Two Plus Two Older Archives  

Go Back   Two Plus Two Older Archives > General Gambling > Probability
FAQ Community Calendar Today's Posts Search

Reply
 
Thread Tools Display Modes
  #1  
Old 04-28-2005, 03:50 AM
westside_eh westside_eh is offline
Junior Member
 
Join Date: Apr 2005
Posts: 10
Default Calculating odds on the flop when 2 cards are needed

First post. Found out about the forum from a friend, some pretty good information here.

Anyway, I know the trick to calculating odds where after the flop you multiply your outs by 4, and on the turn you multiply them by 2.

Are there any tricks to calulating the odds after the flop when 2 cards are needed to complete the flush/straight?
Reply With Quote
  #2  
Old 04-28-2005, 08:24 AM
olavfo olavfo is offline
Senior Member
 
Join Date: Jan 2005
Posts: 303
Default Re: Calculating odds on the flop when 2 cards are needed

Welcome to the forums.

Let's say you have a backdoor flush draw with two spades on your hand and one spade on board. You need spade-spade on turn and river.

Before the turn there 47 unknow cards and 10 spades. If a spade comes on the turn, there are now 46 unknown cards and 9 spades.

The chance of hitting spade-spade is

P = (10/47)*(9/46) = 90/2162 = 0.0416 = 4.16%

In odds form:

(1/P -1) : 1 = (1/0.0416 - 1) : 1 = 23 : 1

So the math is simple. Calculate the chance of hitting on the turn and multiply it with the chance of hitting again on the river. (Remember to adjust the number of outs and unknown cards on the river. You have one less out, since the turn card hit, and you also have one less unknown card.)

olavfo
Reply With Quote
  #3  
Old 04-28-2005, 12:01 PM
binions binions is offline
Junior Member
 
Join Date: Jan 2004
Posts: 4
Default Re: Calculating odds on the flop when 2 cards are needed

[ QUOTE ]
Let's say you have a backdoor flush draw with two spades on your hand and one spade on board. You need spade-spade on turn and river.

Before the turn there 47 unknow cards and 10 spades. If a spade comes on the turn, there are now 46 unknown cards and 9 spades.

The chance of hitting spade-spade is

P = (10/47)*(9/46) = 90/2162 = 0.0416 = 4.16%

In odds form:

(1/P -1) : 1 = (1/0.0416 - 1) : 1 = 23 : 1

[/ QUOTE ]

Correct - backdoor flush draws are always 23:1. Backdoor straight draws are 21.5:1 when you start with 3 in a row open-ended.

Bottom line, each of these should count as 1 to 1.5 outs on the flop. After all, if on the flop, you needed 1 specific card to win (Ace of Spades let's say), you are 23:1 to hit the Ace of Spades by the river. So, backdoor draws are like 1 to 1.5 outs on the flop.

If you miss the turn, the backdoor out disappears. If you catch a suit you need on the turn, now you have 9 outs to the flush. If you catch a straight card you need, you either have 4 or 8 outs to the straight.
Reply With Quote
Reply


Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -4. The time now is 12:35 PM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2024, vBulletin Solutions Inc.