# Poker probability

#### Dae Yeol Yoon

One of hedge funds question:

There are two sets of cards ranging from 1 to 10. (20 cards total - Red cards 1 t0 10 and Blue cards 1 to 10)
Each of two players will have a pair of these cards (distributed randomly by a dealer)
Winning is determined on following conditions in order ( 1st condition vs 2nd condition -> 1st condition wins):
1) two consecutive numbers: e.g. 10-1, 9-10, 9-8 ... (10-1 being the best)
2) same pair: e.g. 9-9, 8-8, ...
3) higher number: e.g. (9-1 > 8-4) or (10-2 > 5-9)
4) when highest numbers are equal, then compare the other pair: e.g. (9-4 < 9-7)

Questions:
A. Probability of winning when I have 'Red 9'
B. Probability of winning when I receive 3 cards from the dealer, while the other player receives only two cards
(I will have to make the best pair out of three cards)

#### Archidamus

Dae,

I know its a year later, but I wonder if the question was more to see 'how you would set up solving' versus 'actually solving' it. As I am going through this, it seems impossible to do within even a 5 minute window, but I'm no Gauss. Intuitively i'd guess just under 80% of the time as you'd be better positions against someone with an 8 or lower, but we that's just gut feeling.

Methodology(A): Take each individual case as to your second card, and find all the hands that would beat it. Example, if your second card was a '10', which you have a 2/19 probability of being dealt that ten, then only an opponent with 10-1 would beat you, and that's a (2/18*1/17)*2 or 4/306. The probability of loss for 10 as your second card is (2*4)/(19*18*17). let that = q(ten) is the loss probability.

Do this for all probabilities ten through one, then sum them to get a total q or probabilities of losses. then 1-q-t= p. where t= the probabilities of a tie. May be the thing they wanted to to pick up one, kind of like a trick question.

Methodology(B): Again repeat the same methodology above but know there are nearly 10 times as many cases to sum up. Example if being dealt a 10 and a 1, then no hands can beat that, and there for q(ten,one) = 0. Very hard to do manually. If you are a good coder maybe you could create a monte carlo in 20-30 minutes and run a few billion iterations.

#### Alex Concac

Hi, guys. Here is what I want to add. If there is a poker program, but it is, the program with which you are sitting online proves it. That program is written with all the requirements, to qualify in various forms. After that, the theory of probability and (+) (-) will all be taken into account and completely redone. Guys, everyone has his hobby and everyone spends time as he likes. What if I like to play https://www.score88poker.bid poker when I have a day off once a week?

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#### Daniel Duffy

##### C++ author, trainer
Hello guys! I think this would be a great game and I definitely would want to play it. I love regular poker but this seems to be even better.
Just don't lose your shirt.

#### anthocleod

Pretty interesting odds you picked up. I think the second case of this sounds very doubtful because there is no chance of such numbers falling at least once in the whole game.
If you choose IPL Betting, then it will definitely be more successful. Have you tried it that way? I've been here more than half a year, but I'm happy with the results because the odds are excellent for a reasonably experienced gambler like me. If you are still interested, you should take note of such a valuable resource. I'm sure it will help you.

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