# Playing Cards Probability

#### Ebiru2387

##### New Member
Hi all,

If you have ever watched the World Series of Poker it is common nowadays to see a percentage indicating the probability of certain cards that will be an "out" to allow a player to make a hand that will defeat their opponent.

I play a game commonly with friends called Oh Hell, which is a lot like Spades. Its a game of taking Tricks and Trump can be any suit determined at random each hand. So if i have a 7 of clubs, and the Ace of Diamonds is the trump card (diamonds suit is trump), i am often curious about the probability of the cards that can beat that 7 of clubs.

So what can beat that 7 of Clubs? If i lead a hand with the 7 (meaning those with clubs must play a club or can play trump if they don't have clubs) then the 8 - A of Clubs and ANY Diamond can beat it (excluding the Ace which is not part of the deck because it is the trump card). This means that 19 cards could beat that 7 club. So if my amateur understanding of probability is correct, that would mean there is a 37.2% chance of a higher card winning the hand (19 / 51). Is this correct?

Taking it one step further, let us say the there are 3 players and each player has 5 cards in their hand. I have the 7 of clubs, and no other clubs or diamonds in my hand. Does the probability change to consider that at MOST there are only 10 cards that could beat my 7 of clubs? (because worst case scenario i am assuming that my opponents have all higher clubs and/or diamond cards). How would i use math to determine that probability? Also how would one determine the average probability? For example it is very unlikely that my other two opponents have higher clubs and diamonds ONLY in their hand. Is there a way for me to understand the average scenario?

Please note i am only doing this for fun. I just want to utilize basic statistics for me to make more educated decisions when i play.

Kind Regards,

- Brad

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#### fed2

##### Active Member
'sup

So what can beat that 7 of Clubs? If i lead a hand with the 7 (meaning those with clubs m...
I reckon the answer is you need to know is the chance that in a deal of 15 cards from a deck of 51 - 5 = 46 they get at least one card that beats yours, or which there are10, i guess. The probability of at least 1 crd beating yorus = 1 - the probability that no card beats yours = 1 - [choose(36,15)/choose(46,15)]. These 'choose' things are funny brackets you see in wikipedia on poker probs. So im putting it at about 2% chance your first lead goes through. That's my answer, but theres a fair chance i messed it up so you'll have to figure that fer yerself, hear? youd think counting was easy, just whip out them fingers and toes, but turns out it can be hard.

Please note i am only doing this for fun.
ditto

well my cat just barfed on the carpet...

#### katxt

##### Active Member
Even the simple case is complicated. You have 5 cards which include 1 club, 0 diamonds, and 4 of other suits. Your opponent has 5 cards. You lead the 7C. What are your chances of winning?
Does you opponent have at least 1 club?
If yes, then what are the probabilities they have exactly 1, 2, 3, 4 and 5 clubs? For each of these, what is the probability that their highest is greater than 7?
If no, what is the probability that they have at least 1 diamond?
Perhaps a simulation would be easier. kat

#### katxt

##### Active Member
Simulation -
against one opponent about 29.4% chance of winning
against three opponents about 1.4% chance of winning

#### katxt

##### Active Member
Perhaps your 2% is a bit of a coincidence. (In Excel COMBIN(36,15)/COMBIN(46,15) is about 1%)
With three hands against you, you lose if there is any club>7 somewhere in the three hands, or if there is any clubless hand (about 20% of the hands) which have any trump at all (about 80% of them).

#### fed2

##### Active Member
Maybe, question suggests that there are

at MOST there are only 10 cards that could beat my 7 of clubs
I take that to mean there are 10 cards in the deck that could beat his.

Who knows what the hell it actually means.