Foxwoods Trip Report

[JTrue]

I went on a trip to Foxwoods from April 14th to the 16th. I had been there once before but only got to play 2-4 for about 5 hours. This time I played around 20 hours in the two days I was there. The first day I sat down for about 4 hours for my first session, the game was very loose/passive with almost no preflop raising. I ended up doing ok and won $20.00 over four hours.
On the same day after lunch I sat down again at the same table played for 6 hours and was up $29.00 by the time I left. The game was a little more agressive since two new players had entered the game but still very loose.
The next day I got up at 10:00am and got to the casino by 12:00. The first session I played around 3 hours and was up $2.00 when I went to eat lunch. This game had a few better players in it but mostly just loose/passive players. One memorable or should I say not so memorable was when I flopped two pair with my A J and a guy called me to the river and won with a backdoor flush.
After lunch I sat down at the same table and proceeded to lose $100.00 to a women who could have been my grandmother. I will not bore you with bad beat stories but let me tell you she was on the rush of her life and I lost about $100.00 in 1 hour. After that I got up and walked around for a bit, I knew I wasn't playing bad, cards were just not going my way.
After I walked around for a bit I sat back down at a table with alot of the same people from the day before, I thought I played very well and was up $170.00 in 6 hours. One memorable hand was when the dealer dealt me my first card, then proceeded to deal me my next card and it got flipped up so everyone saw it. It was the 10 of spades. He mucked that card and dealt me another card. I waited until the action was on me and looked at my cards, wow he had dealt me pocket 10s. Someone had raised before me and I just called on the button. On the flop to my amazement came 2 10 4, wow I hit a 1:47 shot. Needless to say the board ended up pairing up with two 3,s on the the turn and river giving me the boat, two other players were in and one was drawing to an ace high flush, two bad he didn't hit it because I would have really gotten paid off.
I ended up winning $100.00 over the 20 hours I played on my trip.

[daryn]

actually it's only around 15.3-1 against to have hit that ten. i had a long post where i outlined the calculation but i deleted it by mistake /images/graemlins/frown.gif

[Jason Strasser]

Im no math genius, but.

The odds of hitting the 10 is (1/49) + (1/48) + (1/47) which is a little better than 1/16. (.0625/1)

Duke engineering what?

[wontons]

I was gonna start this post by saying....what are you f%ckin retard ...how the hell did u come up with that number stay at 2-4 u dumb bastard...but 2 people already were on top of that...so good job...now go buy yourself a book and study how to calculate pot odds.

[Losing all]

You sir, are a dickhead

[Homer]

The odds of hitting the 10 is (1/49) + (1/48) + (1/47) which is a little better than 1/16. (.0625/1)

No, it is 1 - [(48/49)*(47/48)*(46/47)] = 3/49 = 46:3 = 15.3:1

Another way to do it is C(1,1)*C(48,2)/C(49,3) = 1128/18424 = 15.3:1

Duke engineering what?

I'm hoping you aren't a Duke engineering student being arrogant. Hopefully it's just a joke I don't get.

-- Homer

[Homer]

I was gonna start this post by saying....what are you f%ckin retard ...how the hell did u come up with that number stay at 2-4 u dumb bastard...but 2 people already were on top of that...so good job...now go buy yourself a book and study how to calculate pot odds.

I will try to convince JTrue to stay at 2-4 for a while as long as you promise to never procreate.

-- You are my hero, Homer

[amerksmann22]

JTrue,

You have any regular games that you go to in Rochester. I am in that area and looking for some games.

Thanks,
Mike

[daryn]

you definitely are no math genius /images/graemlins/grin.gif

[JTrue]

Sorry guys meant there was only one card that could hit out of 47 cards, stated it in the wrong terms. What would x/1 be on the turn and river? 22.5:1 and 45:1 respectivley? I think my calculations are right but not sure.

[daryn]

well say you didn't hit the ten on the flop:

now you have seen 6 cards, so 46 remain unseen. you need one of them, so on the turn of course it's 45:1 against. if you miss again on the turn, there are 45 unseen cards, and you need 1, so you're then 44:1 against hitting on the river.

after the river is dealt if you've still missed, the probability goes to zero /images/graemlins/wink.gif



edit: if what you meant was, if i miss on the flop what are the odds i hit on the turn or river, that would be 1-[(45/46)*(44/45)] = 22:1 against

[JTrue]

Post by Wontons:
[ QUOTE ]
I was gonna start this post by saying....what are you f%ckin retard ...how the hell did u come up with that number stay at 2-4 u dumb bastard...but 2 people already were on top of that...so good job...now go buy yourself a book and study how to calculate pot odds.

