Poker… Texas holdem.
If you are playing 9 handed and get an Ace and a non Ace.
The chance of another player having an Ace is 1:36
If an Ace hits the flop (first 3 cards), the chance of another player with an Ace drops to what? (Meh, does it change…)
If there is an Ace within the final 5 community cards… Same question?
How could you work these things out simply? On the fly like…
Your time starts now…
Two hours and thirty-one minutes, and counting…..
75% chance it’s the future Harkness
Bump for Pm2ring…
Mr Ironic said:
Bump for Pm2ring…
Sorry. I’ve never played Texas hold’em, and I don’t understand all the jargon in the OP. And I have no desire to learn.
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Well firstly….
Boo
It is only 52 cards of 4 suits… for the life of day it is only 13 cards of interest…
Really? There is no need to count cards, it’s just odd’s v’s instinct…
Well at the end of the day it’s not that difficult to play, but the question becomes how often will certain hands hold up?
Poker is, to my mind, intractably complex because it involves human psychology.
Working out the dry odds that someone else has a better hand is easy.
Working out the odds that this particular player has a better hand GIVEN THAT he has raised $25 whereas the previous time he raised $50 and had nothing but there was that one time he only raised $25 but already had a straight … etc etc etc … is very tricky.
Working out the dry odds that someone else has a better hand is easy
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Yes the dry odds, please.
Leave the bullshit to me.
Poker is, to my mind, intractably complex because it involves human psychology.
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Yes, but the simplest of basics gives you an over hand…
I’ll have a stab (note: I don’t play texas hold ‘em!):
>>If you are playing 9 handed and get an Ace and a non Ace.
>>The chance of another player having an Ace is 1:36
Maybe 1:8? (52 cards, 2 revealed to you, one ace in your hand = 3 aces in 50 remaining cards, each player gets two chances to draw an ace: 3/50 + 3/50 = 6/50 = 1:8.3)
>>If an Ace hits the flop (first 3 cards), the chance of another player with an Ace drops to what? (Meh, does it change…)
1:12? (2 aces left in 47 cards, each player has two chances to draw an ace 2/47 + 2/47 = 4/47 = 1:11.75)
>>If there is an Ace within the final 5 community cards… Same question?
1:11? (2 aces left in 45 cards, 2/45 + 2/45 = 4/45 = 1:11.25) I took that as one ace in the 5 community cards.
>>How could you work these things out simply? On the fly like…
Practice. Learn from a Master…
Good luck!
(warning: my probability calcs could be total shite)
podzol said:
fly like…Practice. Learn from a Master…
Good luck!
(warning: my probability calcs could be total shite)
and you couldn’t read my poker face while my brows confuse you.
roughbarked said:
podzol said:
fly like…Practice. Learn from a Master…
Good luck!
(warning: my probability calcs could be total shite)
and you couldn’t read my poker face while my brows confuse you.
Probability 100%, RB.
Someone ought to check my working, but
A/
You start with 1 ace. You can see 2 cards. There are 50 cards you can’t see. There are 3 hidden aces. There are 47 hidden non-aces. There are 16 cards in other players’ hands. There are 34 cards in the deck.
So what are the chances that none of the aces are in the 16 cards?
(47! / (16! 31!)) / (50! / (16! 34!))
(34 × 33 × 32)/(50 × 49 × 48) = 30.5%
ie There is about a 69.5% that one of the eight other players has at least one ace.
B/
There’s an ace in the flop.
Now there are 47 cards unseen. There are 2 hidden aces. There are 45 hidden non-aces. There are 16 cards in other players’ hands. There are 31 cards in the deck.
(45! / (16! 29!)) / (47! / (16! 31!))
= (31 × 30)/(47 × 46 )
There’s a 43% chance that there are no aces in the 16 cards: ie, there is a 57% chance that one of the other players has an ace.
C/
There’s one ace altogether in the window cards.
45 cards unseen, 2 hidden aces, 43 hidden non-aces, 16 cards in the other hands, 29 cards in the deck.
(43!/(16!27!))/(45!/(16!/29!) = 29 × 28 / (45 × 44)
There’s 41% chance that there are no aces in the 16 cards, and a 59% chance that one of the other players has an ace. Note that this is a higher % than in scenario B, because we have revealed two more non-aces.
So how to estimate this on the fly?
It might be simpler to just remember a few dozen stats than try to do these calculations in your head during the few seconds of thinking time you have during a game of poker. But you might think this way:
A/ Three aces, about two thirds of the cards are in the deck. Any given ace has about a two thirds chance of being in the deck, so the chance of them all being in there are 2/3 ^3, ie 8/27 ,about 30%, so there’s a 70% chance one of the enemy has an ace.
