# I think you are right Mr. B...

[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]

Posted by Subhotosh Khan on October 09, 2002 at 11:02:46:

In Reply to: Question, Mr K... posted by Denis Borris on October 08, 2002 at 22:16:49:

: : : A poker had of 5 cards is drawn from standard 52 card deck w/o replacement.
: : : What is the probability we get 4 cards of one kind?

: : : If do not get 4 of a kind on 1st draw, we keep the highest for which we have several of a kind (if none, keep only the highest card) then throw away the others and replace w/fresh cards. Can only do this once.
: : : What is the probability we get 4 of a kind if we follow this strategy?
: : *************************************
: : Probability of drawing fist card (any) = 52/52
: : Probability of drawing second card (same as first) = 3/51
: : Probability of drawing third card (same as first) = 2/50
: : Probability of drawing fourth card (same as first) = 3/49
: : These are "mutually exclusive" events. Find combined probability.

: The art of probability
: Is sure not my forte.
: SO in all PROBABILITY(!) a stupid question:
: wouldn't the probability of getting 4-same simply be 13 / (all combos) ?
***************************************
I see error in my ways.

I was calculating chances of 4 of a kind on consecutive draw.

Even at that the third line is wrong. It should be:

Probability of drawing fourth card (same as first) = 1/49

[that's what happens to lazy-bones trying to cut and paste]

Name:
E-Mail:

Subject: