Stop-n-Go question

theslick

Quote from

http://www.paddypowerpoker.com/ps_stopandgo.php

"The flop is more likely to come without a card over T then it is to have an over card"

That doesn't seem right.

If you have TT and he has 2 overcards, then there are 48 unknown cards left and 32 of those are <T then the probability of all 3 flop cards <T is

(32/48) * (31/47) * (30/46) = 27%

which is way below 50%

brianmc

theslick wrote:

Quote from

http://www.paddypowerpoker.com/ps_stopandgo.php

"The flop is more likely to come without a card over T then it is to have an over card"

That doesn't seem right.

If you have TT and he has 2 overcards, then there are 48 unknown cards left and 32 of those are <T then the probability of all 3 flop cards <T is

(32/48) * (31/47) * (30/46) = 27%

which is way below 50%

You're correct.

spectre

theslick wrote:

Quote from

http://www.paddypowerpoker.com/ps_stopandgo.php

"The flop is more likely to come without a card over T then it is to have an over card"

That doesn't seem right.

If you have TT and he has 2 overcards, then there are 48 unknown cards left and 32 of those are <T then the probability of all 3 flop cards <T is

(32/48) * (31/47) * (30/46) = 27%

which is way below 50%

"Without an over card", this implies <=T
Therefore you should be using 34 good cards.
The probability is 35% which is still well short though.

NickyOD

Who on earth wrote that!!!!?????

RoundTower

The shop is coming soon where you will be able to buy great value things with your Paddy Points!

RoundTower

NickyOD wrote:

Who on earth wrote that!!!!?????

lol I just checked.

cardshark202

RoundTower wrote:

lol I just checked.

I looked but I couldn't see a name anywhere on that page.

EDIT: Haha I just found the culprit. lol

RoundTower

If you liked that article, you may enjoy more by the same author. Here is one about playing aces preflop when people go all in before you.

And did you ever wonder how much of an underdog you are if you have two random cards and your opponent has AK? Look no further...

Or how often do you flop a set if you see the flop with a pocket pair? This guy has a degree in mathematics

NickyOD

RoundTower wrote:

If you liked that article, you may enjoy more by the same author. Here is one about playing aces preflop when people go all in before you.

And did you ever wonder how much of an underdog you are if you have two random cards and your opponent has AK? Look no further...

This one is my favourite.

http://www.paddypowerpoker.com/ps_commonlyaj.php

"AJ is a “glass cannon”. It’s either a huge favourite, a huge underdog or 50/50".

Huzzahh!!!

RoundTower

I can't decide which one I like best, but anyone who plays tournaments on Tribeca has this guy to thank.

PPP's Poker Expert wrote:

[AK] is a hand that plays similarly to AA when deep-stacked or early in a tournament...Raise, Raise, always Raise! Unless the pot has previously been raised of course, then you flat call

NickyOD

RoundTower wrote:

I can't decide which one I like best, but anyone who plays tournaments on Tribeca has this guy to thank.

Good Lord!!! :rolleyes:

Mellor

theslick wrote:

If you have TT and he has 2 overcards, then there are 48 unknown cards left and 32 of those are <T then the probability of all 3 flop cards <T is

(32/48) * (31/47) * (30/46) = 27%

which is way below 50%

You're right. It's less than 50% for all three cards to hit less than 10. But that wasn't the point he was trying to make. (even if it sounded very like it).
He means if you have TT, and your appondant has AK. Pre-flop it's almost a 50/50 coin-flip, so if you re-raise AK, he will most like call.
Instead flat call and if (35% of the time) the flop comes <10. Push all-in. Thye odds are now stacked in your favour and he is unlikely to call. But he made a balls of his point.

Hectorjelly

I love the beach

NickyOD

Hectorjelly wrote:

I love the beach

Why?

Hectorjelly

Sand Sun, not sure really now that you mention it. The beaches here in portugal sure are nicer than in ireland though.

Poker & Pints

BEACHES, BEER and BIRDS in BIKINIS

theslick

spectre wrote:

"Without an over card", this implies <=T
Therefore you should be using 34 good cards.
The probability is 35% which is still well short though.

