Basic Hold'em Question

amtannn · #1 02-02-2003, 12:03 PM

Hi, all. I'm a recreational player looking to have some fun and hopefully improve so I started playing no limit tournaments. I'm sure that this is pretty basic, and probably written in a book somewhere, but yesterday I was playing in a no limit tournament and I made a play I'm not sure was correct.

I came in for $50 on the button with a Q10. Small blind calls and Big blind lets it run. Flop comes KJ7. Small blind checks, big blind checks, middle limper calls, I bet $200. Small blind calls, big blind re-raises all in for $600. Middle limper folds. I've got about $1,500 in chips and I call, small blind calls as well. Turn comes 4. Small blind checks, and I bet $200. Small blind folds. River comes 2. Big blind turns over his cards and has a K7. I lose to two pair.

My question is if I thought he had two pair or trips, which was what I thought he had, was I dumb to call his all in bet? Is an open ended straight worth calling against a two pair? I know that I have 8 outs, and can calculate whether it makes sense given the size of the pot. But how do you calculate the fact that he has 4 cards that can beat my straight? Where can I learn this kind of basic info?

ZManODS · #2 02-02-2003, 01:06 PM

What you want is expected value of you hand. With 8 outs you have about 31.5% chance to improve to straight about 2.17:1 underdog. With his 4 outs he has 16.5% chance to improve or about a 5:1 underdog. You will make your straight about 31.5% of the time, but 1/5th of that time you will lose to his full house making you win 25.2% and losing 74.8% when you dont improve or he makes his full house. Out of 100 times you will win (if i calculated right and neglecting the fact the SB may also may put in 400) 1150chips 25.2 time = 28980. You will lose 400chips 74.8 times = 29920. Your 28980-29920= -94. -94/100 = -9.4. So you expect to lose -9.4 chips every time you make that play. Dont call.

amtannn · #3 02-02-2003, 01:54 PM

Thanks for your help. I figured it was a dumb bet, but wasnt sure how to calculate it. But I'm curious about your math. I'll make a straight 31.5% of the time, but how do you calculate the fact that some of the time I make the straight he will make the full house? You simply multiplied my 31.5% chance by 20%? But why 20%? I thought you said that he only had a 16.5% chance of getting a full house? Shouldnt it be multiplied by 16.5%?

But I have one more problem. Arent the odds of him making a full house & me making a straight even lower than 16.5% since when I make a straight thats one less card available for him to make a full house?

Bozeman · #4 02-02-2003, 03:33 PM

Actually, the times you make your straight, he has to hit one of his 4 outs in only one card (since the str8 making card cannot fill him up). If he definitely has two pair, you have 8 outs out of 45 cards, so about 32.7% chance of making your str8. Thus you have about 32.7%*40/44=29.8% chance of winning. Counting all the boards against k7 if no flushes are possible gives 30.1% because of the rare runner-runner trips. Even against 77 or KK you will win 25.9% of the time.

In this case there is 150+200+200+200+400=1150 in the pot and you have to call 400? Then you are getting 2.875:1 on your call, and you have 2.32:1 odds against middle two pair and 2.86:1 against a set, so it looks like you have a reasonable call.

However, you need to consider whether the small blind is taking any of your outs.

BTW, questions like this should be posed in the tournament forum,

Craig

ZManODS · #5 02-02-2003, 04:26 PM

Then you are getting 2.875:1 on your call, and you have 2.32:1 odds against middle two pair and 2.86:1 against a set, so it looks like you have a reasonable call.

You cant go by pot odds because sometimes he does make his straight he will still lose to a full house.

ZManODS · #6 02-02-2003, 04:43 PM

I messed something up before but ill try again.

Probabilty with 8 outs: (1 - (39/47)*(38/46)) = 31.5%
Probabilty with 4 outs: (1 - (43/47)*(42/46)) = 16.5%

Out of 100 times you will make youre straight 31.5 times and lose when you dont make it 68.5 times. You will win 31.5x1150=36225 and lose 68.5x400=27400. But the 16.5% time he makes his full house you lose even if you make youre straight. So 16.5% of the 31.5% you do make your straight you still lose. So out of the 31.5 times you make your still lose 5.2 times. So you real outcome is (31.5-5.2) 26.3 wins and (100-26.3) 73.7 loses.
Win: 26.3x1150=30245
Lose: 73.7x400=29840
30245-29840= +405 so you can expect to win (expected value) 405 chips each time you make that play. If you can include the extra 400 that the SB would put in also, your EV would be even higher. So it would be smart to call knowing that he had 2 pair.

amtannn · #7 02-02-2003, 06:03 PM

1) Why do you multiply 16.5% by 31.5%? I realize you are trying to adjust for the fact that my straight will lose to his full house, but is there a mathematical rule that says you simply multiply the two numbers together and that results in your odds. So if Y is ahead in a hand but X has a 50% chance of getting trips (and moving ahead of Y), but Y has a 20% of beating trips, does that mean X has a ((.5) - (.5 x .2)) 40% chance of winning?

2) Should the fact that this is a tournament play into my decision? Meaning, do you always play positive EV hands even if they are only marginally so and a loss will end your tournament?

ZManODS · #8 02-02-2003, 06:39 PM

So if Y is ahead in a hand but X has a 50% chance of getting trips (and moving ahead of Y), but Y has a 20% of beating trips, does that mean X has a ((.5) - (.5 x .2)) 40% chance of winning?

Yes im pretty sure thats the way it goes. Think about it, X will get trips 50% of the time. Now Y will get better than trips 20% of the time. So (.5 * .2) 10% of the time when X does get trips he will still lose. You simply neglect the of times when Y improves because X will lose anyway.

And Yes, with tournament play you shouldnt make a bet with little EV. I forgot all about the fact it was a tournament.

Someone please correct me if im wrong... dont think i am though.

Bozeman · #9 02-02-2003, 07:50 PM

ZManODS,

If you read the posts, you'll see that the suspect 2pair is all in, and that I account for the times that a str8 loses. The pot odds I gave work fine.

Craig

Bozeman · #10 02-02-2003, 08:01 PM

ZManODS,

It doesn't work this way for events that are not independent. In this case, the cards that make you a str8 (A, 9) can't make your opp. a full house. Thus if you make a str8 (on the turn for example), the 2pair must hit one of 4 outs on the river, this only happens ~4/40=10% of the time, not the 16% you use.

In the hypothetical example, it is probable that more of Y's improvements will occur when X doesn't improve, so X will win somewhere between 40and50% of the time. Simply looking at X's outs overestimates his win percentage, but your method underestimates it usually.

For seven card stud, the boards are almost independent, so your method is a reasonable approximation, but holdem is quite different.

Craig