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Q: Probability ( Answered 5 out of 5 stars,   1 Comment )
Question  
Subject: Probability
Category: Science > Math
Asked by: sheetwise-ga
List Price: $150.00
Posted: 24 Jan 2003 03:05 PST
Expires: 23 Feb 2003 03:05 PST
Question ID: 147900
Consider a deck of 52 cards with one Joker (which plays wild).  Place
all 53 symbols each of 5 reels on a slot machine, and play (the
equivalent of drawing from a deck WITH replacement, it is possible to
get 5 Ace of spades).  What is the probability of drawing the
following --

1. Five Jokers
2. Straight Flush
3. Five of a kind suited
4. Five of a kind unsuited
5. Four of a kind suited
6. Four of a kind unsuited
7. Full House
8. Flush
9. Straight

I have the numbers, but want to see the calculations.

Request for Question Clarification by mcfly-ga on 24 Jan 2003 08:16 PST
Hi Sheetwise,

I've been working on your question for a while now and have calculated
the figures you request.  To ensure that you are happy with the
outcome, I am posting the probabilities here for you to check against
your expected figures.  If they match your answers I will post a full
answer including all my calculations.

Kind regards,

mcfly

1) 1/418195493 = 2.39x10^(-9)
2) 16/7890481 = 2.03x10^(-6)
3) 833/418195493 = 1.99x10^(-6)
4) 32501/418195493 = 7.77x10(-5)
5) 110505/418195493 = 2.64x10(-4)
6) 6891060/418195493 = 0.0164
7) 35721/7890481 = 4.53x10^(-3)
8) 2151293/418195493 = 5.14x10^(-3)
9) 625/7890481 = 7.92x10^(-5)

Clarification of Question by sheetwise-ga on 27 Jan 2003 07:23 PST
It is important to  calculate the unique probability, in order
of rank 1-9.  A winning flush (#8) would need to be reduced by 
the number of flushes in #2,3,5 and 7.  A combination which could
produce 4 of a kind suited (with wild cards) may also produce
a straight flush, which has precedence.  No winning combination
should be counted twice.

Your answer for five of a kind suited (#3) is:

3) 833/418195493 = 1.99x10^(-6) 

The answer to this is clearly 52((2^5)-1) = 1612 : 53^5

There are two symbols on each reel = 2^5, less the one combination
of 5 jokers which satisfies #1. There are 52 winning symbols.

For #2 I have observed 44,661 : 53^5 straight flushes as a subset of
#'s 8 and 9. 

8)2,052,880 : 53^5
9)2,806,080 : 53^5


Gregg

Clarification of Question by sheetwise-ga on 27 Jan 2003 07:31 PST
CORRECTION:

8)2,053,680 : 53^5 
9)2,694,240 : 53^5 
 
Gregg

Clarification of Question by sheetwise-ga on 27 Jan 2003 20:43 PST
In consideration of the clarification I have made, I will raise the
value of the answer to $100.  When considering the probability of
drawinging a specific poker hand, the result would serve no use if you
did not discount those combinations which satisfy a hand which beats
the target hand.  To make these adjustments might double the work.

While I have run a combinatorial analysis, I am stumped on the (maybe
not so) subtle changes introduced by a wild card.

Gregg

Request for Question Clarification by bobby_d-ga on 28 Jan 2003 00:30 PST
I love probability, and I would like to attempt your question, but I
would just like to clarify the meaning of "suited" and "unsuited".  My
assumption is 5 of a kind "suited" means all 5 cards have the same
face value and the same suit (or, the five cards are all exactly the
same, bar any wildcards).  "Unsuited" leaves the suits irrelevant.

Is this correct?

