[1]

That is assuming that all 10 outs are live.
If otherwise:
 Dead     Odds 
    0   2.81:1 
    1   3.33:1 
    2   4.08:1 
    3   5.19:1 
    4   6.94:1 
    5   9.95:1 
    6  15.81:1 
    7  30.03:1 

[1] It is actually a little bit better than that, since we only looked at combinations that had two of our outs, yet there are several more combinations with two, three, four or even 5 of our outs.

Originally posted by VorpalF2F

AA vs KK vs QQ vs AK preflop?

Let's start with two hands and work our way up
I don't think it matters whether we are 6-max or full ring.

There are 6 combinations of AA
There are 1326 two-card hands possible. 1320 of them aren't AA, so the odds of gettting AA are 1320:6 or 220:1 against.
Stated as a percentage the chances of getting AA are 6/1326 or 0.4525%
Once that hand is dealt, there are 1325 combinations left, and 6 of those are KK so the chance of the second player getting KK is 6/1325 or 0.4528%
For QQ it is 6/1324 or 0.4532%
Since there are 2 aces and 2 kings gone, there are only 2 of each left. That would yield 6 combos, but two of them are the other pairs, so there are 4 combinations of AK, and only 1323 possible combos left.
So for AK it is 4/1323, or 0.3023%
For all of these to happen on the same hand, just multiply the percentage probabilities together:
0.4525% x 0.4528% x 0.4532% x 0.3023% = 0.00000002807414320%

Stated as odds, the odds of this happening are 3,561,996,505:1
I am a bit surprised.

I remember having AA > AA, QQ and K9o AIPF (Don't know what the K9 guy was thinking) in a 45 manner years ago. The QQ won with a flopped set.

In a Stud tourney, I had 4s rolled up. Every street was capped. Villain had rolled up Ts. Even more amazingly, neither of us improved on our starting hands.

This was suggested by this thread https://forums.pokerstrategy.com/forum/thread.php?postid=3373722#post3373722

The numbers 1 to 36 are assigned the colours as found on an American roulette wheel, 37 and 38 are used for 0 and 00

script output — the numbers are the maximum streak for that colour
[11100] INFO roulette.main 94 iterations: 10,000,000,000 interval: 1,000,000,000
[11100] INFO roulette.main 95 Red:32, Black:30, Green:8
[11100] INFO roulette.main 96 End

So out of 100 billion "spins" the maximum streak was 32.
I was surprised by this, since I'm sure I had done a coin-toss simulation once and streaks far longer in fewer trials.

So what are the odds?
First the coin toss:
The probability of flipping heads (or tails) is 0.5
Two consecutive heads would be 0.5 × 0.5 which equals 0.25. Stated as odds, this is 3:1 against
Three would be 0.5 × 0.5 × 0.5 which equals 0.125, or 7:1 against
and so on.
To hit a streak of 32 consecutive heads (or tails) the probability is (0.5)³² which is 0.00000000023283064365386962890625
Stated as odds, that is 4,294,967,295:1[1]

Now let's do the red/black roulette
probability of either red or black: 18/38 = 0.47368421052631576 <<< somewhat less than 50%
probability of 32 consecutive: 0.00000000004127100756144647852104
Stated as odds: 24,230,084,484:1
Wow! that slight difference from 50/50 makes a huge difference in the outcome.

Without the 00, what difference would it make?
Odds now are 10,321,314,386:1
Quite surprising.

Now we know why casinos make money — but of course we already knew that

[1] Clearly my memory of a far longer streak with far fewer flips is incorrect. I'm pretty sure my simulation did not involve 4 billion flips. Either my memory was faulty, or my simulation program was. This was for a school assignment in the '70s so either is likely.

Where's the graph of it? Can you do one with 48% vs 52%? And supply graph?

Originally posted by dangermowse3
Where's the graph of it? Can you do one with 48% vs 52%? And supply graph?

Those are direct calculations. not a simulation, so there is nothing to graph.

If you want to find the odds of a streak of a certain length with those probabilities, just follow the pattern and do it.

Speaking of simulations, I ran the python script through 10 billion spins, and managed to hit a streak of 34.

>roulette -i 10000000000                             
iterations:10,000,000,000 interval:1,000,000,000     
15:36:17               0 Red:  0 Black:  0 Green:  0 
15:57:14   1,000,000,000 Red: 31 Black: 28 Green:  6 
16:19:14   2,000,000,000 Red: 31 Black: 29 Green:  6 
16:39:50   3,000,000,000 Red: 31 Black: 29 Green:  7 
17:01:17   4,000,000,000 Red: 34 Black: 29 Green:  7 
17:23:58   5,000,000,000 Red: 34 Black: 29 Green:  7 
17:45:45   6,000,000,000 Red: 34 Black: 29 Green:  7 
18:08:10   7,000,000,000 Red: 34 Black: 29 Green:  7 
18:29:21   8,000,000,000 Red: 34 Black: 29 Green:  7 
18:50:13   9,000,000,000 Red: 34 Black: 29 Green:  7 
19:12:14  10,000,000,000 Red: 34 Black: 30 Green:  7 
03:35:56 (12956 seconds)                             

10 billions yes I think that's somewhere close to the number of games I need to see EV.

Originally posted by dangermowse3
10 billions yes I think that's somewhere close to the number of games I need to see EV.

That was for the roulette simulation.
Betting red 10 billion times the same amount each time would result in a tremendous loss
That might be a fun simulation to try.

I added a function to the python script that keeps track of the streak lengths
this is for 100,000 iterations

 len    red     black    green
    1    13005    13169  4713
    2     6266     6185   256
    3     2914     2913    16
    4     1438     1357     2
    5      679      723     0
    6      292      254     0
    7      171      156     0
    8       84       64     0
    9       22       35     0
   10       18       18     0
   11        8       12     0
   12        0        3     0
   13        4        2     0
   14        0        1     0

and for 1 billion spins

 len      red         black      green
    1    131222526    131218977 47234499
    2     62148689     62146213  2487037
    3     29439567     29442881   130189
    4     13948765     13945736     7042
    5      6604099      6606039      379
    6      3130285      3131803       17
    7      1481526      1481501        0
    8       703075       701187        1
    9       332987       332216        0
   10       156782       158060        0
   11        74829        74561        0
   12        35434        35482        0
   13        16824        16846        0
   14         7947         8091        0
   15         3703         3748        0
   16         1728         1686        0
   17          871          784        0
   18          394          420        0
   19          180          199        0
   20           95           97        0
   21           49           40        0
   22           16           26        0
   23            7            6        0
   24            3            8        0
   25            3            1        0
   26            2            0        0
   27            0            0        0
   28            0            0        0
   29            0            0        0
   30            0            0        0
   31            0            0        0
   32            0            0        0
   33            1            0        0

I just started the run of 10 billion, which should take about 4 hours