What are the odds?

100 % and that is actually a mathematically solid answer.

Just over one in 4 billion if i can still remember how to count.

More like 1 in approx. 17000. I knew that 1 in 4 billion was no good

The probability of the deal that we want can be expressed as the following expression:

Probability { First hand dealt is Ax AND second hand is Ax AND third hand is Ax AND fourth hand is Ax}

This can be written as Prob {Ax dealt} x P{Ax dealt second time given that Ax already dealt once} x P{Ax dealt third time given that Ax already dealt twice} x
P{Ax dealt fourth time given that Ax already dealt thrice}

Inserting numbers:
4x48 / (52C2) x 3x47 / (50C2) x 2x46 / (48C2) x 1x45 / (46C2) where 52C2 =52! / (50! x 2!) etc.

This simplifies to 4! x (2!)^4 x 48! / 52! or more meaningfully as about 1 in 17000 in terms of odds.

Originally posted by Nunki
More like 1 in approx. 17000. I knew that 1 in 4 billion was no good

The probability of the deal that we want can be expressed as the following expression:

Probability { First hand dealt is Ax AND second hand is Ax AND third hand is Ax AND fourth hand is Ax}

This can be written as Prob {Ax dealt} x P{Ax dealt second time given that Ax already dealt once} x P{Ax dealt third time given that Ax already dealt twice} x
P{Ax dealt fourth time given that Ax already dealt thrice}

Inserting numbers:
4x48 / (52C2) x 3x47 / (50C2) x 2x46 / (48C2) x 1x45 / (46C2) where 52C2 =52! / (50! x 2!) etc.

This simplifies to 4! x (2!)^4 x 48! / 52! or more meaningfully as about 1 in 17000 in terms of odds.

that's a decent information, thanks
some of the terms I have to google to understand it but however

You hold a pocket pair

This hand isn't from Hold'em, but the odds are the same (with a slight variation -- see below):
It happened to me today -- and I'm just thankful it was limit!
Feral Cow Poker Hand Converter
PokerStars Limit 5 Card Draw $0.50/$1.00 - 4 players

Button: $17.31
SB: $14.99
BB: $26.08 (Hero)
UTG: $35.41

Dealing Hands: ($0.75) 8♥:8♦:5♥:A♣:5♠: (4 players)
UTG raises to $1, 2 folds, Hero calls $0.50

First Draw: ($2.25) (2 players)
Hero discards 1, UTG discards 3,
8♥:8♦:5♥:5♠: || 3♦:
Hero checks, UTG bets $1, Hero calls $1

Hero mucked 8♥:8♦:5♥:5♠:3♦:, two pair, eights and fives
UTG showed K♥:K♠:K♦:7♠:7♣:, a full house, Kings full of Sevens
UTG won $4.06
(Rake: $0.19)

In 5-Card Draw, the odds of drawing 3 and making a boat are 97.3:1 against.

We have 5 known cards, so there are 47 unknowns.

Needed: 1 of the 2 cards that match our pair
we threw 3 unpaired cards away -- so three ranks of the remaining 12 have only 3 pairs, the other 9 ranks have 6 pairs each
so in the unknown cards, 9 pairs from the first category, and 54 from the other, so 63 pairs possible to match with the two cards that match our pair.
So a total of 126 combinations total.

But that is only one way to make a full house.
We could also draw 3 cards of the same rank.
Since we threw away 3 cards, there is 1 combo each of 3 matching cards for each rank we threw.
of the other 9 ranks, there are 4 ways to make 3-of-a-kind for each rank.
3 + (9*4) = 39
126 + 39 = 165 possible ways to draw a full house holding a pair

Since there are 5 known cards, there are 16,215 3-card combinations, and 165 of them give us a full house
16,215-165 = 16050
16050:165 = 97.27:1 <<<Odds against drawing a full house in 5-Card Draw when drawing 3 to a pair

It is a bit different in Hold'em, because we know only the two cards in our hand.
But the calculations are the same:

First, 1 card to match our pocket pair, and another pair.
Now it is easy: There are 12 ranks, and 6 ways to make a pair from each rank -- 72 combinations of pairs,
So 144 3-card flops give us a boat this way.

There are also 4 ways to make trips from each rank, and 12 ranks, so another 48 combos give us a full house
144 + 48 is 192.