Thanks,

bobby_d

Clarification of Question by sheetwise-ga on 28 Jan 2003 09:17 PST
booby_d,

Correct.  4 suited would be any combination of wild cards and matching
cards (value and suit).  The total number of winning 4 suited hands
would be all combinations LESS those combinations which could also be
interpreted as straight-flush or five of a kind suited. Using a
'Joker' as a wild card, the hand 10H-Joker-Joker-AH-Joker could be
interpreted as 4-AH suited, 4-10H suited, or as
Ace-High-Straight-Flush.  This hand would only be counted once, as a
straight-flush.

Gregg

Request for Question Clarification by mathtalk-ga on 06 Feb 2003 09:22 PST
Hi, sheetwise-ga:

I just wanted to confirm you are still interested in answer to your
question, showing the methods of solution for each part?  Your last
clarification said you intended to raise the list price for the
question, but it appears this has not yet been done.  Reply to this
request for clarification and I will be notified and can quickly
respond.

regards, mathtalk-ga

Clarification of Question by sheetwise-ga on 06 Feb 2003 10:29 PST
mathtalk-ga,

Yes, I would still like the answers.  I cannot modify the question
because it is being answered -- but yes, I will raise the price to
$100.  Beware, I have found this is much more complicated than it
looks.  As in the kennyh-ga comment -- where straight flushes were
calculated -- the number of card combinations which will create a
straight flush rises exponentially with each wild card in the hand
(The correct answer is 43,520).  Good luck.

Gregg

Request for Question Clarification by mathtalk-ga on 06 Feb 2003 10:55 PST
Gregg,

Oops, my bad.  I went ahead and opened an answer box.  I've unlocked it now.

regards, mathtalk-ga

Clarification of Question by sheetwise-ga on 06 Feb 2003 14:43 PST
mathtalk-ga,

I have made the changes.

Thanks,
Gregg

Clarification of Question by sheetwise-ga on 06 Feb 2003 14:50 PST
On 27 Jan 2003 07:23 PST I posted that straight flushes would result
in 44,661 winning combinations.  Later that day, 27 Jan 2003 07:31
PST, I corrected the number of flushes and straights -- but did not
correct the number of straight flushes -- which should be 43,520.  I
was reading off of a table generated with a programming error.

I mention this because I posted the number 43,520 today.

Gregg

Clarification of Question by sheetwise-ga on 06 Feb 2003 15:27 PST
I have raised the price to $150 and introduced one more change.  % of
a kind suited should precede thge straight flush.  This is consistent
with the numbers I have listed previously.  The only change would be
the subset of 5 of a kind which included four wild cards and could be
considered as a straight flush.

To completeley clarify -- I am interested in the calculations which
would determine the UNIQUE frequency of EACH of the following hands,
drawn WITH replacement, from a 53 card universe, where the Joker is
wild.

1. Five Jokers 
2. Five of a kind suited 
3. Straight Flush 
4. Five of a kind unsuited 
5. Four of a kind suited 
6. Four of a kind unsuited 
7. Full House 
8. Flush 
9. Straight 

Gregg

Request for Question Clarification by mathtalk-ga on 07 Feb 2003 22:56 PST
Hi, Gregg:

Quick comment.  Under your revised ordering I count:

1. Five Jokers (1) [agree]
2. Five of a kind suited (1612) [agree]
3. Straight flush (38720) [disagree]

Note that my last figure is exactly 4800 below the figure you
previously cited for number of straight flushes (which happens to be
the count for straight flushes with no Jokers).  I'm not sure what to
make of this discrepancy, if that's what it is.  I'd be happy to go
ahead and post details of that computation if you like.

regards, mathtalk

Clarification of Question by sheetwise-ga on 08 Feb 2003 08:49 PST
mathtalk,

It is interesting that the difference is 4800.  I have checked the
logic of the program, and reviewed the results -- which are straight
flush combinations. I would like to see the calculations -- and I
would be happy to send you the code I used to calculate the results
(VB, but the code takes four hours to run on a machine with dual 2GH
processors, 2GB RAM).  If there are also discrepancies in the number
of straights and flushes -- that may shed light.