I didn't include the 2 Ts cause I reckon if you hit one of those for your set then you'd be ditching the 2nd stage of stop-n'-go and doing the check/raise thing instead.

NickyOD

Hectorjelly wrote:

Sand Sun, not sure really now that you mention it. The beaches here in portugal sure are nicer than in ireland though.

Oh you mother*******. I love Portugal!!

TacT

Can I go to portugal too, it sounds like fun?

peaderfi

theslick wrote:

Quote from

http://www.paddypowerpoker.com/ps_stopandgo.php

"The flop is more likely to come without a card over T then it is to have an over card"

That doesn't seem right.

If you have TT and he has 2 overcards, then there are 48 unknown cards left and 32 of those are <T then the probability of all 3 flop cards <T is

(32/48) * (31/47) * (30/46) = 27%

which is way below 50%

the probability that there is no overcard on the flop to a pocket pair is as follows

Total number of possible flop combinations

50C3 (50 choose 3) = 19,600

Total number of combinations of no overcards coming on the flop

(4x - 6)C3

where x is the rank of your card ie from 3 up to K (where J=11 , Q = 12 , and K = 13 - please note the A and 2 cannot be used in this equation (A beacuse there are no overcards to it and the 2 because all the cards apart from the other twos are overcards and since you are holding pocket 2's there are only 2 other 2's in the deck and since 3 cards come on the flop...well as you can see an overcard is going to come))

aside:
heres how i came to that formula
overcards to your pocket pair:
(14 ranks of cards (A counts as 1 and 14) - all the ranks including 
and below your cards)*4 suits of cards

total number of cards left in the deck minus your hand is 50

so number of cards which are not overcards to your cards is

(50 – ((14 – x)*4))

but (50 – ((14 – x)*4)) = 50 - 14*4 + 4x  
= 4x - 6

now the probability that there are no overcards to a pocket pair on the flop is as follows (provided you don't know what your opponents cards are)

(4x - 6)C3 divided by 50C3

so in our case this is

(4 * 10 - 6)C3 = 5984
divided by

50C3 = 19,600

now 5984/19,600 = 30.5%

so the probability that an overcard will hit is 100% - 30.5% = 69.5%

the above two figures are for the general case where you have know idea what your opponent has (as in you don't have a read on him)

but in our case we know what the opponents cards are so it changes our formula a bit because we can now discount the two overcards our opponent holds for coming down on the flop so the formula for calculating the number of overcards now becomes

((14 - x)*4) + 2

which then affects our formula for no overcards hitting as follows

(50 – ((14 – x)*4) +2) = 48 - 14*4 + 4x = 4x - 8

so our probability formula now becomes

(4x - 8)C3 divided by 48C3

Note:
48C3 = 17296

For Pocket 10s this becomes (4*10 - 8)C3 = 4960

and 4960 divided by 17296 = 28.677%

so the probability now that any overcard hits on the flop is
100% - 28.677% = 71.32%

so paradoxically by knowing that he has any 2 overcards we actually increased the percentage chance of overcards hitting on the flop and we are now in a worse situation than before. Go Figure

but if we look at the hand he describes (assuming AKo) in more detail the only overcards that we are really afraid of on the flop are Aces or Kings or the 119 combinations which would give him some sort of a non flushed straight (((4*(Q+J+1/2 of the 10s))C3 - 1 straight flush) = (4*(2.5))C3 - 1 = 119

so we can discount all the other overcard combinations and we do this as follows

no of combinations where an A or a K comes down on the flop is as follows:

where at least one A or one K comes on the flop

is 47C2 = 1081

^^^^
all the other combinations of Aces and Kings are included in this figure so

no of combinations where an A or a K comes down on the flop is 1081

now
48 cards left minus 3 Aces and 3 Kings leaves us with 42 cards

so 42C3 = 11480

and there are 119 of those combinations which we do not like...see above

(meaning we are left with 11361 "niceish" flops)

so the total combinations which are definitely not good flops for us are

17296 - 11480 + 119 + 1081 = 7016

(this figure changes a lot if his cards are suited because then overcards don't really count if the board is suited...i'm gonna assume for simplicity that they aren't...if people do want figures for the flush possiblities i'll stick them in their own section at the bottom of this post so as not to confuse the casual reader)

now 7016 out of 17296 possible combinations gives a 39.14% chance that his cards will develop...meaning there is a 60.86% chance that they won't and that your cards will hold up on the flop

so even though the person writing the article is technically wrong about his general statement of

