blonde poker forum

If you are dealt in SB on a 10 handed table and recieve an ace
what is the chance another ace has been dealt out before it comes back to you?

I'm struggling to work it out I can't seem to think of the right equation

the first ace is always 13/1 but as cards are dealt the fraction changes to the position you're sat in through card elimination

for example if cut off is dealt an ace aswell that is the 9th card with 3 aces left which gives you 3/44

so in the 10 cards dealt out before you receive your second card, chances an ace has already been dealt out is....?

Unless you know what those cards are, I wouldn't think it affects the probability of you being dealt a second ace.

In theoretical physics terms, if seat 5's card isn't known by someone, it doesn't have a value.

If memory serves, that is...

As Tal suggests - if you don't know what the cards are then they don't affect the probability.

Drawing a red card from a pack is 26/52 =1/2

If beforehand you had already taken 40 cards away then the odds of drawing a a red would still be 26/52 = 1/2 (if you don't know which 40 cards you'd removed).
The fact that there's only 12 cards  left in the pack you're drawing from is irrelevant.

The answer to both your questions is it is irrelevant.  The chance of a second ace is 1 in 17 wherever you are sat.

Probabilty shmobabilty! Can't remember the last time I had 2 aces at the same time. Must be storing them up for the next deep stack.

Why does it matter?

You always seem to get that bloody deuce? :) :) :)

Or even worse the  4d which for second you think is the other Ace.  :(

Good point. Live poker is rigged. Everyone knows that.

Moaning this one in like a champ!

Rest.

Odds on getting any suited broadway in hold em (A-T) assuming 9 handed

Odds on getting a pair in 6 card Omaha (assuming 6 handed)

Odds are probably smaller than beating Alex in a race.

The amount of players at the table makes no difference to the odds of being dealt a specific hand.

Ok so calculate the above pls

First Broadway card is 5/13. Second card being the same suit is 12/51.

60/663 is about 1/11 or 9%

Have I read that correctly or are you asking for both cards to be Broadway cards?

Suited broadways are 3% of all hands.

The 6O question it is relevant whether you're asking for hands with at least a pair, or only hands with just 1 pair.

At least one pair

card 1: 1
card 2: 48/51 (3 cards pair us)
card 3: 44/50 (6 cards pair us)
card 4: 40/49 etc
card 5: 36/48
card 6: 32/47

multiply together to get the probability of being dealt an unpaired hand = 34.5%

therefore the probability of at least 1 pair in 6 cards (or 2 pair or trips or a full house or quads, but ignoring straights and flushes) is 65.5%

if you took out any ace, then deal out 10 cards from a shuffled deck afterwards what is the probability one is an ace?
it's approx 20% of the deck with 3 aces in, it must be worked out! :) ty

The way I'm seeing the original question, if your first card is an ace, of the 51 remaining cards 3 are aces.
9 cards are dealt before it comes back to you and the question is what are the chances of one of them being an ace.
each card has a 5.88% chance of being an ace (3/51*100)
multiplied by 9 shots, that's 52.92% chance that someone elses first card is an ace.

look right?

This

chances of card 1 being a suited broadway card 20/52
chances of second card being the same suit and aslo 10+ 4/51

(20/52) x (4/51) = 80/2652 = 3.02%

that's a different question and the answer isn't 20%
not sure if I'm awake enough to work out what it is though.

the way you've phrased it, from a 51 card deck, containing 3 aces, what are the chances of 1of the next 10 being an ace (not 2 or 3)

I got to assume it surely can't be odds in favour another ace is dealt out.

think possibly you will have to add all fractions up 3/51----- 3/43 and multiply by chances you not receiving an ace, in a tree diagram possibly?

