1)  What is the probability that 3 flushes of hearts will be dealt to any
of 8 players at a table within 33 hands (that is, anyone can show a
flush on the table, but count only one flush per hand)?

2)  What is the probability that the same player at a table of 8
players will be dealt 2 flushes of hearts within 30 hands?

Rules:

52 cards in a deck.
8 players at a table.
Each player is dealt two cards.
Then, 5 "community" cards are laid on the table.
The best 5 card hand wins (choosing the best 3 out of 5 cards laid on the table.)

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Clarification of Question by vaulting_dmg-ga on 16 Dec 2005 11:44 PST

The issue is not just any flush, but a flush of the same suite three
times in a very short period of time.

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Request for Question Clarification by byrd-ga on 19 Dec 2005 10:46 PST

Hi valuting_dmg-ga,

Could you please clarify your question on a couple of points? They are:

1. Under the rules portion of your question, you state that the hand
must be made from your two hole cards plus any three from the
community cards. Holdem rules say that you make the best hand by using
the five best of the total seven cards, which may include two, one, or
none of your hole cards. Would you like the calculation based on the
restriction that you have stated, or actual holdem rules?
 
2. In your clarification, you state that the flushes must be of the
same suit.  Would you like the answer based on getting only heart
flushes, or flushes in any suit, as long as the flushes are all of the
same suit? (If the flushes can be of any suit, the answer will be four
times as great than if the flushes must be in hearts).

Note that you can post a clarification while the question is locked.
Thanks much, looking forward!

Best,
Byrd-ga

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Clarification of Question by vaulting_dmg-ga on 19 Dec 2005 11:54 PST

Thank you VERY much for offering to respond.  In answer to your
request for clarification:

1)  Use Hold'em rules.

2)  Not any flush, but a flush of hearts.

Thank you.

vaulting_dmg

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Clarification of Question by vaulting_dmg-ga on 19 Dec 2005 11:58 PST

Also, if you don't mind, could you show the mathematical calculation.

Thanks.

vaulting_dmg

---

Clarification of Question by vaulting_dmg-ga on 01 Jan 2006 14:52 PST

It's been 2 weeks.  Is anyone there?

The rules I have seen on Texas Hold'em allow you to use any
combination of hole cards and community cards.  You don't necessarily
have to use both hold cards in your five card hand.

http://www.thepokerforum.com/texasholdem.htm

This link says you could even use the five community cards to make
your hand and none of your hole cards.

What cards you can use to make your hand will affect the probabilities.

I made a few assumptions with my calculations:

1. The player with the flush is the last one dealt a card
2. The player was dealth a heart as his second "hole" card
3. No "burn" card was a heart
4. The 4 other needed Hearts were community cards

The calculations go as follows...

# of Hearts: 13
# of Cards in a Deck: 52

Odds of getting a Heart as 2nd hole card: 13/37 = 35.135%
(37 being cards left at this point in the hand)

Odds of 4th community card being a Heart: 12/34 = 35.29%

"      " 5th "                         ": 11/33 = 33.33%

"      " 6th "                         ": 10/31 = 32.25%

"      " 7th "                         ": 9/29 = 31.03%

To get the odds of this happening you just multiply all the odds
together and it gives you overall odds of: .413%

That covers the odds of it happening at any point in time. 

Over 33 hands...

If it happens .413 times in ever 100 hands then we divide 100/3.03 = 33
And divide .413/3.03 = .136%

Now odds for it happening twice in 30 hands...
100/3.33 = 30
.413/3.33 = .1239%
.1239 x .1239 = .015%

So to summarize;
the odds of getting a flush of any specific suit in 33 hands is: .136%
the odds of getting 2 flushes of any specific suit in 30 hands is: .015%

Sorry, dougw, but that's basically a pile of... garbage. Even apart
from your expressed assumptions, which by themselves make your answer
more or less meaningless, you're also implicitly assuming that none of
the other players *ever* get dealt a heart, which is ridiculous.

I'm currently compiling an answer to this question - it's much tougher
than dougw's analysis would indicate, but solvable nonetheless. 8-) In
the meantime, here's the answer to a subtly (but significantly)
simpler question: what's the probability of a specific player getting
2 flushes of hearts within 30 hands?

The good thing here is that we don't need to worry about whether the
cards are community ones or not. So all we're doing in each hand is
choosing seven cards at random without order from the deck and asking
how likely it is that at least five will be hearts. I'll use (a b) to
denote "a choose b".