Regards,
Gregg

Request for Question Clarification by mathtalk-ga on 08 Feb 2003 10:37 PST
Hi, Gregg:

I'm pretty sure I've accounted for the discrepancy.  Here are details
anyway.

I broke the straight flush outcomes into cases by number of Jokers. 
Note that an Ace in my interpretation may be either the high or low
card (but otherwise no "wrapping around" is allowed in forming a
straight).

In each case the suit is well determined (since five Jokers has been
excluded).  Therefore I will omit until the end the factor of four
corresponding to choice of suit.

a) no Jokers

The low card of the straight is one of ten possible values (Ace up to
10).  The five cards are all distinct so there are 5! arrangements of
these on the reels:

10 * 5! = 1200

b) one Joker

The four non-Joker cards must either be a run of four in sequence, or
else have a missing "inside" card filled by the Joker.  We can diagram
and enumerate the possibilities by choices for low card:

c1 c2 c3 c4     [11 choices for low card, Ace through Jack]
c1 __ c3 c4 c5  [10 choices for low card, Ace through 10]
c1 c2 __ c4 c5  [10 choices for low card, Ace through 10]
c1 c2 c3 __ c5  [10 choices for low card, Ace through 10]

So there are 41 possibilities for the cards (all distinct), and 5!
arrangements of these on the reels:

41 * 5! = 4920

c) two Jokers [this is the case where I made an error]

Again we can diagram and enumerate the possibilities according to if
and where the Jokers are used to "fill" an inside straight:

c1 c2 c3        [12 possible low cards, Ace through Queen]
c1 __ c3 c4     [11 possible low cards, Ace through Jack]
c1 c2 __ c4     [11 possible low cards, Ace through Jack]
c1 __ __ c4 c5  [10 possible low cards, Ace through 10]
c1 __ c3 __ c5  [10 possible low cards, Ace through 10]
c1 c2 __ __ c5  [10 possible low cards, Ace through 10]

Since the two Jokers are interchangeable, there are 5!/2 = 60
arrangements on reels:

64 * (5!/2) = 3840

d) three Jokers

With these diagrams:

c1 c2           [13 possible low cards, Ace through King]
c1 __ c3        [12 possible low cards, Ace through Queen]
c1 __ __ c4     [11 possible low cards, Ace through Jack]
c1 __ __ __ c5  [10 possible low cards, Ace through 10]

The three Jokers are interchangeable, so there are 5!/3! = 20
arrangements on the reals:

46 * (5!/3!) = 920

Combining these case by case numbers we match your cited number (after
multiplying by the 4 possible suits):

4 * [1200 + 4920 + 3840 + 920] = 43520

regards, mathtalk
Answer  
Subject: Re: Probability
Answered By: mathtalk-ga on 08 Feb 2003 20:11 PST
Rated:5 out of 5 stars
 
Hi, Gregg:

You seem to have a good grasp of the basic methods for counting
permutations and combinations, so I've marshalled the arguments
below in a somewhat brisk fashion.  Please let me know if some or
all of the arguments require more discussion.

I'll present the results as frequency counts within a universe of
53^5 possible outcomes.  See the table at bottom for a summary of
the various results.

I've assumed the Ace can be used either in a high or low position
within a straight, as is the case here:

https://winnersplay.com/casino/jokerswild.html

regards, mathtalk

<<<<<<<<<<<<<<<<<< BEGIN Analysis <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

1. Five Jokers : 1 (obvious)

2. Five of a kind suited : 1612

There are 31 proper subsets of the five reels (not all five reels) 
which could be occupied by Jokers, and 52 choices for an identical
non-Joker card to occupy any remaining reel. 52 * 31 = 1612

As Gregg has observed, all four Joker hands are classified here.

The total can be expanded into cases by number of Jokers, so:

52 * (C(5,0) + C(5,1) + C(5,2) + C(5,3) + C(5,4))
 = 52 + 260 + 520 + 520 + 260

where C(5,k) is combinations of 5 things taken k at a time, shows us
for the case of k Jokers how many five a kind suited outcomes there
are.