Because we have 50 unknowns, not 47 as in 5-Card Draw, there are 19,600 possible flops
So 19,600 - 192 = 19,408
19408:192 = 101:1 <<<Odds against flopping a full house in hold'em when holding a pocket pair

This is the start of a look at the odds in this hand:
https://forums.pokerstrategy.com/forum/thread.php?postid=3209676#post3209676

We'll look at the situation from three perspectives:
:diamond:1  Direct odds of making a hand in 3 ranges: 8-lo, 98-lo and T8-lo
:diamond:2  Odds of beating a pat 9 or a pat T
:diamond:3  Additional considerations considering that this is a PKO

:diamond:1  Direct odds
The Easy Part
I have A♥:8♦:2♠:6♣:4♠:
This is a superb starting hand for 2-7 Single Draw.
The absolute worst 8-lo is the 18th-ranking lowball hand, and only 17,340 hands beat it.
The odds of making an 8-lo are easy: We have 3s, 5s, and 7s as outs -- so 12 outs
We can see 5 cards, so 12 out of 47 means that our odds are 35:12 against hitting -- or 2.9:1
If we think that a 98642 will win, then add 4 outs -- 31:16 or 1.9:1 against
If we think that a T8762 will win, then add 4 more -- 27:20 or 1.35:1 <<nearly 42% chance of success

In the next post we'll look at how our draw stacks up vs 1, 2 and 3 other hands.

Conditions:
 We have a one card draw to an 8-lo with no chance of a straight
 Two players have raised ahead of us, and we assume at least one has a pat T

From the previous exercise, we see that we are almost a flip: our odds are 1.9:1
But there are two players to contend with.
There are only 32 cards between the ranks 9-2 inclusive
We have 4 of 'em, the pat guy has 5 of 'em and the other guy as 4 of 'em.

So our chances of getting one of our outs is reduced.
we might not be missing any outs
we might be missing 4 of 'em
So instead of 16 outs to hit our 9-lo, we either have:
12 (4 missing outs) odds of 35:12 == 2.9:1 (26% chance)[1]
13 (3 missing outs) odds of 34:13 == 2.6:1 (28% chance)
14 (2 missing outs) odds of 33:14 == 2.4:1 (30% chance)
15 (1 missing out) odds of 32:15 == 2.1:1 (32% chance)
16 (0 missing outs) odds of 31:16 == 1.9:1 (34% chance)

If we simplify, and consider that our chance of success is 30%, or chance of missing is therefore 70%
This applies also the other other player drawing as well.
So the chance that both miss is 0.7 x 0.7 or 0.49 -- you can see where this is going:

Major Lesson #1:
Holding a pat Ten, you are a very slight favourite vs two players each drawing 1 card

However, you have a 30% chance of tripling up if you hit -- and if you add bounties, it is even better

Major Lesson #2:
A one card draw to an 8 gives decent odds to play vs two other hands.

[1] To convert a probability expressed as "odds" to percent, add the two numbers together and divide by the smaller one:
example: 4:1 odds = 20% chance

To convert a percentage chance to odds, subtract the expressed % from 100, then divide the answer by the original %
example 20% chance -- 100 - 20 = 80 so odds are 80:20 or 4:1

Let's say, -- 3 players all hold QJ and turn the nut straight.

We'll ignore the board for now.
We know that in a 52-card deck there are 1326 possible hold'em hands.
If we only consider Qs and Js, there are 8 cards, so 28 different combinations of those two cards
12 are pocket pairs - 6 of each rank leaving 16 combos of QJ
4 combos are suited -- we include those because suitedness doesn't matter for straights.

One player has QJ (probability 1) It is certain that one player has the target
Three players do not have QJ by definition
(we assume that order is irrelevant)

Here is where I went wrong -- corrected thanks to Nunki (see below -- the old incorrect calculation is in the spoiler)

Now that one player has QJ, there are only 3 each of Queens and Jacks left.
So in those 6 cards, there are 15 2-card combos. 6 are pocket pairs, so 9 combos of QJ (offsuited or suited)
Of the 1326 combinations of 2 cards in a 52-card deck, 1 is gone (the target hand)
The probability that the next player does have QJ is 9/1325 = 0.006792453

Now there are only 2 of each left (assuming player 2 has it)
So of those 4 cards, there are 6 combinations -- but 2 are pocket pairs, leaving 4 combinations of the target hand
, so the probability that player 3 also has the target hand is 4/1324 = 0.003021148

Thus the probability that both have the target hand is 0.006792453 * 0.003021148 or 0.00002052101
which is 48730:1

Wanna throw the last hand in?
Just multiple 0.00002052101 by 1/1323
The odds of 4 people all holding the same hole card is therefore 64470524:1

I am confused as to what you are trying to calculate here. Is it the probabilty of three players in a six handed game of holdem being dealt QJ? Is it the probabilty of two players in a six handed game of holdem being dealt QJ given that one player has already been dealt QJ?

Hi Nunki,
It is the probability that 3 players have the same hole cards (ignoring suits)
In order for that to happen, you designate one player as having the "target" hand -- the probability of that is 1
Also, 3 other players must not not have that hand -- and the probability of that happening is also 1

So what remains is the probability of the other two having the target hand at the same time.

The is similar to the classic birthday problem. <<It is totally different from the classic birthday problem, since the birthday problem deals with an unspecified target, and an unknown number of people

btw -- I'm no mathematician, so if I screwed up the calculation, feel free to correct me. <<Thanks Nunki!

Cheers,
VS

Hmmm, I am still a little confused.

The target hand will possibly not have a probability of 1 as it can't be a pocket pair, also for example there are only 9 combos of QJ if it is given that one player has QJ etc.

"The hardest thing to do in mathematics is to count." A. Osbaldestin

Originally posted by VorpalF2F

It is possible for all 6 players in a 6-handed game to all have the same hand.

Surely this can can only be true on the river when all six players are playing the board ( eg. AAAAK). I am still confused.

For example, if in HU holdem I get dealt QJ the probability of my opponent holding QJ is 9/(50C2) if we ignore suits.

If I get dealt a pocket pair (HU) then the probability of my opponent holding the same pocket pair is 1/(50C2).

You're right
I see where I went wrong.

I'll redo the original after today's session.
If one player has QJ, then there only 3 of each left -- I was still using 4 DOH!

Thanks for pointing that out!

So it is only possible for 4 players to have the same hole card rank -- after that, there are no more left.
UG: Q♠:J♥:
MP: Q♥:J♦:
CO: Q♦:J♣:
BU: Q♣:J♠:

OK I redid the original post, and hid my explanation for the incorrect derivation

Thanks for your help!

Well this should be fairly easy.
Here is the hand we're considering:

There are 13 ranks, and every rank there are 6 pairs, so there are 78 possible pairs in our 52-card deck
However, we threw 2 cards away and since they were different ranks, 2 ranks now have only 3 combos that make pairs for those two ranks, and of course the 1 rank where we have trips has no possible pairs.

We assumed that our opponent had a pat hand -- this is most likely a straight or a flush, so in that cases there are 5 more ranks where there on only 3 pairs for each rank.
That leaves:
1 rank with 0 pairs
7 ranks with 3 pairs each == 21 pairs
5 ranks with 6 pairs each == 30 pairs.
Total then of 51 pairs

We know 5 cards, so there are 47 unknown cards, and in those 47 cards are 1081 two-card combos
Thus there are 1081-51 "bad" combos and 51 "good" combos
So the odds of drawing a pair are 1030:51 or 20.2:1

What if opponent has a full house instead of a straight, flush (or -- heaven forbid -- a straight flush):
2 ranks with 0 pairs
1 rank with 1 pair
5 ranks with 3 pairs = 15
5 ranks with 6 pairs = 30
48 pairs total
So 21.5:1

I'm sure that there is some way to combine those two numbers based on the number of flushes, straights and full houses possible, and the odds of drawing a pair will be different if opponent is not pat, but for now, this will do.

As always, comments and corrections are appreciated

Easy? It depends which level of accuracy you are looking for. Which question should you ask?

The question asked is 'what are the odds of boating up after the draw, given that you hold a set and Js2c when your opponent stands pat with a proper hand'.

If you are looking for the exact answer you have to allow for the cases when one of your discards has the same rank as either zero, one, two or three of villains hole cards. eg. 52 pairs are possible when your opponent has a J in his straight (or flush) etc. If you take every every possibility you have a whole bunch of possible counts. You can combine all of these with weighted averages and you will have the right answer. Mind boggling but possible. Counting is often difficult.

Far more easy to answer is: 'what is the probability of boating up if we draw two when holding trips?'. The answer to this question would be very close to the answer to the first question: 66/47C2 . There will be very little difference. Just like there is very little difference with 51/42C2.

Some crazy preflop action there and you were almost beat even after you nailed the draw!

Originally posted by Nunki
Some crazy preflop action there and you were almost beat even after you nailed the draw!

Considering pat hands include straight flushes, it is almost never a sure think in 5-Card Draw
However, for pat hands there are only 40 combinations of straight flushes (including royals)
 3744 full houses
 5108 flushes and
10200 straights
Of those, I have the best full house.