- Ways to get exactly five hearts: (13 5).(39 2) = 953,667
- Ways to get exactly six hearts:  (13 6).(39 1) =  66,924
- Ways to get exactly seven hearts:       (13 7) =   1,716
So we get a total of 1,022,307 ways out of (52 7) possible
combinations, for a probability of 0.76414% (to 5 sf) that the given
player will get a flush of hearts in a single hand.

Probability of not getting a flush in a single hand is 99.23586%, so
P(no flushes in 30 hands) = 0.9923586^30 = 0.79444 (5 d.p.)
P(exactly one flush) = (30 1) (0.9923586)^29 (0.0076414) = 0.18352 (5 d.p.)
So P(at most one flush) = 0.97796 and therefore P(at least two
flushes) = 0.02204 or roughly 2.2%. Note that this is something like
150 times as much as the figure dougw came up with.

Now, you might think that at this point all we have to do is multiply
by 8 to get the answer to your second question as 17.63%. This should
in fact be pretty close, but it won't be quite right because one
player getting a flush affects the probability that other players get
one in the same hand. (The reason it should still be close is that
their chances of getting a flush are pretty small anyway, so the
effect won't be large.) I'll post a fuller answer later.

OK, let's look at the first question. The fundamental thing we need to
calculate is the probability that at least one player will get a flush
of hearts in a given hand. Unlike my previous comment, we do now need
to distinguish between community cards and each player's hole cards; a
community card that is a heart will improve the chances of somebody
getting a flush much more than a player's hole card being a heart
will. We will look at the community cards first.

Obviously we need at least 3 hearts in the community cards, and we
could have as many as 5. Of course if there are 5 hearts in the
community cards then everyone gets a flush!

If there are 4 hearts in the community cards, then the probability
that nobody gets a flush is
(38 2).(36 2). ... .(24 2)    38! 31!
-------------------------- = --------
(47 2).(45 2). ... .(33 2)    47! 22!
since there are 38 non-hearts remaining (one is in the community
cards) and 47 cards remaining in total. So the probability that at
least one person gets a flush of hearts is 1 - 38!31!/(47!22!) =
0.985205 (6 dp).

If there are 3 hearts in the community cards, things get tricky. Don't
say I didn't warn you!

==========Assuming 3 hearts in the community cards===========
There are now 10 hearts and 37 non-hearts remaining. For a player to
get a flush of hearts, both their hole cards must be hearts; so up to
5 players may be dealt a flush.

P(5 flushes) = (8 5) (10 2)(8 2)(6 2)(4 2)(2 2) (37 2)(35 2)(33 2)
               ---------------------------------------------------
                  (47 2)(45 2)(43 2)(41 2)(39 2)(37 2)(35 2)(33 2)

 = (56.10! / 32) / (47! / (32.37!)) = 56.10!37!/47!
 = 1.08148 * 10^-8

For four flushes it gets a bit harder. We can start in a similar way,
but every hand with five flushes will also be counted (5 4) = 5 times,
so we have to subtract these off.

P(4 flushes) = (8 4) (10 2)(8 2)(6 2)(4 2)(39 2)(37 2)(35 2)(33 2)
               ---------------------------------------------------
                 (47 2)(45 2)(43 2)(41 2)(39 2)(37 2)(35 2)(33 2)

 - 5.(1.08148 * 10^-8)

 = 9.96318 * 10^-6

For three flushes, we have to subtract all the 4-flush hands 4 times,
and the 5-flush hands (5 3) = 10 times. In the big fraction here I'll
omit the cancelling terms at the end - you should have seen by now how
they go.

P(3 flushes) = [(8 3) (10 2)(8 2)(6 2)] / [(47 2)(45 2)(43 2)]
- 10.(1.08148 * 10^-8) - 4.(9.96318 * 10^-6)
 = 1.05526 * 10^-3

Now for two flushes, we need to subtract the 5-flush ones 10 times,
the 4-flush ones 6 times and the 3-flush ones 3 times. So we get

P(2 flushes) = (8 2) (10 2)(8 2) / [(47 2)(45 2)
 - 10(1.08148 * 10^-8) - 6(9.96318 * 10^-6) - 3(1.05526 * 10^-3)
 = 0.0297404

And finally we get 
P(1 flush) = (8 1) (10 2) / (47 2) - 5(1.08148 * 10^-8) 
 - 4(9.96318 * 10^-6) - 3(1.05526 * 10^-3) - 2(0.0297404)
 = 0.270338

So P(at least one flush) = 0.301144 (6 d.p.)
=============================================================

Now remember, all that was assuming there were three hearts in the
community cards. So the overall probability that there is at least one
flush of hearts in a given hand is as follows:

P(5 hearts in CC) * 1 + P(4 hearts in CC) * 0.985205 
  + P(3 hearts in CC) * 0.301144

= (13 5)/(52 5) * 1 + (13 4)(39 1)/(52 5) * 0.985205
  + (13 3)(39 2)/(52 5) * 0.301144

= 0.035622 (6 dp).

