This is a discussion on Rule of 4 and 2/Pot odds? within the online poker forums, in the Learning Poker section; Hello guys. New to this site and have a few questions: I've been playing poker casually over the past 12 years at casinos and never really 


#1




Rule of 4 and 2/Pot odds?
Hello guys. New to this site and have a few questions:
I've been playing poker casually over the past 12 years at casinos and never really took into account calculating exact pot odds during the hand. I just kind of estimated the size of the pot and and the size of the bet....didn't really use the rule of 4 and 2. Was this a huge misstep?!! Anyways, over the past few months I've been looking at the rule of 4 and 2 a bit more. How accurate is this method? I mean, the way I've been calculating my odds in the past few months goes something like this  say for example you have an open ended straight draw on the flop (8 outs) and hitting the straight would give you the nuts; therefore your chance of winning is 8/47 or approx. 1/6 (17%) meaning out of 6 hands in the same situation, you would win once and lose five. As such, if the pot is six times (break ever) or greater (profit) the size of the bet to be called, I'll make the call. Now here's where I'm a bit confused. After the flop, based on the rule of 4 and 2, the chance to hit the straight is 32% (8*4). Why do your odds suddenly double to ~1/3 (32%)? There's only 8 cards out of 47 on the turn (8/47 = 17%) that will help you make your straight. If the turn misses, there's still only 8 cards that will help you (8/46) which is still approx. 17%. Going by the rule of 4 and 2, saying you have 32% chance kind of seems like saying there are 16 cards that can help you out? Let's compare this to the lotto. Say the odds to hit the lotto is 1:100M. If you buy $2, now you have two chances to win (same as the turn and river?), but the odds for each $1 line is still 1:100M. Is this the same scenario or am I approaching this the wrong way. Thoughts and comments please. PS  Sorry for the long post. 
#2




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It's 32% after the flop because you have 2 chances to hit your straight. 1 on the turn, and 1 one the river. This is what I do in my head to see if I should call. 8 X 4= 32 ~~33% Then I see how much it is to call and how much is in the pot. If I am getting more than 2:1 then usually I call. 2:1 = 2/3=.666 Complement of .666 is .333 
#3




It is much more useful to use the rule of 2 in MOST places. The reason is that we usually care about our odds for only the next street, since it isn't usually a good assumption that all betting stops after the turn, for example.
Times when the rule of 4 is useful:  Someone goes all in on the flop and you must decide whether or not to call with a draw  You have a read that suggests that action will slow down on the turn, giving a "free" river card  Any time you're pretty sure you'll get to see streets 4 and 5 due to putting in the current bet As you can imagine, these scenarios don't actually come up that often. The only time I ever find myself (usually) using the rule of 4 is in tournament play, when someone shoves the flop, I have a big stack (won't be crippled by calling), and I must decide whether a call is profitable. Let's say I'm headsup against an opponent with a much smaller stack. Let's say I have the nut flush draw, and I will win if I hit my draw. My opponent has just shoved his 10,000 chip stack into a pot of 20,000 chips, making the pot 30,000 total. Now I must call 10,000 chips into a pot of 30,000, so my pot odds are 3:1. We have 9 outs to our nut flush draw. Since our opponent is already all in, we know that there will be no further betting, and if we do call, we will get to see TWO streets. So we use the rule of 4: our odds to hit is about 9 outs x 4 = 36% So we are getting a little better than 2:1 on a call. Our pot odds are bigger than 2:1 (3:1 in fact), so assuming this call wouldn't cripple our stack (for tournament play), then this is a great spot. Let's now consider the SAME situation, except that now our opponent is NOT all in. He has bet 10,000 into a pot of 20,000 again, but he has 100,000 left to work with. Now there is a chance that after we call the flop bet, he will ALSO bet the turn. So we are only guaranteed to see one street with a call. Now we use the rule of 2. With 9 outs, we see that we only have about a 9 outs x 2 = 18% chance of hitting on the turn. This gives us odds of about 4:1 on a call. Since our pot odds are 3:1, this is no longer a profitable call. Hope this makes sense. If not, you can ask a clarifying question, or feel free to send me a PM. 
#4




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Say for example you have 2 chances to roll a six on a die. The first chance (turn), the probability is 1/6 (~17%). If that misses, the probability is still 1/6 on the second roll (river). Now put this scenario into poker. Say that rolling a six represents the 8 outs you have for an open ended straight. On your first roll (turn), why would your chances to roll that six now become 1/3 or ~33%? Also, am I correct in saying that when the odds are 8/47 = ~1/6, that there should be at least 6 times the bet amount in the pot for it to be a correct call? 
#5




re: Poker & Rule of 4 and 2/Pot odds?
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8 divided by 48 is equal to 1/6, so yes, 8/47 is about 1/6. But this means your odds of hitting are 1:5 because you have 1/6 chance of hitting and a 5/6 chance of not hitting. So you only need odds of 5:1 or better to break even and profit respectively. Hope that clears things up a bit? 
#7




