Poker Hand
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Which set contains proportionately more flushes than the set of all possible poker hands?
(1) Hands whose first card is an ace
(2) Hands whose first card is the ace of spades
(3) Hands with at least one ace
(4) Hands with the ace of spades
Answer
An arbitrary hand can have two aces but a flush hand can't. The average number of aces that appear in flush hands is the same as the average number of aces in arbitrary hands, but the aces are spread out more evenly for the flush hands, so set #3 contains a higher fraction of flushes.
Aces of spades, on the other hand, are spread out the same way over possible hands as they are over flush hands, since there is only one of them in the deck. Whether or not a hand is flush is based solely on a comparison between different cards in the hand, so looking at just one card is necessarily uninformative. So the other sets contain the same fraction of flushes as the set of all possible hands.
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Comments
dewtell
Oct 24, 2002
 Nice one, though the explanation is a bit
hard to follow. You can get a little intuition
from the inclusionexclusion principle. If there
were only two suits in the deck (say spades and hearts),
then #(hands with As OR Ah) = #(hands with As) + #(hands with Ah) 
#(hands with As AND Ah). Both of the first two terms have the normal
proportion of flushes, and the last term being subtracted contains
no flushes, so the proportion of flushes in the whole expression must be
higher than normal. You can extend this to decks with 4 suits, but the computation
gets more involved.

adamant545
Oct 24, 2002
 HUH
I DONT GET IT 
adamant545
Oct 24, 2002
 CONFUSEDDED 
Quax
Nov 07, 2002
 How about this:
The odds of getting a flush are the same no matter what your first card is  the other four have to match it.
This applies for (1) Hands whose first card is an ace, (2) Hands whose first card is the ace of spades. Since the order of the cards doesn't matter, this also applies to (4) Hands with the ace of spades. As for (3) Hands with at least one ace, it's essentially the same as (1), since you'd already have weeded out any hands that have two aces in them.

dewtell
Nov 07, 2002
 Quax's comment is true for #1, #2, and #4,
which are all the same. But #3 is different
from #1. To see this, try looking at the number
of twocard hands that can be drawn from a
sixcard deck consisting of the A, K, and Q
of hearts and spades. There are 15 possible
hands if order is not important, of which
6 are flushes (2/5ths). If order is important
(i.e., if we distinguish between As Ks and Ks As),
then there are 30 possible hands of which 12 are
flushes. If you start with the As as your
first card, there are 5 possible hands of which
two are flushes. If you just know you have
the As in your hand, you have 10 possible
ordersensitive hands, of which 4 are flushes.
All these give you the same answer (40%) for the
probability of a flush. But if you look
at all the hands that contain an ace, there
are 9 orderindependent hands, of which 4
are flushes (44%). The reason there are only
9 hands, not 10 (5+5) is that AsAh is counted
both among the 5 hands that contain As and the
5 hands that contain Ah. Now the 5 hands containing
As and the 5 hands that contain Ah each have
40% flushes, so if we just added them together
without removing the duplicate, we would get a group of 10
hands that were also 40% flushes. But in removing
the duplicate to get the actual number of distinct
hands, we are removing a hand that we know is *not*
a flush, because it contains two aces. So the proportion
of flushes in the remaining group (the nonduplicated
hands) has to go up when we remove the duplicate.
The same idea applies (on a much larger scale) if
we are looking at 5 card hands from a 52 card deck, but the
numbers are much larger and harder to verify manually.
Does this make it clearer? If not, try making a list
of all the 2 card hands you can draw from that 6 card deck,
and see if it makes more sense.

LimnShicks2
Nov 18, 2002
 This is my second attempt and I didn't get to the part of sloving it because it was confusing 
jonnyonline
Jan 17, 2003
 it's obviously #3
just eliminate each set, one at a time. 
Jake
Jun 08, 2004
 I still dont get it. Since order doest matter, all 4 hands have 1 or more aces. Doesnt this make all of them the same? 
cnmne
Mar 30, 2005
 I agree with Jake. Each hand is known to have at least one ace. Therefore, we know that each hand has an ace, regardless of position or suit. The odds that the other four cards are of the same suit are the same. If #3 (which contains at least one ace) actually has two aces, then there is no chance of being a flush. 
Eshootzi_scrs
Apr 25, 2007
 Could that be any more confusing?
There is no definite answer for the question given the information.
Are we playing Omaha, holdem, 7 card, or draw. Obviously in Omaha having two Aces would be good since they would not be suited, therefore you could make two seperate flushes.
The answer of #3 makes no sense because it says at least one A. If it said hands containing only one A then it would seem to be correct.
Though hands containing five of the same suit would seem to be the answer.
And in what game does order matter? Perhaps 7 stud, in which case having the first card be any suit doesnt matter but the second card suited would be more probable in making a flush. 
Gale
Jan 13, 2009
 Here are a few stats.
5,108 flushes in a deck
2,598,960 five card hands in a deck
ratio .001966
1,980 flushes with an ace in them
886,656 hands with at least one ace in them
ratio .00234
778,320 hands with exactly one ace in them
ratio .002544 
Gale
Jan 13, 2009
 Slight correction, there are 5,148 total flushes counting the 40 straight flushes.
ratio .0019808 
Gale
Mar 17, 2009
 Don't let all these facts confuse you. All the hands with exactly one ace do in fact include a larger ratio of flushes than do all the hands in a deck. In fact, all the hands with at least one ace do too.
But that doesn't mean that aces attract flushes or that flushes attract aces. 
Zag24
Mar 20, 2009
 Great puzzle. Added instantly to my favorites list. 
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