"The flop is more likely to come without a card over T then it is to have an over card"

In the case where you know your opponent has AKo (by osmosis, ESP, X-Ray goggles, someone standing behind him showing you what he has with another deck of cards ) 60.86% of the time the flop will not come with overcards that will help your opponent he is correct, if the sentence is changed to:

"The flop is more likely to come without a card over T that will help your opponent then it is to have an over card"

so in answer to your original post yes the guy who wrote the article is a bit inept

aside:
now onto the flush possibilities which would help the guy in *our very 
particular case* of him having AKs providing you know his cards in addition to 
the 7016 there are now another 165 flops of which you are mortally afraid...so 
7181 total meaning a 41.5% chance [b]of him flopping cards which will lose you 
the hand as it stands then[/b]...so even with the extra flush possiblities you are 
still a 58.5% favourite to win the hand if the turn and the river card are 
magically not dealt and you could win just based on the flop...anywho i'm 
rambling...for anyone wishing to know how much of a favourite you are for the 
whole hand (flop turn and river being dealt) just change all the *C3 in the 
above calculations to *C5

and yes i do have to agree that the guy who writes these articles is a moron...it seems as though he half reads up on something and then pulls his own opinions out of thin air!!!

ah well what a pity for anyone who reads him and takes it to heart...not so much for anyone who doesn't tho

cardshark202

peaderfi wrote:

and yes i do have to agree that the guy who writes these articles is a moron...it seems as though he half reads up on something and then pulls his own opinions out of thin air!!!

ah well what a pity for anyone who reads him and takes it to heart...not so much for anyone who doesn't tho

If you knew who wrote those articles you may regret typing that....

peaderfi

nope not really cause i've read the articles on paddy power poker written by him and from what i can see he doesn't pay very much attention to the theoretical side of poker, he made some very big mistakes in many of them which are just going to mislead novices who read them, a prime example is the one which was highlighted in this very thread on which i was commenting.

He may be a very good player and someone who i would not want to play against because he could very well take all of my money but he is not a good poker theorist and therefore should not be writing articles which contain complicated poker theory and as such i feel justified in calling him a moron (purely in the sense of being a poker theorist) for writing articles which do contain complicated poker theory.

If it was only one article which he had made such a mistake in i would mark it as unusual and not put it down to an apparent lack of knowledge, but the fact that such mistakes appear in many of his articles written for ppp means one of two things...either he is not a very good theorist in which case he should not be authoring these articles on poker theory, or he is deliberately putting these mistakes in his articles to mislead people (amazingly unlikely i know and i don't think he would do this no matter who he is because he would get caught out) in which case he should definitely not be writing such articles

wayfarer

peaderfi wrote:

the probability that there is no overcard on the flop to a pocket pair is as follows

Total number of possible flop combinations

50C3 (50 choose 3) = 19,600

Total number of combinations of no overcards coming on the flop

(4x - 6)C3

where x is the rank of your card ie from 3 up to K (where J=11 , Q = 12 , and K = 13 - please note the A and 2 cannot be used in this equation (A beacuse there are no overcards to it and the 2 because all the cards apart from the other twos are overcards and since you are holding pocket 2's there are only 2 other 2's in the deck and since 3 cards come on the flop...well as you can see an overcard is going to come))

aside:
heres how i came to that formula
overcards to your pocket pair:
(14 ranks of cards (A counts as 1 and 14) - all the ranks including 
and below your cards)*4 suits of cards

total number of cards left in the deck minus your hand is 50

so number of cards which are not overcards to your cards is

(50 – ((14 – x)*4))

but (50 – ((14 – x)*4)) = 50 - 14*4 + 4x  
= 4x - 6

now the probability that there are no overcards to a pocket pair on the flop is as follows (provided you don't know what your opponents cards are)