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I hope you guys understand what I'm trying to say here, but maybe I'm just thinking too much into it? 
#8




Yeah, you're overthinking this. For the rule of 4 and 2, the "4" comes in on your odds to make your hand BY THE RIVER after you've just seen the flop. The "2" part is the approximate odds to make your hand ON THE NEXT STREET.
What Scourrge is saying is deadon accurate, though. People quickly, and often incorrectly, apply the rule incorrectly thinking that due to the "4" part they have better odds than they actually do.....They forget there is likely another bet coming after the turn. Unless you have some reson to believe that you'll be getting a free card somewhere along the line, it's usually more correct to think of your outs after the flop as getting the "2" part twice. 
#9




You roll two dice:
Odds of getting a six on first roll: 1/6 Odds of getting a six on second roll: 1/6 Odds of getting a six on both rolls: 1/6 x 1/6 = 1/36 Odds of getting a six on JUST ONE roll: {(1/6 chance of getting a six on the first roll) x (5/6 (any nonsix on second roll))} + {(1/6 chance of getting a six on second roll) x (5/6 (any nonsix on second roll))} = 5/36 + 5/36 = 10/36 ≈ 27.78% (Make sure that that makes sense!) If we want to include the possibility of getting a six on both rolls, we just add 1/36 to the previous calculation. Odds of a six on AT LEAST ONE roll: (Odds of getting a six on JUST ONE roll) + (Odds of getting a six on BOTH rolls) = 10/36 + 1/36 = 11/36 ≈ 30.56% Let's compare this to the odds of getting a six when we only get one roll: 1/6 ≈ 16.67% 11/36 ≈ 30.56% Note that 2 times the odds of getting a six with one roll is 1/3 (equal to 12/36), so ≈ 33.33% So although it is not perfect, multiplying by 2 is a good approximation when comparing odds with just one roll versus odds when you have 2 rolls. Does this make sense? The odds on each individual roll do not change  you are correct about that. But when we have two chances to get the same outcome, it is more likely to occur at least once. Hope this helps. If there is still confusion, I'd be happy to move to PM and possibly direct you to some sites that might make things a bit easier to understand as well. 
#10




re: Poker & Rule of 4 and 2/Pot odds?
the odds of hitting heads if you flip a coin twice is %100, but nothing with probability can reach 100%(relative frequency estimate of a probability) so you would need >1:1 odds to be profitable. Same applies with 4&2. If you give yourself odds of 1/1 to hit heads on two flips and brick your first flip, the odds just went down to 1/2 on your last flip.

#11




Actually, since you brought it up, here's a simpler example with the coin.
Odds of getting heads on first flip: 1/2 Odds of getting heads on second flip: 1/2 Odds of getting heads on BOTH flips: 1/4 Odds of getting heads on JUST ONE flip: Odds of getting heads ONLY on first flip + Odds of getting heads ONLY on second flip = {(1/2 chance of heads on first flip)x(1/2 chance NOT getting heads on second flip)} + {(1/2 chance of getting heads on second flip)x(1/2 chance of NOT getting heads on first flip)} = 1/4 + 1/4 = 2/4 = 1/2 Odds of getting heads on AT LEAST ONE flip: (Odds of getting heads on JUST ONE flip) + (Odds of getting heads on BOTH flips) = 2/4 + 1/4 = 3/4 In this example, the factor of 2 that comes in play is not as useful, because there are fewer outcomes possible. Since in a deck of cards there are many possible outcomes, we approach a perfect factor of 2. (This part is less important, so long as you understand the previous things.) One last thing I'll say: Someone tells you "I will give you $1,000 if I flip this coin once and it comes up heads. But if it comes up tails, you will have to pay me $10,000." Pretty bad deal. Now (assuming a fair coin), let's say someone else offers this instead: "I am going to flip this coin 100 times, and if it comes up heads at least once, I will give you $1,000. If after 100 flips, it has not come up heads at least once, you will have to pay me $10,000." Now this is a pretty damn good deal, isn't it? Because you only need the coin to land on heads ONCE. Although this is an extreme example, it illustrates how having more chances IS very important when all you need is a single outcome (such as a heart falling on the river to complete a nut flush). 
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