B/ Two aces now, still about two thirds of the cards are in the deck, so 2/3 ^ 2 = 4/9, 44%, about 56% chance someone else holds an ace.
C/ Same as above, except the total number of cards has gone down by about a twentieth, so the chance of each ace has gone down by a twentieth, so the chance of both aces has gone down by a tenth, about 40%, so 60% chance someone else is holding an ace.
dv said:
Someone ought to check my working, but
/snip/
chance someone else is holding an ace.
you forget.. my brows suggest that I hold the three other aces.
Mr Ironic said:
the chance of another player with an Ace drops to what? (Meh, does it change…)
Does it change? It reminds me of the Monty Hall problem where intuition incorrectly suggests that the probability of the prize being behind the originally chosen door increases from 1/3 to 1/2, but in fact the probability doesn’t change at all. That doesn’t mean that the conditional probabilities never change when additional information is revealed, but rather one must be careful and examine how the additional information alters the ensemble of possible dealt hands.
However, like PM 2Ring, I have little interest in calculating the probabilities of poker hands.
However, like PM 2Ring, I have little interest in calculating the probabilities of poker hands.
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Well it is a good thing one of us is onto it.
A/ Three aces, about two thirds of the cards are in the deck. Any given ace has about a two thirds chance of being in the deck, so the chance of them all being in there are 2/3 ^3, ie 8/27 ,about 30%, so there’s a 70% chance one of the enemy has an ace.
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This is where I get confused…
70%?
If 18 cards are dealt, in a perfect world, one of each would take the first 13 spots…
Leaving 5 chances of other numbered cards to double up, thats 5 in 13…
Leaving a 30% chance your ace is paired with an opponent?
I don’t really understand your objection. Probably best that you stick with the precise figures in the first part of my response.
Wow, BOM wasn’t very close with it’s forecasting for today. Predicted a max of 21, and it only made it to 15.4.
Sorry, wrong thread.
You’re thinking of BOM odds
47! / (16! 31!)) / (50! / (16! 34!))
(34 × 33 × 32)/(50 × 49 × 48) = 30.5%
ie There is about a 69.5% that one of the eight other players has at least one ace.
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Well because we have similiar figures but you reversed them…
I don’t see why.
I worked out the odds for the other aces to be in the deck, rather than in the other hands.
Then I subtracted that from 100% to give the chance that it is in one of the hands.
I worked out the odds for the other aces to be in the deck, rather than in the other hands.
Then I subtracted that from 100% to give the chance that it is in one of the hands.
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Hmmmm…
If 2/3rds, of all Aces, are still in the deck…
Where is the 4th Ace?
Where is the 4th Ace?
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In “my” hand.
In “my” hand.
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Nup, you must have 2/3rds of an Ace.
30ish percent.
Whoops 1/3rd…
I still don’t understand your point, really. I know that I have exactly one ace in my hand.
Like I said … maybe just forget about the BOTE tips and go to the precise calculations in the first part of my post.
go to the precise calculations in the first part of my post.
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Yeah but no…
I think you have them arse about tit.
I think you haven’t read the question right.
Part of the premise is that I know I have one ace and one non-ace in my hand.
Part of the premise is that I know I have one ace and one non-ace in my hand.
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Yes. One of four.
And you said , roughly, that 2/3rds of the cards are still in the deck.
Going on that…
There is 100% chance an opponant has another…
(shrugs) I can’t explain it any more clearly.
Fair enough.
We will call it a draw.
Until you see the error of your ways…
“Yes. One of four.
And you said , roughly, that 2/3rds of the cards are still in the deck.
Going on that…
There is 100% chance an opponant has another…”
I can sense myself become rude and I don’t want to do that but seriously that just doesn’t make any kind of sense.
I’d be interested in calculating the odds if I understood the question.
Is there a statement of the question in plain English somewhere?
(Or has dv answered it already?)
that just doesn’t make any kind of sense.
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Well maybe I forgot the smiley face…
But if you have 1/4 of the cards in question and the deck has 2/3rds…
Thats 3/12ths plus 8/12ths… 11/12ths
So an 8% chance another has an Ace…
:)
Hey, Alex, are you a Kennett supporter?
Mea culpa, wrong thread.
Is there a statement of the question in plain English somewhere?
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OK
Consider this.
52 cards, 4 suits of 13 cards. The Ace considered the highest value.