(4x - 6)C3 divided by 50C3

so in our case this is

(4 * 10 - 6)C3 = 5984
divided by

50C3 = 19,600

now 5984/19,600 = 30.5%

so the probability that an overcard will hit is 100% - 30.5% = 69.5%

the above two figures are for the general case where you have know idea what your opponent has (as in you don't have a read on him)

but in our case we know what the opponents cards are so it changes our formula a bit because we can now discount the two overcards our opponent holds for coming down on the flop so the formula for calculating the number of overcards now becomes

((14 - x)*4) + 2

which then affects our formula for no overcards hitting as follows

(50 – ((14 – x)*4) +2) = 48 - 14*4 + 4x = 4x - 8

so our probability formula now becomes

(4x - 8)C3 divided by 48C3

Note:
48C3 = 17296

For Pocket 10s this becomes (4*10 - 8)C3 = 4960

and 4960 divided by 17296 = 28.677%

so the probability now that any overcard hits on the flop is
100% - 28.677% = 71.32%

so paradoxically by knowing that he has any 2 overcards we actually increased the percentage chance of overcards hitting on the flop and we are now in a worse situation than before. Go Figure

but if we look at the hand he describes (assuming AKo) in more detail the only overcards that we are really afraid of on the flop are Aces or Kings or the 119 combinations which would give him some sort of a non flushed straight (((4*(Q+J+1/2 of the 10s))C3 - 1 straight flush) = (4*(2.5))C3 - 1 = 119

so we can discount all the other overcard combinations and we do this as follows

no of combinations where an A or a K comes down on the flop is as follows:

where at least one A or one K comes on the flop

is 47C2 = 1081

^^^^
all the other combinations of Aces and Kings are included in this figure so

no of combinations where an A or a K comes down on the flop is 1081

now
48 cards left minus 3 Aces and 3 Kings leaves us with 42 cards

so 42C3 = 11480

and there are 119 of those combinations which we do not like...see above

(meaning we are left with 11361 "niceish" flops)

so the total combinations which are definitely not good flops for us are

17296 - 11480 + 119 + 1081 = 7016

(this figure changes a lot if his cards are suited because then overcards don't really count if the board is suited...i'm gonna assume for simplicity that they aren't...if people do want figures for the flush possiblities i'll stick them in their own section at the bottom of this post so as not to confuse the casual reader)

now 7016 out of 17296 possible combinations gives a 39.14% chance that his cards will develop...meaning there is a 60.86% chance that they won't and that your cards will hold up on the flop

so even though the person writing the article is technically wrong about his general statement of

"The flop is more likely to come without a card over T then it is to have an over card"

In the case where you know your opponent has AKo (by osmosis, ESP, X-Ray goggles, someone standing behind him showing you what he has with another deck of cards ) 60.86% of the time the flop will not come with overcards that will help your opponent he is correct, if the sentence is changed to:

"The flop is more likely to come without a card over T that will help your opponent then it is to have an over card"

so in answer to your original post yes the guy who wrote the article is a bit inept

aside:
now onto the flush possibilities which would help the guy in *our very 
particular case* of him having AKs providing you know his cards in addition to 
the 7016 there are now another 165 flops of which you are mortally afraid...so 
7181 total meaning a 41.5% chance [b]of him flopping cards which will lose you 
the hand as it stands then[/b]...so even with the extra flush possiblities you are 
still a 58.5% favourite to win the hand if the turn and the river card are 
magically not dealt and you could win just based on the flop...anywho i'm 
rambling...for anyone wishing to know how much of a favourite you are for the 
whole hand (flop turn and river being dealt) just change all the *C3 in the 
above calculations to *C5

and yes i do have to agree that the guy who writes these articles is a moron...it seems as though he half reads up on something and then pulls his own opinions out of thin air!!!

ah well what a pity for anyone who reads him and takes it to heart...not so much for anyone who doesn't tho

This is probably the most annoying and unnecessarily long post that I have read on boards in a long time. Partly because what has already been said in the thread was enough.

theslick wrote:

(32/48) * (31/47) * (30/46) = 27%

This equation is all that you need, and the point was made. (it doesnt really matter that he's counting the 10s as over-cards, only that the result is <50%) .....

