#1
|
|||
|
|||
Up date of what I think I learned regarding pot odds and breaking even
I looked in Phil Gordon's book and I came up with this:
Pot odds = pot/what I'm asked to call. Example pot 7 a bet of 2 so i would get 9/2 or 4.5 to 1 on my money. To break even I am told = 1/pot odds+1 or in this case 1/4.5+1 or 1/5.5 18% I need one win for every 5.5 loses. 1 win 11 5.5 loses 11. I was trying to say this in my posts before on pot odds and stuff (or I meant to say it) just got some things mixed up. I hope this one is correct |
#2
|
|||
|
|||
Re: Up date of what I think I learned regarding pot odds and breaking even
I`m just gonna offer a friendly word of advice. what is it that you want to achieve, are you really going to sit there and crunch numbers constantly whilst playin? Are you playing at such high levels where this is even important?
Not that i`m saying it isn`t but generally all you need to know are your odds of completeing your hand against the odds you are getting from the pot to call, alright the i need one win for every five losses ...........teah great but do you really care about that. Seriusly you just seem worried about posting somethin now mate, theres plenty of people on here posting complex mathematical solutions to the wierd and wonderful world of odds, but just concentrate on playing cards and enjoying yourself... but seriously though if you haven`t seen this http://www.pr0crast.com/2+2.NL.Anthology.v1.htm Then its definitely worth a look |
#3
|
|||
|
|||
Re: Up date of what I think I learned regarding pot odds and breaking even
[ QUOTE ]
To break even I am told = 1/pot odds+1 or in this case 1/4.5+1 or 1/5.5 18% I need one win for every 5.5 loses. 1 win 11 5.5 loses 11. I was trying to say this in my posts before on pot odds and stuff (or I meant to say it) just got some things mixed up. I hope this one is correct [/ QUOTE ] this part is wrong. it is 1 win 11 for every 5.5 losses of 2 .. which makes it an EV of zero. It costs you 2 to call not 11. SO if you have better than a 1 in 5.5 chance of winning a call is profitable. Example - on the turn you know you have 12 outs. You have a 12/46 chance in hitting your card, so a call is profitable. 12 times you win $11 = +$132 34 times you lose $2 = -$68 so in 46 times you win +$64 $64/48 = $1.33 so you win $1.33 every time you make this call |
#4
|
|||
|
|||
Re: Up date of what I think I learned regarding pot odds and breaking even
My point is 1 to 5.5 is a break even ration, if in the example above I win 1 time and loses 5.5 times I break even (1x11=11 5.5x2=11). So true any draw that offers better then 1 to 5.5 would be a profitable call, or @18% and better.
|
#5
|
|||
|
|||
Re: Up date of what I think I learned regarding pot odds and breaking even
If nothing else I admire your persistance.
Lets look at 4:1 odds to see if we can determine where we break even. First, what this means is if you put in 1 you are getting 5 back in return (you'll net 4, but you get back 5 which includes the 1 you put in.). Lets just look at one event where you win. You put in 1 and get back 5. Simple, right. Lets expand this to two events, win one lose one. We already know when you win you pay 1 and get 5 back. In the second event you pay 1 and get nothing. Now lets combine the two events: You paid 2 and got back 5. Three events: win one, lose two. This results in paying 3 and getting back 5. So far in all these event you are ahead as the payout of 5 has exceeded the amounts you put in. What about Four events: Win one, lose 3. We get 5 and we put in 4. Still ahead Five events : win one, lose four. Put in 5 and get 5 back. Let me repeat that - Put in 5 and get 5. Looks like we've reached out break even point. It's not a coincidence. The break even point IS EQUAL to the odds. 4:1 pots odds means lose 4 times:win once. This is the break even point. What if you want to convert our 4:1 odds to a percent. Or to rephrase the question; How often do I need to win if the pot is offering me 4:1 odds. The answer is NOT 1/4 for 25%. 1/4 means I won once out of four events. Four is a specific set of outcomes, the bad ones, not the total. For percents we wnat the total outcomes. (See More info below) As we've discovered above it takes a total of 5 events, of which winning one, gets us to a break even point. Winning one out of 5 is 20%. Remember odds are the two outcomes of an event expressed againt each other. Lets use a simple coin flip. Heads or tails -whats the odds? 1:1. It's either heads or tails. What about percent. Also refered to as chance. What's the chance of me flipping a coin and getting heads? Odds are 1:1. Can we divide 1/1? Well that gives us 1? Is that 100% then? Obviously not. We can no more divide 1/1 now, than we could divide 4/1 earlier. Percent is the relationship of a specific outcome occuring to the total outcomes possible in an event. Flipping the coin is the event. It has two possible outcomes - heads or tails. If we compare the specific outcome of heads to the total outcomes we get 1 to 2, or 50% Converting 4:1 odds to percent we need to apply the same principles we used when flipping the coin. 4 is the number of bad outcomes and 1 is the number of good outcomes (review my odds definition two paragraphs above if this doesn't make sense). Bad outcomes plus good outcomes equal total outcomes. So 4+1 =5. We get 5 total outcomes of which 1 is good. 1/5=20%. To use Phils math equation I would get 1/(4+1) or 1/5. The +1 is just getting the denominator to equal total outcomes, rather than just one specific set of outcomes. Sorry for any grammatical errors. Assuming anyone reads it all and notices of course. |
#6
|
|||
|
|||
Re: Up date of what I think I learned regarding pot odds and breaking even
I appreciate you appreciate my percistance. But If you would. Use the 7 pot 2 bet 2 to call example for continuity, and then I'll read your post to see how you show me wrong.