3. Straight flush : 43520

Let's break the analysis into cases by number of Jokers.  Note that
an Ace in my interpretation may be either the high or low card, but
otherwise no "wrapping around" is allowed in forming a straight.

In each case there's a factor of 4 for choice of suit as all non-
Joker cards must be of the same suit here.

a) no Jokers

The low card of the straight is one of ten possible values (Ace to
10).  The five cards are distinct so there are 5! arrangements of
these on the reels:

10 * 4 * 5! = 4800

b) one Joker

The four non-Joker cards must either be a run of four in sequence,
or else have a missing "inside" card filled by the Joker.  We can
diagram and enumerate the possibilities by choices for low card:

c1 c2 c3 c4     [11 choices for low card, Ace through Jack]
c1 __ c3 c4 c5  [10 choices for low card, Ace through 10]
c1 c2 __ c4 c5  [10 choices for low card, Ace through 10]
c1 c2 c3 __ c5  [10 choices for low card, Ace through 10]

So there are 41 possibilities for the four non-Jokers.  With five
distinct cards there are 5! arrangements of these on the reels:

41 * 4 * 5! = 19680

c) two Jokers 

Again we can diagram and enumerate the possibilities according to if
and where the Jokers are used to "fill" an inside straight:

c1 c2 c3        [12 possible low cards, Ace through Queen]
c1 __ c3 c4     [11 possible low cards, Ace through Jack]
c1 c2 __ c4     [11 possible low cards, Ace through Jack]
c1 __ __ c4 c5  [10 possible low cards, Ace through 10]
c1 __ c3 __ c5  [10 possible low cards, Ace through 10]
c1 c2 __ __ c5  [10 possible low cards, Ace through 10]

So there are 64 possibilities for the three non-Jokers.  Since the
two Jokers are interchangeable, there are 5!/2 = 60 arrangements on
reels:

64 * 4 * (5!/2) = 15360

d) three Jokers

With these diagrams:

c1 c2           [13 possible low cards, Ace through King]
c1 __ c3        [12 possible low cards, Ace through Queen]
c1 __ __ c4     [11 possible low cards, Ace through Jack]
c1 __ __ __ c5  [10 possible low cards, Ace through 10]

So there are 46 possibilities for the two non-Jokers.  The three
Jokers are interchangeable, so there are 5!/3! = 20 arrangements on
the reels:

46 * 4 * (5!/3!) = 3680

Combine these case by case numbers to get the total straight flushes:

4800 + 19680 + 15360 + 3680 = 43520

Also, keep the diagrams above in mind when we count the "simple"
straight outcomes.

4. Five of a kind unsuited : 39000

We will handle this by showing that the total number of five of a
kind hands, whether suited or not, is 13 * (5^5 - 1).  Then we just
subtract the previously found number 1612 of five of a kind suited
hands to get the number 39000 of five of kind unsuited hands.

Since we exclude the five Joker hand from this count, any five of a
kind hand, whether suited or not, has a well defined denomination.
There are 13 possible choices for the denomination.  Given that 
choice of denomination, any particular reel not occupied by a Joker
must have one of four possible cards.

If you consider the number k of Jokers ranging from 0 to 4, then an
expression which counts the number of five of a kind outcomes is:

13 * [4^5 + 4^4*5 + 4^3*C(5,2) + 4^2*C(5,3) + 4*C(5,4)]

Here the first term represents using no Jokers, the second using one
Joker, and so forth.  The combinations C(5,k) for k Jokers is of
course consistent with choosing k of 5 reels to be Jokers; thus by
comparison with binomial expansion of (4+1)^5, the count is:

13 * (5^5 - 1) = 40612

After subtracting five of kind suited's, 40612 - 1612 = 39000.


5. Four of a kind suited : 183520

The case analysis by number of Jokers is a little more complicated,
but we are able to close out the rest of the three Joker hands.

a) no Jokers

Pick one card (suit and denomination) to be the four of kind, and
any different denomination card to be the fifth (since otherwise the
fifth card having equal denomination would give five of a kind).  