But also because

peaderfi wrote:

but in our case we know what the opponents cards are so it changes our formula a bit because we can now discount the two overcards our opponent holds for coming down on the flop so the formula for calculating the number of overcards now becomes

((14 - x)*4) + 2

which then affects our formula for no overcards hitting as follows

(50 – ((14 – x)*4) +2) = 48 - 14*4 + 4x = 4x - 8

so our probability formula now becomes

(4x - 8)C3 divided by 48C3

Note:
48C3 = 17296

For Pocket 10s this becomes (4*10 - 8)C3 = 4960

and 4960 divided by 17296 = 28.677%

so the probability now that any overcard hits on the flop is
100% - 28.677% = 71.32%

so paradoxically by knowing that he has any 2 overcards we actually increased the percentage chance of overcards hitting on the flop and we are now in a worse situation than before. Go Figure

All this is wrong! If you really feel the need to post needlessly long maths at least get it right. I stopped reading after this but I really should have stopped long before.

Hitman Actual

wayfarer wrote:

This is probably the most annoying and unnecessarily long post that I have read on boards in a long time.

I dont know, there's not half enough maths threads on here these days.

Tel you what Peader, if you can do this I'll really be impressed: If just one overcard to the ten flops, say a K, can you use Bayes Theorem to calculate the probabilty of opponent having a King, assuming he definately has overcards. (Not sure that there's enough info here for Bayes, but what the hell!).

wayfarer

lenny_leonard wrote:

I dont know, there's not half enough maths threads on here these days.

If it was sound suitable maths I'd be a happy chappy but sound suitable maths this is not. Everything he had said had already been said but in much simpler terms

btw I'll have a stab at the bayes theorem one for a larf before somebody who actually knows what they are talking about comes along. I don't know the equation but anyway.

If he holds a hand with 2 over (unpaired) cards then he either has AK, AQ, AJ, KQ, KJ, QJ. There are 16 different combinations of each but since a K comes on the flop, that means that there are only 12 combinations of AK, KQ and KJ, ie.

AK - 12
AQ - 16
AJ - 16
KQ - 12
KJ - 12
QJ - 16

Then it's just the number of hands holding a king divided by the total number of hands

(12 + 12 + 12) / (12 + 16 + 16 + 12 + 12 + 16)

= 43%

ta-da! (could very easily be wrong)

Lplated

wayfarer wrote:

All this is wrong! If you really feel the need to post needlessly long maths at least get it right. I stopped reading after this but I really should have stopped long before.

Sorry to be stupid, any chance you could explain why it was wrong, Wayfarer?

spectre

lenny_leonard wrote:

I dont know, there's not half enough maths threads on here these days.

Tel you what Peader, if you can do this I'll really be impressed: If just one overcard to the ten flops, say a K, can you use Bayes Theorem to calculate the probabilty of opponent having a King, assuming he definately has overcards. (Not sure that there's enough info here for Bayes, but what the hell!).

Ok here's my solution:
Event A: opponent has a King
Event B: Flop comes {King, a, b} where a,b are in the range {2,3,...,9,10}

Note: We assume opponent has overcards to our 10s but NOT a pocket pair

Pr(A|B) = Pr(B|A)Pr(A)/Pr(B)

Pr(B|A) = [ 3 x 34C2 ]/[ 48C3 ]

Pr(A) = (4 x 12)/( 16C2 - 24 ) = 0.5......4 Kings x any A,Q,J/ (all cominations of unpaired Aces Kings Queens or Jacks)

Pr(B) = [ 4 x 34C2 ]/[ 48C3 ]


so Pr(A|B) = Pr(B|A)Pr(A)/Pr(B) = 3/4 *0.5 = 0.375

Answer 37.5%

Comments???

wayfarer

Lplated wrote:

Sorry to be stupid, any chance you could explain why it was wrong, Wayfarer?