Now, the probability of getting at least three hands containing a
flush of hearts in 33 hands is relatively simple to work out. The
probability of not getting a flush of hearts in a given hand is
0.964378, so:

P(0 flushes) = (0.964378)^33 = 0.302109
P(1 flush) = (33 1) (0.964378)^32 (0.301144) = 0.368253
P(2 flushes) = (33 2) (0.964378)^31 (0.301144)^2 = 0.217638

for a total of 0.888000; so the chances of getting three or more hands
containing flushes of hearts in 33 hands is almost exactly 11.2%.

--------------------------------------------------------------------

Well, time for me to go home for the weekend now, so I'll leave the
second question until later. It looks a bit trickier to get the
correct answer - hopefully that doesn't put you off! At least I'll
have plenty of time to think about the best strategy for getting
there....

On another note, there is a potentially simpler way to get the above
answer, if you have a spreadsheet to do it with (and I used this to
check my answer). For the section with three hearts in the community
cards, we can directly compute the probability that nobody gets a
flush. If nobody gets a flush they must all have 0 or 1 hearts in
their hole cards (remember there are 10 hearts left, so there's enough
for everyone to get one). There are therefore 256 possibilities for
whether each of the 8 players gets 0 or one hearts; you can, with a
bit of care, set up an array to compute combinatorial values for each
element (noting that the values for each player will depend on how
many hearts were dealt to the previous players). Then you can add up
all those values and divide by (47 2)(45 2)(43 2)(41 2)(39 2)(37 2)(35
2)(33 2) to get the total probability that nobody gets a flush, and
the rest of the question goes as above.

I'm afraid that I actually made the last question harder than it
needed to be. A better approach - again, for the section where we
assume three hearts in the community cards - would go like this:

In the case that nobody gets a flush, in each hand every player must
receive either 0 or 1 hearts for his hole cards. Consider the number
of players who get a heart (from 0 to 8). We will determine the number
of combinations in each case. This will be a product of the following
factors:
 - ways to choose the players receiving a heart
 - ways to choose the heart received by each player who gets one, in turn
 - ways to choose the non-heart for each player who got a heart, in turn
 - ways to choose two non-hearts for each remaining player, in turn
Thus:
0: (8 0) (37 2) (35 2) (33 2) (31 2) (29 2) (27 2) (25 2) (23 2)

1: (8 1) (10 1)(37 1) (36 2) (34 2) (32 2) (30 2) (28 2) (26 2) (24 2)

2: (8 2) (10 1)(37 1) (9 1)(36 1) (35 2) (33 2) (31 2) (29 2) (27 2) (25 2)

and so on, down to 

8: (8 8) (10 1)(37 1) (9 1)(36 1) (8 1)(35 1) (7 1)(34 1) (6 1)(33 1)
(5 1)(32 1) (4 1)(31 1) (3 1)(30 1)

 = (8 8) (10.37) (9.36) (8.35) (7.34) (6.33) (5.32) (4.31) (3.30)

You can add these up and divide by (47 2)(45 2)(43 2)(41 2)(39 2)(37
2)(35 2)(33 2) as in my last comment, to get the same result rather
quicker.

However, if we'd done it that way, the calculations I made would have
had to be done anyway for the second question (more on this later), so
at least it wasn't wasted effort.

As I hinted last time, the first step to answering the second question
is going to be determining the probability distribution of the number
of flushes of hearts in a single hand. In other words, for one hand,
how likely is it that there are 0, 1, 2, ..., 8 flushes of hearts?
[Incidentally, for this comment, and indeed my other comments, any
time I refer to a flush I specifically mean a flush of hearts, even if
I don't say so.]

Now the answer is trivial in the cases where there are 0, 1, 2 or 5
hearts in the community cards, and we've worked them out for the case
where there are 3. So we just need the case where there are 4 hearts
in the community cards. I'll spare you the details, but for example
for 7 flushes we have three cases: 2 players get two hearts for their
hole cards, 5 get one heart and one gets none; or the respective
numbers could be 1, 6 and 1, or 0, 7 and 1. (Note that there are only
9 hearts remaining.) Each such combination (there are 29 of them) has
a probability formula that can be written out and summed. These can
then be combined with the overall probability of getting 3, 4 or 5
hearts in the community cards to begin with, as we did for the first
question. So without further ado, the overall probability distribution
for the number of heart flushes in a single hand is (all to 6 sf):

0  0.964378
1  0.0232036
2  0.00543082
3  0.00366753
4  0.00211570
5  6.19484 * 10^-4
6  8.47831 * 10^-5
7  4.66636 * 10^-6
8  4.95269 * 10^-4

You will note that the chance of getting 8 flushes is somewhat
surprisingly about 100 times greater than that of getting 7 flushes.
This is because the chance of 5 hearts in the community cards is about
22 times less than that of 4 hearts, but when it happens we always get
8 flushes, whereas if there are 4 hearts in the community cards we
only get 7 flushes roughly 1 in 2300 times.

The chances of getting the indicated number of flushes, with a
specific set of players, are the same as the above table but divided
by (8 n) where n is the number of players getting flushes:

0  0.964378
1  0.00290045
2  1.93958 * 10^-4
3  6.54916 * 10^-5
4  3.02242 * 10^-5
5  1.10622 * 10^-5
6  3.02797 * 10^-6
7  5.83295 * 10^-7
8  4.95269 * 10^-4

Now, we want to compute the probability that nobody gets two flushes
in the 30 hands. We will break this up into cases according to the
total number of flushes received by all players, which will therefore
be from 0 to 8. Each case will be broken up into sub-cases as
required, based on the distribution of flushes among hands (you'll see
what I mean by this as we go along). In each case we will have to
consider:
 - distribution of hands containing flushes among the 30 hands
 - distribution of the players involved in each among the 8 players

0 flushes: only one possible case, nobody ever gets a flush
: 0.964378^30 = 0.336838.

1 flush: has to be 29 zeros and 1 one, which I will indicate simply as
[1]. There are (30 1) ways of allocating the flush among the 30 hands.
In this case we could just use the general probability: (30
1).(0.964378)^29.(0.0232036)^1 = 0.243136. The usual process, however,
will be to use the specific probability and count the number of player
combinations. In this case that would give us
(30 1).(8 1).(0.964378)^29.(0.00290045)^1 = 0.243136.

2 flushes: two possible cases now.
[2]:   (30 1).(8 2).(0.964378)^29.(1.93958 * 10^-4)^1 = 0.0569064
[1,1]: (30 2).(8 1).(7 1).(0.964378)^28.(0.00290045)^2 = 0.0742221

3 flushes:
[3]:     (30 1).(8 3).(0.964378)^29.(6.54916 * 10^-5)^1 = 0.0384299
[2,1]:   (30 1).(29 1).(8 2).(6 1).(0.964378)^28.(1.93958 * 10^-4)^1.
(0.00290045)^1 = 0.0297802
[1,1,1]: (30 3).(8 1).(7 1).(6 1).(0.964378)^27.(0.00290045)^3 = 0.0125008

4 flushes:
[4]:       (30 1).(8 4).(0.964378)^29.(3.02242 * 10^-5)^1 = 0.0221691
[3,1]:     (30 1).(29 1).(8 3).(5 1).(0.964378)^28.(6.54916 * 10^-5)^1.
(0.00290045)^1 = 0.0167592
[2,2]:     (30 2).(8 2).(6 2).(0.964378)^28.(1.93958 * 10^-4)^2 = 0.00248932
[2,1,1]:   (30 2).(28 1).(8 2).(6 1).(5 1).(0.964378)^27.
(1.93958 * 10^-4)^1.(0.00290045)^2 = 0.00626965
[1,1,1,1]: (30 4).(8 1).(7 1).(6 1).(5 1).(0.964378)^26.(0.00290045)^4 = 0.00126891

You can continue on in the same way - let me know if you want me to
fill in the details. (There are 7 sub-cases for 5 flushes, 11 for 6
flushes, 15 for 7 flushes and 20 for 8 flushes.) If you want to work
on the rest yourself, here are the subtotals for each case, correct to
6 s.f. (note that I've been rounding to 6 sf in all figures I've
shown, but calculating with much higher precision; if you use the
rounded figures, expect minor variations in these results):
0 flushes: 0.336838
1 flush:   0.243136
2 flushes: 0.131128
3 flushes: 0.0807109
4 flushes: 0.0489562
5 flushes: 0.0208864
6 flushes: 0.00568656
7 flushes: 0.000967431
8 flushes: 0.00526515

Total probability: 0.873576.

Thus, the probability that some player does get two flushes of hearts
in the space of 30 hands is 0.126424 or 12.6%.

Thatnks for joining us for this week's episode of "Why Probability
Calculations Can Be Harder Than You Think." Be sure to tune in next
week, when we will calculate the probability of a randomly selected
Google Answers member spending far too much time answering a question
when he should be working.