Four cards are interchangeable, so for the outcomes we have five 
possible reel arrangements:

52 * 48 * 5 = 12480

For future reference we'll note that one-quarter of these turn out
to be flushes if the suit of the fifth card agrees with the others.

b) one Joker

Even with a wildcard the suit and denomination of the four of a kind
are identifiable.  Choose that, and the fifth card of a different
denomination, and we have three interchangeable cards.  

Thus the number of reel arrangements is 5!/3! = 20:

52 * 48 * 20 = 49920

Again we'll note for future reference that one-quarter of these are
also flushes.

c) two Jokers

We choose the suit and denomination for an identical pair and some
fifth card with different suit or denomination.  The Joker pair and
non-Joker pair are each interchangeable, so we have 5!/(2^2) = 30 as
the number of reel arrangements:

52 * 48 * 30 = 74880

Again we'll note for future reference that one-quarter of these are
also flushes.

d) three Jokers

With three Jokers and any two non-Joker cards, one gets four of a 
kind suited or better.  In fact we can compute the outcomes in this
case simply by subtracting from the number of all three Joker "hands"
(outcomes) those which have already been counted above as straight
flushes or five of a kinds (suited or unsuited).

There are C(5,3) = 10 choices for reels to hold three Jokers, and 
52^2 possibilities for the other two reels in an outcome with exactly
three Jokers:

10 * 52^2 = 27040

Previously we counted 4*920 = 3680 of these as straight flushes, and
within five of a kind categories we counted 13 * 4^2 * C(5,3) three
Joker hands (see the expression under five of a kind unsuited for
details).  Thus the residue of three Joker hands which will only
qualify as four of a kind suited is:

27040 - (3680 + 2080) = 21280

Collecting the results from our case analysis:

12480 + 49920 + 74880 + 21280 = 158560

gives the total number of four of kind suited outcomes.

6. Four of a kind unsuited : 1759680

As we did with the five of a kind unsuited category, we come count
the four of a kind unsuited outcomes by subtracting the suited four
of a kind outcomes from the number of all four of a kind outcomes.

It is perhaps a little easier to follow when details are broken down
into cases by numbers of Jokers:

a) no Jokers

Choose a denomination 13 ways, and then a different denomination for
the fifth card in 12 ways.  The reel/position for this fifth card is
chosen in 5 ways.  Arbitrarily choosing suits for all five reels:

13 * 12 * 5 * 4^5 = 798720

We subtract from this the number 12480 of four of a kind suited with
no Joker outcomes, and we have:

798720 - 12480 = 786240

four of a kind unsuited with no Joker outcomes.

b) one Joker

Choose a denomination 13 ways, and then a different denomination for
the fifth card in 12 ways.  The reel/position for this fifth card is
chosen in 5 ways and for the Joker in 4 remaining ways.  Arbitrarily
choosing suits for all four non-Joker reels:

13 * 12 * 20 * 4^4 = 798720

which, by a quirk of arithmetic, is the same number as we got above
with no Jokers.  We subtract from this the number 49920 of four of a
kind suited with one Joker outcomes, and we have:

798720 - 49920 = 748800

four of a kind unsuited with one Joker outcomes.

c) two Jokers

Choose a denomination (for the pair) 13 ways, the other denomination
in 12 ways, the positions of the Jokers in C(5,2) = 10 ways and of 
the "fifth" card in 3 ways.  The suits are assigned arbitrarily to
all three non-Joker reels:

13 * 12 * 30 * 4^3 = 299520

Subtract from this the number 74880 of four of a kind suited with two
Jokers outcomes, and you have:

299520 - 74880 = 224640

four of a kind unsuited with two Jokers outcomes.

Collecting the results of these three cases gives:

786240 + 748800 + 224640 = 1759680

Note that because these are the unsuited outcomes, none are flushes.