[/ QUOTE ]

Actually Wontons these would not be pot odds, they would be odds on improving. And although you do need to calculate your odds to improve in order to calculate pot odds it is totally different. Pot odds is when you compare your chances of improving to how much you have to put in the pot to win a certain amount. ie you have an openended straight draw, have 8 outs therefore you are getting 2.2:1 odds on the turn to improve and 4.8:1 odds on the river to improve. If you have to put in 40 to win 50 then you are not getting proper odds, now if you have to put in 40 to win 160 on the flop before the turn you are getting sufficient odds to call. One thing you should also be aware of are implied odds which is sort of a questimate on how many more calls will be made in further betting rounds, and how much you stand to win if your hand does hit. Thanks WONTONs for your nice post. Hope you learn something.

[JTrue]

Thanks daryn

[JTrue]

Go to wnypoker.com, they have a forum there with a calander of games, also a few bars have some games. Texas Hold'em Club of Rochester, Jerimiahs, Tipsy Mcstaggers, Baird Road Pub. Thats all I can think of right now. The number for Texas HOld'em Club of Rochester is 585-734-3531.

[CountDuckula]

[ QUOTE ]
I was gonna start this post by saying....what are you f%ckin retard ...how the hell did u come up with that number stay at 2-4 u dumb bastard...but 2 people already were on top of that...so good job...now go buy yourself a book and study how to calculate pot odds.

[/ QUOTE ]

People post here to learn and to share their knowledge. Insults and name-calling do not contribute to either of those purposes; they only serve to inflate your own ego and to alienate the target (which, in turn, leads to rejection of the proffered information). It's good to criticize when you see incorrect information (that's how we learn), but it's not constructive to be obnoxious about it.

-Mike

[BottlesOf]

Is there a book I could buy that would teach me how to have done that myself? I was ready to accept the 1/49 + 1/48 explanation, although it looked a little odd.

Hopefully, the answer is some 2+2 book and not a high school text book /images/graemlins/tongue.gif

[daryn]

eh.. all homer really did was calculate the odds of the ten NOT coming, then just subtract from 1.


the other method used was just to take the total number of flops not containing the ten, call it X, and the total number of flops containing the ten, call it Y, and express it as X:Y against flopping the ten.

to get those numbers, you just figure there are 49 unseen cards, since you've seen 3 tens. the total number of flops possible is just 49*48*47 = 110,544.

but this number counts flops like A /images/graemlins/heart.gif5 /images/graemlins/diamond.gif2 /images/graemlins/spade.gif and 5 /images/graemlins/diamond.gif2 /images/graemlins/spade.gifA /images/graemlins/heart.gif as 2 different flops, when they're not.. so you have to alter that number. you do it by dividing that number by x!, where x is the number of cards.. in other words 3 on the flop, 3! = 6, so 110,544/6 = 18,424 unique flops.

now you just need to know the total number out of 18,424 that contain the T, and the number that don't. well the number of flops that contain a T are just (48*47)/2! = 1,128.

now just subtract 1,128 from 18,424 to get the number of flops that don't contain a T.. 17,296.

thus the odds against hitting the T are 17,295:1,128


or 15.33:1 against.

[Homer]

Is there a book I could buy that would teach me how to have done that myself?

Most basic probability books will explain how to do this calcuation.

I was ready to accept the 1/49 + 1/48 explanation, although it looked a little odd.

I'd have to whip out my books to come up with a decent explanation as to why this is wrong, but for now think about this example:

Say you are going to roll a die six times and want to know the probably of getting at least one six. If you did the calculation the wrong way, like above, you would get 1/6 + 1/6 + 1/6... = 7/6. Obviously, the probably of getting at least one six cannot be greater than 100%. It is actually:

P(>=1 six) = 1 - P(0 sixes)

= 1 - (5/6)^6

Hopefully, the answer is some 2+2 book and not a high school text book

Sorry... /images/graemlins/tongue.gif

[Homer]

eh.. all homer really did was calculate the odds of the ten NOT coming, then just subtract from 1.

shhhhh....

[felson]

You guys have made this way way WAY too complex.

There are 49 unseen cards. You have 3 chances to hit your card. 3/49.

[Homer]

You guys have made this way way WAY too complex.

No.

It is important to understand the general rules, because your method will only work in this specific situation in which there is one object that must be chosen once.

In other situations, such as when you have a pocket pair and want to know the probably of flopping a set, using your example will make the answer 6/50. If you flop a four-flush and want to know how often you'll get there by the river, using your example will make the answer 18/47. Both of these answers are incorrect. You have to use one of the methods I showed.

-- Homer

[felson]

I'm aware that this method doesn't work when there's more than one T out. I just like to use the simplest method to solve a problem. If you want to teach the more general method, then I agree with the approaches you showed.

[Homer]

I just like to use the simplest method to solve a problem.

You're no fun. /images/graemlins/grin.gif