I was going to leave it to peader to look at it so that he could spend a while figuring it out like I had to! Nevermind..

In that part, there is the same number of combinations of undercards in the deck as there is in the first part ie. according to his formula '(4x - 6)C3'. So the probability of a flop of undercards when you are holding TT and your opponent has overcards is:

(4x - 6)C3 / 48C3 = 34.6%

'(4x - 8)C3 / 48C3' would be correct if, for whatever reason, you wanted to look at a situation where your opponent was holding undercards.

peaderfi wrote:

so paradoxically by knowing that he has any 2 overcards we actually increased the percentage chance of overcards hitting on the flop and we are now in a worse situation than before. Go Figure

I honestly can't believe he typed this out and didn't give a second thought to it!

Hitman Actual

spectre wrote:

Ok here's my solution:
Event A: opponent has a King
Event B: Flop comes {King, a, b} where a,b are in the range {2,3,...,9,10}

Note: We assume opponent has overcards to our 10s but NOT a pocket pair

Pr(A|B) = Pr(B|A)Pr(A)/Pr(B)

Pr(B|A) = [ 3 x 34C2 ]/[ 48C3 ]

Pr(A) = (4 x 12)/( 16C2 - 24 ) = 0.5......4 Kings x any A,Q,J/ (all cominations of unpaired Aces Kings Queens or Jacks)

Pr(B) = [ 4 x 34C2 ]/[ 48C3 ]


so Pr(A|B) = Pr(B|A)Pr(A)/Pr(B) = 3/4 *0.5 = 0.375

Answer 37.5%

Comments???

I'll take your word for it! I just remember Bayes from my college days, and was reading a bit on 2+2 about it's application to poker, and was wondering if it could be used here.

nh, wp.

RoundTower

spectre wrote:

Ok here's my solution:
Event A: opponent has a King
Event B: Flop comes {King, a, b} where a,b are in the range {2,3,...,9,10}

Note: We assume opponent has overcards to our 10s but NOT a pocket pair

Pr(A|B) = Pr(B|A)Pr(A)/Pr(B)

Pr(B|A) = [ 3 x 34C2 ]/[ 48C3 ]

Pr(A) = (4 x 12)/( 16C2 - 24 ) = 0.5......4 Kings x any A,Q,J/ (all cominations of unpaired Aces Kings Queens or Jacks)

Pr(B) = [ 4 x 34C2 ]/[ 48C3 ]


so Pr(A|B) = Pr(B|A)Pr(A)/Pr(B) = 3/4 *0.5 = 0.375

Answer 37.5%

Comments???

you know I'm sure wayfarer's answer has to be right but I'm not sure what's wrong with this one.

wayfarer

spectre wrote:

Comments???

I was looking at it there and I got 43% again. Using the same symbols as you did

P(A) = ... = 0.5
Like you have

For the other two I worked it out a different way coz I hate that 'C' thing, and it doesnt work for google. It ain't pretty but its effective.

P(B) = 0.5*[3*(3/48)*(34/47)*(33/46)] + 0.5*[3*(4/48)*(34/47)*(33/46)] = 0.1135
You forgot to take into account the fact that half the time your opponent won't have a K and that it will be in the deck

P(B|A) = 3*(3/48)*(34/47)*(33/46) = 0.0973
Same as your anwser

P(A|B) = P(B|A)*P(A) / P(B) = (0.0973*0.5)/(0.1135) = 42.863%

42.857% was the other one but the difference is more than likely just from rounding off the numbers.

spectre

wayfarer wrote:

I was looking at it there and I got 43% again. Using the same symbols as you did

P(A) = ... = 0.5
Like you have

For the other two I worked it out a different way coz I hate that 'C' thing, and it doesnt work for google. It ain't pretty but its effective.

P(B) = 0.5*[3*(3/48)*(34/47)*(33/46)] + 0.5*[3*(4/48)*(34/47)*(33/46)] = 0.1135
You forgot to take into account the fact that half the time your opponent won't have a K and that it will be in the deck

P(B|A) = 3*(3/48)*(34/47)*(33/46) = 0.0973
Same as your anwser

P(A|B) = P(B|A)*P(A) / P(B) = (0.0973*0.5)/(0.1135) = 42.863%

42.857% was the other one but the difference is more than likely just from rounding off the numbers.