7. Full House : 2196480

a) no Jokers

Pick the denomination for three of kind (13 ways) and another for a
pair of different denomination (12 ways).  Pick the two reels which
the pair will occupy in C(5,2) = 10 ways.  Assign suits arbitrarily:

13 * 12 * 10 * 4^5 = 1597440

Note for future reference that 1/256 of these are flushes.

b) one Joker

The non-Joker cards must consist of two pairs, since otherwise we'd
use the Joker with the natural three of a kind to get four of a kind.
Pick the denominations of the two pairs in C(13,2) ways.  Pick the
reel for the Joker in 5 ways and the reels for the higher of the two
pairs (counting Ace high for this purpose) in C(4,2) = 6 ways.  Suits
are assigned arbitrarily to the non-Jokers:

78 * 5 * 6 * 4^4 = 599040

Note for future reference that 1/64 of these are flushes.

c) two Jokers

This can't happen.  With two Jokers we'd still need a least a pair 
among the non-Jokers to construct a full house, since you cannot have
three different denominations in a full house hand.  But then we'd 
put the two Jokers with the pair to get four of a kind instead.

Combining results gives 1597440 + 599040 = 2196480 full houses.

8. Flush : 2053680

Briefly we reduce the total number of flushes by counts previously
covered in preceding categories.

a) no Jokers

Pick a suit (4 ways) and then any of the 13 cards in that suit for 
each reel, ie. 4 * 13^5.  However we must reduce this total by any
flushes with no Jokers already accounted for above.  This includes
all of the five of a kind suited and straight flushes with no Joker,
one quarter of the four of a kind suited with no Joker, and 1/256
of the full houses with no Joker.

4*13^5 - 52 - 4800 - 12480/4 - 1597440/256 = 1470960

b) one Joker

Pick a suit (4 ways), then any of the 13 cards in that suit for each
of four reels, and multiply by 5 for the choice of reel for a Joker
to occupy, ie. 4 * 13^4 * 5.  This total must be reduced by those 
flushes with one Joker previously accounted for as five of a kind 
suited, straight flushes, four of a kind suited, or full houses.

4*13^4*5 - 260 - 19680 - 49920/4 - 599040/64 = 529440

c) two Jokers

Pick a suit (4 ways), then any of the 13 cards in that suit for each
of three reels, and multiply by C(5,2) = 10 for choices of two reels
for the two Jokers to occupy, ie. 4 * 13^3 * 10.  This total is then
reduced by flushes with two Jokers previously counted as five of a 
kind suited, straight flushes, or four of a kind suited.

4*13^3*10 - 520 - 15360 - 74880 = 53280

Combining these cases gives the total outcomes counted as flushes:

1470960 + 529440 + 53280 = 2053680

9. Straight : 2654400

We reduce the total number of straights by previously recorded counts
of straight flushes.

a) no Jokers

As with straight flushes, the lowest card's denomination is one of 10
values.  Assignment of suits to cards is arbitrary, and the distinct
denominations give 5! arrangements of cards on the reels.  Subtract 
the previous count of straight flushes with no Jokers:

10*4^5*5! - (10*4*5!) = 1224000

b) one Joker

As with straight flushes, there are 41 possibilities for non-Joker 
cards' denominations.  Assign suits to non-Joker cards arbitrarily,
but with five distinct cards we still have 5! arrangements of the 
cards on the reels.  Subtract our previous count of straight flushes
with one Joker:

41*4^4*5! - (41*4*5!) = 1239840

c) two Jokers

As with straight flushes, there are 64 possibilities for non-Joker
cards' denominations.  Assign suits to non-Joker cards arbitrarily,
and with two interchangeable Jokers we have 5!/2 arrangements of the
cards on the reels.  Subtract our previous count of straight flushes
with two Jokers:

64*4^3*(5!/2) - (64*4*5!/2) = 230400

Combining these cases gives the total outcomes counted as straights:

