probability question

[AdamM]

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

[doubleup]

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

[david3103]

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

Click to see full-size image.

[atdc21]

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

[youthnkzR]

Why does it matter?

You always seem to get that bloody deuce?

haha too true! never the same suit ever

[Oxford_HRV]

your questions are all over the place.

Each of them has a 3/51 chance of getting an ace,

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

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.

[Oxford_HRV]

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

Click to see full-size image.

good link TY

[Cf]

your questions are all over the place.

Each of them has a 3/51 chance of getting an ace,

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

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.

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.

[skolsuper]

your questions are all over the place.

Each of them has a 3/51 chance of getting an ace,

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

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.

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.

[Oxford_HRV]

your questions are all over the place.

Each of them has a 3/51 chance of getting an ace,

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

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.

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.

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

[youthnkzR]

i get a first  around 50% of the time, a second  follows a further 50% of the time