Re: Poker Problem - 10/08/02

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Posted by Soroban on October 15, 2002 at 13:11:08:

In Reply to: Re: Poker Problem - 10/08/02 posted by Paul on October 15, 2002 at 00:30:06:

: Posted by Paul on October 08, 2002 at 16:05:00:

: A poker hand 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?
: ------------------------------------------------------

: could someone tell me if im on the right track for the 2nd part of the prob

: there would be 4 cases where you'd get 4 of a kind on next draw
: 1) Keep the highest card from 1st draw, then get 4 cards of same kind on next draw (eg. card you kept was a Queen, then u drew 4 aces)
: 2) Keep the highest card, then get 3 cards same as one you kept and 1 other card
: 3) Keep highest pair, then get 2 cards same as pair you kept and 1 other card
: 4) Keep triple, get remaining card same as triple and 1 other card

These ARE the possible outcomes. Well done!

: i worked of the probability of the 4 cases
: so would the probability of getting 4 of a kind using the stratgegy
: be the sum of the 4 probabilities above?

Yes, we would add them...

: or would i have consider that i didnt get 4 of a kind on the 1st draw?
: if so, how would i incorporate that into the answer?

But this requires a re-reading of the problem itself. If it said that
it is GIVEN that we didn't get 4-kind on the 1st draw, we're done.
But if it said something like, "What is the probability that we don't
get a 4-kind on the first try, and succeed in the second try?", then
we'd multiply by the probability of failing on the first try.

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