I was saying 10 cards is approx 20% of 51 cards, not really written well :s

that's the question I would like answered, I said 10 cards instead of 9 because I'm including the second card that would be dealt to me

There are 12777711870 combunations of 10 cards in 51 and 6540715896 combinations of 10 cards in 48 (the 3 aces removed).  So the probability of no aces in the 10 cards dealt is 51% (6540715896/12777711870) and therefore of an ace being dealt 49%.

your questions are all over the place.
In your OP you're asking what are the chances of one of your opponents getting an ace as their first card, knowing that one of them is in your hand.
Each of them has a 3/51 chance of getting an ace, and there's 9 shots.
3/51*100=5.88% * 9 shots = 52.94%

try it by taking an ace out of the dec and dealing 9 cards in a row
doesn't seem unreasonable that there'd be an ace around half the time

your original question specifically says what are the chances of another ace being dealy BEFORE it comes back to you.
that's 9 cards

Do you want to know if you have an ace as either of your cards, what are the chances of 1 or mor other aces being held by one of the other 9 players?

doubting my method now :)

I'm not sure he knows, as that question has been answered by doubleup (well he included all ten players as requested, but the method is there).

not quite, my suggested question includes any of the 18 cards we haven't seen.
We know that 1 of our cards is and ace and the other is a random non-ace
that means 3/50 cards are an ace
what are the chances that if we deal 18 out, 1, 2 or 3 of the remaining aces are dealt

Using my shitty shorthand, I get a 100%+ answer, which is why I think I might be doing it wrong.
I'm just calculating the way I would count outs before pot odds based call on the turn

You have to use combinations to work out these kind of problems.  Your method is ok for a quick approximation of  a simple scenario but the errors get magnified in complicated calculations.

http://en.wikipedia.org/wiki/Combination

Loads of help and some answers here

http://en.wikipedia.org/wiki/Poker_probability_%28Texas_hold_%27em%29

confirms the probability of suited broadway cards as 0.0302

and although this isn't directly the question Oxford asked, this table gives some clues

(http://i1236.photobucket.com/albums/ff442/pokerpops/ProbabilityofholdingdominatedAce.jpg)

I have lost track of what the OP was asking or why, but that is a well handy table David, ty.

haha too true! never the same suit ever :(

sorry, I have too many questions to ask!

the way I came to this scenario was by thinking, if I get a specific value card and the person next to me receives the same, what are the chances I pair up when it gets back to me?
simply 1/13 1/17 1/21 multiplied

I am always SB being dealt first in these situ's

then I wanted to take into account the variables, some one at random may actually receive the same card as me, which lead me to the OP, after my workings out failed to satisfy my belief I was correct.
the first ace is GTD and you get 1/13 but say you are on BTN 10 handed, (no aces known to be out) you technically have better odds of receiving an ace as you cannot account for what has been dealt out you can only assume your chances of an ace is 4/43

it's this which makes it unable for everyone to have a 3/51 chance as the deck gets shorter, so to take into account the variables here I wanted to know the probability of a 3 ace 51 card deck having ten cards dealt off top (effectively any 10 removed) one being an ace.

good link TY :)

You simply have to ignore what other people have. Let's use the 2 aces example. I'll show why even taking dealing to others into account you still have a 3/51 chance of receiving it. However, i'm going to simplify the problem a bit to illustrate the point:

You are being dealt to first. The first card is always the ace. (this is just to make the numbers easier to work with, it'd also work if we assumed you had a 4/52 chance of being dealt an ace in the first place).

There are only two aces in the deck so only one remains. (again, makes the numbers easier to work with without changing the logic behind the problem)

We are dealing to 9 people before you.

So now you have a 1/51 (1.96%) chance of receiving the second ace.

And we're going to deal the cards face up so they're no longer unknown.

The chances of no-one else being dealt the ace are: 50/51 * 49/50 * 48/49 * 47/48 * 46/47 * 45/46 * 44/45 * 43/44 * 42/43 = 14/17 (82.35%)

So the odds have changed for us now we know if it's been dealt or not.

17.65% of the time we will have a 0% chance of receiving it.
82.35% of the time we will have a 1/42 (2.38%) chance of receiving it. Note that this is higher than the 1.96% of the time we have if the cards are unknown.

So sometimes we now have no chance at all, but sometimes our chance has increased. Let's find the average of these chances...

17.65% * 0 = 0
82.35% * 2.38% = 1.96%

Add 'em up and we get 1.96% (1/51) which is what the chance would be if we consider the cards unknown.

This works for any number of players. So it is completely irrelevent how many players sit at a table for the purposes of determining odds of which cards will be dealt to you, nor does it matter where you are sat at the table.

Well put.

thanks bud, that's awesome! well that's sorted that one out for me! ;)

i get a first  Ahrt around 50% of the time, a second  Ad follows a further 50% of the time  ;sark; ;sark;