In my probability and statistics class, we have an activity where we find the probability of certain poker hands while using a pinochle deck.
For those who don't know what a pinochle deck is, there are 48 cards. Take a regular deck from 9 through Ace and double it. There are still 4 suits, but two of every card (8 Aces, 2 Ace of Spades).

I was able to do the first few, straights, flushes, etc. (although in class with some help...) but I'm kind of confused on the probability of having a best hand of "high card." (I'm not very good at statistics...)

For a regular hand, it says to 13c5 for 5 distinct cards, then exclude straights by subtracting 10. Then it says any pattern of suits except flush, so 4^5-4. So basically you get ((13c5)-10)((4^5)-4)

Translating this to a pinochle deck, I get ((6c5)-__)((4^5)-__) I don't know where the 10 and 4 they subtracted came from, so I don't know what to do with them =/

Anyway, I figured out there should be 642,176 ways to get this hand by finding total ways and subtracting everything else, but I think we're supposed to do it the way the example is given.

Thanks for any help....

/e That 642,176 isn't right, I just realized I haven't done one pair yet, so have to subtract that as well.

This post was edited by furbyjs on Sep 1 2013 09:15am

since it's common i won't bother typing it all out

http://en.wikipedia.org/wiki/Poker_probability#Derivation_of_frequencies_of_5-card_poker_hands

Thank you, but I already found that page and it doesn't help much.

Using a Pinochle deck, which creates many different possibilities, including a 5 of a Kind.

The main problem I'm having is accounting for Flushes in other hands such as 2 Pair, 1 Pair, where it is still possible to have a Flush.
When using a regular 52 card deck, this isn't the case as you can't have a Flush if you have two of the same cards. In Pinochle, there are two King Hearts, two King Spades, two King Clubs, etc.

This post was edited by furbyjs on Sep 1 2013 10:25am

I wouldn't worry about flushes that contain pairs, for 2 reasons

1. flush beat 1 or 2 pair anyway
2. since there are no rules (as far as i know) about poker with such a deck, why not just ignore them.

So unless you are being asked specifically to find the probability of getting a "2 pair flush", forget it.

/e
to rephrase:
you said you want to use poker hands:
there is no such things as a 5 of a kind in poker, or a flush with pairs, or pure pairs (say like a pair of ace of spades vs ace of spade + ace of diamonds), so unless you have been asked to calculate these hands specifically, forget them.

This post was edited by Azrad on Sep 1 2013 12:12pm

I was asked to calculate 5 of a Kind, which was easy enough.

We have to find the total # of ways to get a _____ as a best hand. So the total # of ways I can have a full house, or total number of ways to have 2 Pair.

I don't have to find the probability of a 2 Pair + Flush, but if I do have a 2 Pair + Flush, then it's just a Flush because Flush > 2 Pair.

Therefore, I need to find the probability of 2 Pair that isn't a flush. (Since some hands that are 2 Pair can also be a Flush, I need to find the total number of ways to get a 2 Pair being my best hand, i.e not also a Flush)

It's kind of hard to explain, especially since I don't know how to do it....

/e Basically, if I were to calculate the # of ways to get pairs, some of those ways would include a Flush, so pair wouldn't be my best hand.

This post was edited by furbyjs on Sep 1 2013 01:05pm

Up....

well you could count the number of ways to get 2 pair = A, then count the number of ways to get 2 pair w/ flush = B.

Then the chance to get 2 pair w/o flush would be (A - B)/(total number of different hands)

This post was edited by Azrad on Sep 3 2013 08:15am

I'll explain the problem and how to solve it the way the problem is trying to get you to solve it (the high card part anyway, lmk if you need help on others)...


the reason they subtract 10 for straights is that, if we disregard suite (only look at the card values), there are 10 different straights

A 2 3 4 5
2 3 4 5 6
3 4 5 6 7
...
10 J Q K A

There would be only 9 given the 1-13 range, but ace acts as both 1 and 14 for straights.


the reason they subtract 4 is that, diregarding values of cards, there are only 4 different flushes, one for each suite



Translating this to your weird-ass deck, there are only 2 straights if we disregard suite
9 10 J Q K
10 J Q K A

Now for suites, I think we should treat this deck as if there are 8 suites. This gives us

(8^5) - number of flushes
viable combinations of suites.

Now, there are 8 single-suite flushes, but now since there are 4 same-suite pairs, there are additional "multi-suite" flushes (yes this is worded horribly, deal with it lmao)
For 2 of the same suite, you can have 4 1, 3 2, 2 3, 1 4. That's 4 different ways of mixing two same suites for a flush. Now multiply this by 4 (since there are 4 suite-pairs) and you get 16.
Total flushes equals 16+8 (when blind to value, please keep in mind we are disregarding value here).


((6c5)-2)((8^5)-24)

yes, please check my math
did this at work while trying to be productive so i could've easily overlooked something

This post was edited by Segfault on Sep 3 2013 09:56am

Yeah, I don't know why I didn't think of doing that sooner =/

Even if I did, I'm incredibly bad at these kinds of statistics (combinations and permutations for card hands), so I don't know if it would have helped any....




Thanks, I appreciate the in depth answer.

I can sort of follow along with your reasoning, the number of straights makes sense, but when you start getting into the number of flushes, it starts to confuse me xD
Nonetheless, it helps to have the numbers explained.

Do you get why the original deck only has 4 types of flushes disregarding card value?
If we disregard card value, we only see spades, diamonds, etc. So there are only 4 different types of flushes. One flush of only spades, one flush of only diamonds, etc.


now let's label the suites in your weird deck
spades1
spades2 (the set of "copies" of the spades cards, since there are two of every card)
diamonds1
diamonds2
etc.

So spades1 has one flush by itself (a flush of all spades1). So do all the other suites. That makes 8 flushes disregarding card value.
However, we can also mix spades1 and spades2 for a flush.

1 spades1, 4 spades2
2 spades1, 3 spades2
3 spades1, 2 spades2
4 spades1, 1 spades2

That's 4 "mixed" flushes. This goes for the other suite pairs as well, making a total of 16 "mixed" flushes.
16+8 = 24 flushes disregarding card value

I treat the spades1 and spades2 as seperate suites to reflect that it is easier to draw flushes than it would've been if there was only 1 copy of each card.