|
#7
|
|||
|
|||
Re: Up date of what I think I learned regarding pot odds and breaking even
On 4.5 :1 odds break even is 4.5:1. Meaning you lose 4.5 times for every 1 win. How many TOTAL events is this? 4.5 losses + 1 win = 5.5 total events.
Pot is 9 you put in 2 and win. How much does the dealer shove your way? 11. He shoves 11 chips your way. So how many times must we put in 2 before we equal 11. 5.5 TOTAL times. How many must we then lose to hit our break even. 5.5 total minus the 1 win. or 4.5 times. We lose 4.5, win 1. 4.5:1. If you want to look at this from an amount won perspective then in the above example how much do you win if the pot is 9 and you put in 2 and win? 9. How many times must you put in 2 before losing 9 ? 4.5 times. Lose 4.5 times win 1 for break even. 4.5:1. How often do you win with odds of 4.5:1. In both scenarios above you lost 4.5 times and won once. So out of a total of 5.5 tries you acheived 1 win. or 18%. 4.5:1 odds is equal to an 18% chance of winning. As I've previously explained we can also expand the 4.5:1 odds so that when adding both sides together they equal 100. In this form the actual numbers also represent their corresponding percentages. Since we said 4.5:1 equals 18% lets multply each side by 18 and see what happens. 4.5*18=81. 1*18=18. So we get 81:18. Thanks to rounding it only equals 99.....lol. Looks like 4.5:1 is pretty close to 18% to me Bottom line. If you have 4.5:1 odds to hit your hand and the pot offers you better than 4.5:1 then you have whats known as positve expected value(+EV)as the pot is offering you more money than the opportunity to hit your hand offers |
#8
|
|||
|
|||
Re: Up date of what I think I learned regarding pot odds and breaking even
You took up a lot of room to say you need 18% chance or better to profit, which my fomular gave you in a shorter time. However 4.5:1 as your bottom line states is 22% not 18%
|
#9
|
|||
|
|||
Re: Up date of what I think I learned regarding pot odds and breaking
The lot of room is taken up with helpful explanations. I'd recommend you re-read his first big post because you are still converting odds to percentages incorrecctly.
4.5:1 is not the same as 1/4.5 = 22% It is winning 1 time, losing 4.5. So you won 1 out of 5.5 So 1/5.5 = 18% EDIT: Also glad you stuck around. It can be confusing to begin with, but if you stay around and read and absorb what people here say you can learn a lot and almost definately become a winning player. Good luck [img]/images/graemlins/smile.gif[/img] |
#10
|
|||
|
|||
Re: Up date of what I think I learned regarding pot odds and breaking even
[ QUOTE ]
I looked in Phil Gordon's book and I came up with this: Pot odds = pot/what I'm asked to call. Example pot 7 a bet of 2 so i would get 9/2 or 4.5 to 1 on my money. To break even I am told = 1/pot odds+1 or in this case 1/4.5+1 or 1/5.5 18% I need one win for every 5.5 loses. 1 win 11 5.5 loses 11. [/ QUOTE ] Is this not what I said at the beginning? |
|
|