PDA

View Full Version : Simple question for the flop

Shandrax
08-14-2005, 02:40 AM
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
B is correct

Shandrax
08-14-2005, 03:25 AM
Thanks!

pzhon
08-14-2005, 03:28 AM
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
[ 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
[ 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/images/graemlins/spade.gif is in the first 3 cards is 3/50.

BruceZ
08-14-2005, 12:35 PM
[ 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).