1224000 + 1239840 + 230400 = 2694240

>>>>>>>>>>>>>>>>>>>> END of Analysis >>>>>>>>>>>>>>>>>>>>>>>>>>

SUMMARY OF RESULTS

                                 # of Jokers
 Outcome Type
                     0       1      2      3     4      5     ALL
------------------------------------------------------------------
1. Five Jokers        0       0      0      0     0     1        1
2. 5 of kind/S       52     260    520    520   260     -     1612
3. Strgt Flush     4800   19680  15360   3680     -     -    43520
4. 5 of kind/U    13260   16380   7800   1560     -     -    39000
5. 4 of kind/S    12480   49920  74880  21280     -     -   158560
6. 4 of kind/U   786240  748800 224640      -     -     -  1759680
7. Full House   1597440  599040      0      -     -     -  2196480
8. Flush        1470960  529440  53280      -     -     -  2053680
9. Straight     1224000 1239840 230400      -     -     -  2694240

Request for Answer Clarification by sheetwise-ga on 09 Feb 2003 10:35 PST
mathtalk,

I'm impressed.  I told you it would be harder than it looked ;)

All of the numbers look good -- except the four of a Kind Suited, and
four of a kind unsuited.  We have the same total, but a different
distribution.

I show 4x Suited as 21,280, and 4x Unsuited as 1,896,960 (Total
1,918,240), and you show 4x Suited as 158,560 and 4x Unsuited as
1,759,680 (Total 1,918,240).

I'll check my results against the formula today -- but thought I'd
give you a heads up.  I'm most impressed by a correct calculation for
full house.  What's your background anyway?

Regards,
Gregg

Request for Answer Clarification by sheetwise-ga on 09 Feb 2003 12:04 PST
mathtalk,

Your analysis pointed me to a mistaken assumption in my software.  I
have made changes, and am running now -- it should be complete in
about 2 hours, but I have no doubt the numbers will match.  I will
confirm then.

Thank you,
Gregg

Clarification of Answer by mathtalk-ga on 09 Feb 2003 12:10 PST
Hi, Gregg:

Thanks for the kind words.  I will look again at the four of a kind
calculations.  I notice that that figure you give for all four of a
kind suited hands is equal to the figure I have for just those four of
a kind suited hands with three Jokers.

We caught a break of sorts with the full houses, in that there cannot
be a full house with two Jokers.  So the analysis, unexpectedly
perhaps, was easier for the full houses than any of the other
categories in the lower part of the list.

I have a Ph.D. in math plus a lot of programming experience.  A big
part of the challenge in this problem was sorting out what overlaps
are possible in the categories, and so I was probably lucky that I
didn't put a whole lot of effort into thinking about it until you'd
settled on your final ordering of the outcomes.

regards, mathtalk-ga
sheetwise-ga rated this answer:5 out of 5 stars
Very straight forward. Good communication. Exactly what I was looking
for.  I'd heap on more praise -- but then, what else is there?

Comments  
Subject: Re: Probability
From: kennyh-ga on 27 Jan 2003 04:38 PST
 
I don't think the values of probabilties are important.
Anyway, this is a simple question of combinatoris.
Totally possible outcomes: 53^5
1. Five Jokers : 1 outcome (= 1^5), prob = 1/53^5
2. Straight Flush : since the smallest card is 10,
   there are 10 (one to ten) i.e. A2345, 23456, 34567,...,10JQKA
   each has 5! permutations and four suits. 
   So, there are 10*5!*4 general outcomes without wild card.
   Then, consider if there is one wild card, two wild cards,..., 5
jokers,
   If one joker (may appear at 5 positions) , and the other 4 form 4
of a kind.
   (4!* 11) so, there are 5*4!*11 if containing one joker)
   If two jokers appear, there are P(5,2) = 20 possibilities, and the
other 3 form a three of a kind, so there are 20* 3!* 12 in this case.
   The other cases are similar. 
   
   Kenny

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