9 players get 2 cards each, unshown to each other, betting aside, there is first 3 communitty cards ‘flopped face up’ then another then another. ie 5 community cards. best hand wins.
So any card, say an Ace, is in your hand. What chance another pllayer has an Ace, before the flop, after the flop or after the 5th card?
So any card, say an Ace, is in your hand. What chance another pllayer has an Ace, before the flop, after the flop or after the 5th card?
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That’s not the problem stated.
The problem stated is that there is one Ace in my hand, ANOTHER in the flop, what is the chance of an ace in the others hands.
Then there is one Ace in the hand, one ace in the community cards altogether, what is the chance of an ace in the others’ hands.
Read the first post, please.
Read the first post, please.
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Fair call.
So with 2 Aces out and 2/3rds of the deck unplayed…
There must be a 1/3rd chance of an opponent with an ace.
Mr Ironic’s objections had me worried that basic permutation theory might be wrong so I ran this through a simulator for 1000000 rounds.
Of the 103798 cases in which “I” get one Ace, the enemy has at least one ace in 43524 cases (70.45%).
Of the 24776 cases in which I get one Ace, and the flop has one ace, the enemy has at least one ace in 14284 cases (57.7%).
Of the 37867 cases in which I get one Ace, and the window cards altogether have one ace, the enemy has at least one ace in 15284 cases (59.6%).
These agree very well with my theoretical probabilities stated in post 329214.
Rev, trying to be as jargon free as possible, here it is.
There are 52 cards in the deck. 4 of these are aces.
I get 2 cards. MYCARDS. Each of 8 other players gets 2 cards (16 cards altogether). Given that the question only turns on whether any of these players has one ace or more, you can put these together in a group called OTHERCARDS.
The dealer puts out 3 cards initially, this is the “flop”. DEALERFLOP. He then puts out two more cards one at a time. DEALERTWO. Together these are the window cards. DEALERWINDOW.
There are also the cards remaining undealt of course. Before the flop, there are 34 cards in the deck. After the flop, there are 31 cards in the deck. After the other two cards, there are 29 cards in the deck.
During actual play, you only get to see your own cards, and then you see the flop and the other two cards, but this is kind of irrelevant to the probabilities…
So there are three questions:
A/ “Before the flop”
Given that there is 1 ace in MYCARDS, what is the probability that there is at least one ace in OTHERCARDS.
B/ “After the flop”
Given that there is 1 ace in MYCARDS, and 1 ace in the DEALERFLOP, what is the probability that there is at least one ace in OTHERCARDS.
C/ “After all the cards”
Given that there is 1 ace in MYCARDS, and 1 ace in the DEALERWINDOW, what is the probability that there is at least one ace in OTHERCARDS.
Of the 37867 cases in which I get one Ace, and the window cards altogether have one ace, the enemy has at least one ace in 15284 cases (59.6%).
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OK.
I get that the odds go up with less Aces revealed.
But HTF is it over 50% when it is 13(*4) different cards on 18 dealt?
But HTF is it over 50% when it is 13(*4) different cards on 18 dealt?
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Forget about the four aces. We are worried about the unknowns, not the knowns.
After the window cards are out, here are the unknowns:
There 45 cards. 2 aces.
There are 16 cards in other players’ hands. There are 29 cards still in the deck.
That’s all that matters, anything else is irrelevant.
My working in the previous post is based on permutational theory and the probabilities are exact.
But if you want to BOTEcalc it, think thusly. For each card, there’s about a 29/45 chance that they’ll be in the player’s hands. That’s 0.64 of a chance. So what’s the chance they are both in there? About 0.64 × 0.64 = 0.41, ie 41%. So there is about a 59% chance that at least one of them is in the other players’ cards.
dv said:
Someone ought to check my working, but
I get the same results.
But if you want to BOTEcalc it, think thusly. For each card, there’s about a 29/45 chance that they’ll be in the player’s hands. That’s 0.64 of a chance. So what’s the chance they are both in there? About 0.64 × 0.64 = 0.41, ie 41%. So there is about a 59% chance that at least one of them is in the other players’ cards.
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Agian, I think you have fipped the odds.
How is it that you turn 41% INTO 59%
And again…
If there is a 41% chance that there are two aces in the deck, then there is a 59% chance that there is at least one ace in the other players’ hands.
And again…
If there is a 41% chance that there are two aces in the deck, then there is a 59% chance that there is at least one ace in the other players’ hands.
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Ah Ha!
