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-   -   I am a genius! (http://archives2.twoplustwo.com/showthread.php?t=394826)

DavidC 12-09-2005 09:16 AM

I am a genius!
 
... but since one or two of you are even more of a genius than I am, I'm going to let you help me with this problem (my work to follow, for the sake of humour).

---

Harrington on Holdem, page 150. You're in a tournament with mostly conservative opponents. SB is shortstacked after taking a few beats, but that was an hour ago, so he's not tilting. Your stack: 2100, his 550, UTG+2: 2100. Your hand, 55, in MP3. Blinds are 50-100, no ante, button has a big stack.

PF: 2 folds, UTG+2 limps, 2 folds, you limp, two folds, SB pushes for 550, BB and UTG+2 fold. Pot 850t, 450t to you.

---

Pot odds: 1.85:1

I'm going to assume that villain has either two overcards or an overpair, to make things easier.

When villain has an overpair, you are a 4.5:1 dog. When villain has overcards, you're a 0.8:1 dog, which is good. (These values are taken from page 126.)

I've been out of school way too long... I'm trying to find out how often villain has to have overcards in order to make this breakeven.

0.8x + 4.5y = 1.88

I've been out of school for a while, so for a nice laugh, I'll show you why I'm a genius.

---

0.8x + 4.5y = 1.88, isolating y in terms of x
==> y = (1.88 - 0.8x) / 4.5
==> 0.8x + 4.5* ((1.88 - 0.8x)/4.5) = 1.88
==> 0.8x + 1.88 - 0.8x = 1.88
==> 1.88 = 1.88

Yep, I'm pretty smart. In fact, I'm SO smart, that I knew this even before writing down the equation. [img]/images/graemlins/smile.gif[/img]

Could someone please show me their "rough work" that shows how they figured out how often villain has to have overcards?

--Dave.

12-09-2005 09:22 AM

Re: I am a genius!
 
Convert the odds to percentages.

DavidC 12-09-2005 10:25 AM

Re: I am a genius!
 
[ QUOTE ]
Convert the odds to percentages.

[/ QUOTE ]

[img]/images/graemlins/smile.gif[/img] I'm pretty sure that I'm still going to get 100=100.

edit:

-----------

55x+18y=100
==> y = (100-55x)/18
==> 100=100 [img]/images/graemlins/smile.gif[/img]

edit: I'm not sure about my equation, though... should I have been going for 100=100, 50=50, or 35=35?

Re-edit: 35=35.

12-09-2005 11:31 AM

Re: I am a genius!
 
To solve for two variables you need two different equations. The different sides of the equal sign will always be equal so you want to end up with x = ? and y = ?.

AdamL 12-09-2005 11:52 AM

Re: I am a genius!
 
I like pizza. Food goes in here.

MexKrax 12-09-2005 12:08 PM

Re: I am a genius!
 
Your opponent either has over cards or he does not, therefore x + y = 1.

DavidC 12-09-2005 12:36 PM

Re: I am a genius!
 
[ QUOTE ]
To solve for two variables you need two different equations. The different sides of the equal sign will always be equal so you want to end up with x = ? and y = ?.

[/ QUOTE ]

Cool.

x+y=100
x = 100-y

55x+18y = 33
55(100-y)+18y = 33
5500-55y+18y=33
5500-37y=33
-37y=-5467
37y=5467
y=5467/37
y=147%? crap! [img]/images/graemlins/smile.gif[/img]

I'm not having an easy time with this.

Edit: It's entirely possible that the answer is 14.7, since I did x+y=100... hmm..

DavidC 12-09-2005 12:44 PM

Re: I am a genius!
 
[ QUOTE ]
[ QUOTE ]
To solve for two variables you need two different equations. The different sides of the equal sign will always be equal so you want to end up with x = ? and y = ?.

[/ QUOTE ]

Cool...
y=147%? crap! [img]/images/graemlins/smile.gif[/img]


[/ QUOTE ]

x+y=1
x=1-y

0.55x+0.18y=0.35
0.55(1-y) + 0.18y = 0.35
0.55 - 0.55y + 0.18y = 0.35
0.2 = 0.37y
y = 0.2/0.37 = 54%

Therefore if your opponent has an overpair less than 54% of the time, you're making money on this call.

Edit: I have no idea why this worked and my first x+y=100 didn't work. [img]/images/graemlins/frown.gif[/img] However, I think this is the right answer.

Thanks guys. (If I'm wrong, please let me know.)

re-edit: A quick check:
0.55x+0.18y=0.35
0.55(0.46)+0.18(0.54)=0.35?
Indeed it does... sweet.


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