This applies also to hold'em with slight alterations.
It tells you the odds of having no pair at all on the flop, however in the case of lowball, three cards have been thrown away, so the available pool of cards is smaller

Scenario:
Dealt 5 cards, we throw away 3. and get 3 more.
We have seen 5 cards, so there are 47 unknown cards.
What cards we throw away affects what the drawing odds are, so we'll look at one single situation:
 we hold two low cards, and high trips. 6 cards in the deck will pair one of our existing cards.
We will draw them one at a time.
Card #1: Of the 47 cards, 41 are "safe" so the chance of not drawing a pair are 41/47 == 0.8723
Card #2: there are 46 cards left, but since we now hold 3 cards, there are now 9 cards that can make a pair [1] so the probability of not drawing a pair are 37/46 = 0.80435
Card #3: There are only 45 cards left, and 12 of them would pair one of our existing cards, so the probability of not drawing a pair is 33/45 = 0.7333
We now have 5 cards -- multiply those three together == 0.515 or 51.5%

[1] This is an oversimplification, since we threw away 3 cards of the same rank. Normally this wouldn't be the case, so we could sometimes draw a card of the same rank as one we had thrown out previously.

4-way -- everyone gets 7 cards, and a pair of nines holds.

PokerStars, 7 Stud H/L Limit - $0.25/$0.50 ($0.05 ante) - 8 players
Hand delivered by CardsChat

Creeep61: $8.19 (33 bb)
buldakov: $7.19 (29 bb)
nalomi1980: $3.35 (13 bb)
feetz123: $10.15 (41 bb)
VorpalF2F: $18.20 (73 bb)
Mr SUNGAME: $15.49 (62 bb)
mochabeto: $9.90 (40 bb)
pa4kata1: $13.64 (55 bb)

Third Street: ($0.40) Hero (VorpalF2F) is in Seat 8
Xx Xx 3♦: pa4kata1 bring-in $0.10____pa4kata1 calls $0.15
Xx Xx 7♠: Creeep61 calls $0.10____Creeep61 calls $0.15
Xx Xx 8♦: buldakov folds
Xx Xx 7♦: nalomi1980 folds
Xx Xx 5♣: feetz123 folds
A♥:K♣:T♠: VorpalF2F raises to $0.25
Xx Xx T♣: Mr SUNGAME calls $0.25
Xx Xx 8♥: mochabeto folds

Fourth Street: ($1.40) (4 players)
Xx Xx T♣:A♦: Mr SUNGAME checks
Xx Xx 3♦:T♥: pa4kata1 checks
Xx Xx 7♠:Q♦: Creeep61 checks
A♥:K♣:T♠:9♠: VorpalF2F checks

Fifth Street: ($1.40) (4 players)
Xx Xx T♣:A♦:6♥: Mr SUNGAME checks
Xx Xx 3♦:T♥:2♦: pa4kata1 checks
Xx Xx 7♠:Q♦:8♠: Creeep61 checks
A♥:K♣:T♠:9♠:2♠: VorpalF2F checks

Sixth Street: ($1.40) (4 players)
Xx Xx T♣:A♦:6♥:2♥: Mr SUNGAME checks
Xx Xx 3♦:T♥:2♦:K♠: pa4kata1 checks
Xx Xx 7♠:Q♦:8♠:J♠: Creeep61 checks
A♥:K♣:T♠:9♠:2♠:4♣: VorpalF2F checks

Seventh Street: ($1.40) (4 players)
Xx Xx T♣:A♦:6♥:2♥: Xx Mr SUNGAME checks
Xx Xx 3♦:T♥:2♦:K♠: Xx pa4kata1 checks
Xx Xx 7♠:Q♦:8♠:J♠: Xx Creeep61 checks
A♥:K♣:T♠:9♠:2♠:4♣:9♣: VorpalF2F checks

Total pot: $1.40 (Rake: $0.06)

Showdown:
VorpalF2F shows A♥:K♣:T♠:9♠:2♠:4♣:9♣: (HI: a pair of Nines)
pa4kata1 shows 5♥:3♠:3♦:T♥:2♦:K♠:Q♣: (HI: a pair of Threes)
Creeep61 mucks 2♣:4♦:7♠:Q♦:8♠:J♠:T♦: (HI: high card, Queen)
Mr SUNGAME shows 8♣:9♦:T♣:A♦:6♥:2♥:J♥: (HI: high card, Ace)
VorpalF2F wins $1.34