Right you are wayfarer. I was thinking that my answer sounded a bit too low.
Well done

gerire

OK nuf of all this wheres Dev to answer for his sins????????????

Well explained maths btw wayfarer

peaderfi

wayfarer, you are right my post is wrong and this is the first time i've actually had a chance to read over my post since i'm in the middle of exams and have just spent the past two days without internet (it went kaput shortly after i posted because i tripped and pulled my router down onto the ground breaking its case open

in my original post

the plus here

((14 - x)*4) + 2

and here

(50 – ((14 – x)*4) +2) = 48 - 14*4 + 4x = 4x - 8

should have been a minus

so it should really be

(52 - 14*4 + 4x) = 4x - 4

so which means that you also made an elementary adding mistake in your post as follows

wayfarer wrote:

In that part, there is the same number of combinations of undercards in the deck as there is in the first part ie. according to his formula '(4x - 6)C3'. So the probability of a flop of undercards when you are holding TT and your opponent has overcards is:

(4x - 6)C3 / 48C3 = 34.6%

the probability of a flop of undercards when you are holding TT and your opponent has overcards is:

(4x - 4)C3 / 48C3 = 41.28%

but that is neither here nor there because either way though it means that in fact i was wrong and it reduces the chance of overcards falling...i apologise for this

now as opposed to being especially grumpy bout me making a mistake in adding why don't you take into account that it was 4am when i posted it...

now onto your comments about me

wayfarer wrote:

I honestly can't believe he typed this out and didn't give a second thought to it![/wayfarer]

i did think that it was very wierd and did have a good think about it but i couldn't see the mistake of the plus being a minus and thats why i stuck in that sentence...so don't comment on something which you obviously don't know anything about (ie my thought process or the length of time i thought about something because i spent the better part of half an hour looking over thinking that it couldn't be right but like i said i was tired and didn't see my mistake)

in answer to your earlier post yes the original post is enough but i figured that i would go ahead anyway and try do a general post for if you hold any pocket pair what the chances of overcards coming is...and so provide a post which may be useful in general to people, so i was partly answering his post but mainly doing the general thing.

now that i'm done with that bit, wayfarer you are correct bout there being a 43% chance that your opponent holds a king assuming he definitely holds overcards...now back to my studying

wayfarer

peaderfi wrote:

in my original post

the plus here

((14 - x)*4) + 2

and here

(50 – ((14 – x)*4) +2) = 48 - 14*4 + 4x = 4x - 8

should have been a minus

so it should really be

(52 - 14*4 + 4x) = 4x - 4

so which means that you also made an elementary adding mistake in your post as follows



the probability of a flop of undercards when you are holding TT and your opponent has overcards is:

(4x - 4)C3 / 48C3 = 41.28%

So when you're holding TT, the number of undercards (incl. the other two tens) in the deck is, going by your new formula

= 4x - 4, where x = 10
= 4*10 - 4
= 40 - 4
= 36

but 4 twos, 4 threes, 4 fours, 4 fives, 4 sixes, 4 sevens, 4 eights, 4 nines and 2 tens = 34

Can you have a look at it again?



It doesnt really matter to me what time you posted it at or what your thought process was, all I'm looking at is the post. You could have just left it to the morning if you felt that you were getting it wrong.

The reason I was pretty peeved was because I hate maths like this. I find that its not only very easy to make mistakes when doing, but also its quite hard to follow and to check if its correct. And in the end, I don't think the result is worth much when you compare it with other ways of working out the anwser (like the way that was used at the start of the thread). Its much easier to count the number of cards using your fingers and stick a simple calculation into google rather than learning off a couple of peculiar formulae. And in my experience you'll be right more often too.

Another thing is less people who read the board will be able to follow this compared to using other methods. You should be aiming to make posts that will benefit the majority of the readers here. It took me a sec to remember what 'C' stood for and the only reason I went back and had a look over your numbers was because of the bit that I quoted earlier in the thread