So if you have one…
There is a 30% chance that another has another.
“So if you have one…
There is a 30% chance that another has another.30% chance that another has another.
“
No.
You definitely have one, in the parameters of this puzzle.
If you have one, and there is one in the window cards, then there is a 59% chance that at least one other player has at least one.
Please read my previous post carefully.
then there is a 59% chance that at least one other player has at least one.
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Sorry DV.
But you have still to explain why 5/13ths is over 50%
Seriously, how can the odds of 18 cards from 52 ever be more than 18/32?
I’ll give you the second Ace in 21.
That leaves 2 in 21/52… less than half.
FMD I hope you are not building buildings in Tokyo…
I can feel a smackdown coming on.
Skunkworks said:
I can feel a smackdown coming on.
Yep, DV is in Singapore, not Tokyo.
Look, I’ll have one more go, Mr I.
All the cards you can see … there’s one ace in your hand, one non-ace. Four non-aces in the window, one ace.
What matters now is the cards you CANNOT see. 45 cards: 2 aces. 43 non-aces. All of these cards exist in either of two places: 16 of them are in the other players’ hands, and 29 are in the deck.
That’s all that matters.
Now forget about the cards you can see. Your own cards, the cards in the window. Put them completely from your mind. They don’t matter at all for the purpose of working out the probability of another player having an ace.
SO: just think about the 45 cards.
For the purpose of this puzzle (the final part), there are two possibilities: either at least one other player has at least one ace, or none of the other players have aces.
If none of the other players have aces, then both of the aces exist in the deck, still. The chances of both aces still being in the deck is 41%.
So if the chance of both aces still being in the deck is 41%, then the alternative (that at least one ace is in the other players’ cards) is 59%.
FMD I hope you are not building buildings in Tokyo…
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Fuck me dead and call me Deady McDead winner of the all Deadtown Dawn of the Dead contest, Mr Ironic …
Everything I have said in this thread comes out of basic permutation theory they teach in grade 11.
All of these cards exist in either of two places: 16 of them are in the other players’ hands, and 29 are in the deck.
That’s all that matters.
________________
So thats 16/29
(that at least one ace is in the other players’ cards) is 59%.
Ah Ha… One Ace…
So, It is half that, ergo 30%
Bazinga!
FMD.
Okay, I think I’m being trolled. If so, good job. If not, I’ve done my best. Maths theory comes up with a clear answer, Rev’s agreed, modelling 1000000 rounds came up with compatible answers, I’ve explained in layman’s terms and in technical terms. You can’t say fairer than that.
Outie.
yeah I see it now.
Where I thoght you had your maths arse about tit…
It was just a function of 2.
Sorry DV…
Okay, I think I’m being trolled.
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No seriously, I appreicate the debate and more so the input.
Most people would agree with you, I guess. The Rev has it wrong as well…
it’s just I don’t.
That DV is crazy in the coconut.
FWIW I don’t think it is trolling…
>>So if the chance of both aces still being in the deck is 41%, then the alternative (that at least one ace is in the other players’ cards) is 59%.
I read it as “the chance of another player with an Ace drops to what?” (from the OP).
That is: “an” ace, not “one or two” aces.
I think that is where the confusion comes in.
That is: “an” ace, not “one or two” aces.
I think that is where the confusion comes in.
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Even if you change it to “an ace” rather than at least an ace, the percentages are pretty much the same (since the probability of a player getting both of the remaining aces is low).
Ok. I don’t mean to upset anyone…
But why arn’t odds elegant?
In 18 cards you have 18/13ths chance of of an Ace being in a players hand. One and a third.
If you take out an Ace and a non-Ace you have 3 Aces in 50 cards. 16 dealt to others and 2 lots of 17 in the deck.
Therefor the chance of an Ace in the 16 and an Ace in each 17 is considered equal.
So the odds go up from 5/13ths to 1
What is this 70%, because of all the other permutations…the middle ground between the 2 extremes?
Mr Ironic said:
But why aren’t odds elegant?
I wouldn’t say that odds aren’t elegant. I think the “law of large numbers” is quite elegant. The “law of large numbers” basically says that for a large enough sample size, the empirical probability is the same as the theoretical probability. This is not as trivial as it might seem and requires proof. The proof, while not difficult, is quite profound in that it strikingly demonstrates the consistency of probability theory.
- Hey, I am going in the Indianapolis 500 today!
— That’s pretty exciting! … Um … shouldn’t you learn how to drive first?
- Nah, reckon I’ll just go with my gut. These things mostly come down to common sense.