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-   -   Simple question for the flop (http://archives2.twoplustwo.com/showthread.php?t=314815)

Shandrax 08-14-2005 02:40 AM

Simple question for the flop
 
Let's assume I want to know the chances for a specific card (ace of spades) to hit the flop. I have 2 versions, but I don't know which is correct.

A. 1/50 + 1/49 + 1/48 = 6.124%

or

B. 1 - 49/50 x 48/49 x 47/48 = 6%

Which is the one?

Luzion 08-14-2005 03:20 AM

Re: Simple question for the flop
 
B is correct

Shandrax 08-14-2005 03:25 AM

Re: Simple question for the flop
 
Thanks!

pzhon 08-14-2005 03:28 AM

Re: Simple question for the flop
 
B is right if you assume that you have two cards that are not the ace of spades. It's overly complicated in this case, but that technique is often useful.

A is wrong. The probability that the first card on the flop is the ace of spades is indeed 1/50. The probability that the second card on the flop is the ace of spades is also 1/50, not 1/49. If you are given the extra information that the first card is not the ace of spades, then the conditional probability (that is, using the information) is 1/49. You need to add up the actual probabilities, not the conditional probabilities from different contexts.

The simplest correction to A is

A. 1/50 + 1/50 + 1/50 = 6%.

Shandrax 08-14-2005 03:37 AM

Re: Simple question for the flop
 
[ QUOTE ]
The simplest correction to A is

A. 1/50 + 1/50 + 1/50 = 6%.

[/ QUOTE ]

Now that is a surprising solution.

pzhon 08-14-2005 12:22 PM

Re: Simple question for the flop
 
[ QUOTE ]
[ QUOTE ]
The simplest correction to A is

A. 1/50 + 1/50 + 1/50 = 6%.

[/ QUOTE ]

Now that is a surprising solution.

[/ QUOTE ]
Imagine you don't stop after the flop, but you deal out all 50 cards not in your hand. Each position is equally likely. The probability that the A[img]/images/graemlins/spade.gif[/img] is in the first 3 cards is 3/50.

BruceZ 08-14-2005 12:35 PM

Re: Simple question for the flop
 
[ QUOTE ]
[ QUOTE ]
The simplest correction to A is

A. 1/50 + 1/50 + 1/50 = 6%.

[/ QUOTE ]

Now that is a surprising solution.

[/ QUOTE ]

That works because having the As on any of the 3 cards are mutually exclusive (it can only occur on 1 card).


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