Comp Sci Poker test: probability of Full House (confusion)

[shivajikobardan]

Homework Statement

random number generator independence testing

Relevant Equations

poker test probabilities

I'm studying this for poker test. This should not be memorized as this has 3,4 and 5 digit versions. Memorizing all of them isn't possible. So I need a way to calculate them.
I'm trying to learn through this example.

I'm not getting the process(I know math behind it ie permutations, combinations, some basic probability formula of favorable events etc). But I don't get the card game here. I watched how poker is played and they didnt' discuss about it. I've played some cards, but I'm no expert at it. I just play basic games only(no 21 cards Rummy or Marriage Rummy game etc).

So what's the scenario like in game?
a) for say 3 of a kind, the values are from 2-10 and J to A. so a card chosen at random could be any of them so 13C1.

b) Now your hand has 1 card. What you do now? do you need the digits to be same? or just the suit(hearts, diamond etc)? As far as I've read, it's suits same in poker.(But digits should be same in Random number testing).

What's the game scenario like? some say 2 cards are fixed to some already.

Why would it be so easy to choose the rest 3 cards of same suit(same hearts)? (i think 4C3 is easy kind of).

I think it should be (52-1)C3 (ofc I'm wrong, it'd give probability greater than 1)

c) Now you want to choose another type for full house. Now you should have 52-3=49 cards for it. d) Now we need to choose 2 cards that're same suit from 49 cards..?

What'd be the best way for me to learn the choices in this game? I tried free online poker but the scenario is different there.

Any learning simulator for card probabilities?

 

[pbuk]

https://en.wikipedia.org/wiki/List_of_poker_hands

 

[FactChecker]

Homework Statement:: random number generator independence testing
Relevant Equations:: poker test probabilities

I'm studying this for poker test. This should not be memorized as this has 3,4 and 5 digit versions. Memorizing all of them isn't possible. So I need a way to calculate them.
I'm trying to learn through this example.
View attachment 305223
I'm not getting the process(I know math behind it ie permutations, combinations, some basic probability formula of favorable events etc). But I don't get the card game here. I watched how poker is played and they didnt' discuss about it. I've played some cards, but I'm no expert at it. I just play basic games only(no 21 cards Rummy or Marriage Rummy game etc).

So what's the scenario like in game?
a) for say 3 of a kind, the values are from 2-10 and J to A. so a card chosen at random could be any of them so 13C1.

b) Now your hand has 1 card. What you do now? do you need the digits to be same?

Yes. There will be 4 cards of that rank ("digit"). The 4 cards of equal rank will have one card of each suit of hearts, diamonds, spades, and clubs.

or just the suit(hearts, diamond etc)?

No. These must be different.

 

[pbuk]

Why would it be so easy to choose the rest 3 cards of same suit(same hearts)? (i think 4C3 is easy kind of).

I think it should be (52-1)C3 (ofc I'm wrong, it'd give probability greater than 1)

You are looking at this the wrong way round. We know that there are $\binom{52}{5} = 2,598,960$ different poker hands and what we are trying to do is work out how many of them are a full house.

• There are $\binom{13}{1}$ different kinds of card (aka types or ranks) for the 3-of-a-kind.
• There are $\binom{4}{3}$ different combinations of cards making a 3-of-a-kind.
• ...

 

[shivajikobardan]

https://en.wikipedia.org/wiki/List_of_poker_hands

I understand the full house, three of a kind etc meaning. I'm not getting the game and card distribution.

 

[shivajikobardan]

Yes. There will be 4 cards of that rank ("digit"). The 4 cards of equal rank will have one card of each suit of hearts, diamonds, spades, and clubs.

No. These must be different.

Are we choosing all 5 cards or just 1 card, 2 card? PS I don't understand why 13C1? we're choosing a digit from 52 cards, it could be any digits so we have more options. because if we use A in hearts, we can also use A from diamond or club or others.

 

[shivajikobardan]

You are looking at this the wrong way round. We know that there are $\binom{52}{5} = 2,598,960$ different poker hands and what we are trying to do is work out how many of them are a full house.

• There are $\binom{13}{1}$ different kinds of card (aka types or ranks) for the 3-of-a-kind.
• There are $\binom{4}{3}$ different combinations of cards making a 3-of-a-kind.
• ...

can you give 1 example of this.

Say I want to make this full house.

First I need 6. Probability=

Then I need 3 6es. Probability=

They all should be of different color, probability=

I need king

Probability=

I need to choose different heart and club type, so probability=

?

 

[FactChecker]

Are we choosing all 5 cards or just 1 card, 2 card? PS I don't understand why 13C1?

The statement of the first step is misleading. You are actually choosing one rank out of 13, not a card. The only reason I say that is that the next step is to choose 3 out of 4 cards with that rank, so there must not be a card chosen in this first step, only a rank.

 

[shivajikobardan]

The statement of the first step is misleading. You are actually choosing one rank out of 13, not a card. The only reason I say that is that the next step is to choose 3 out of 4 cards with that rank, so there must not be a card chosen in this first step, only a rank.

hmm i think now I get it.

I'm trying to find full house for 5 digits(not poker card now, but different name same thing tbh).
We've pool of 10K random numbers and we've 5 digits of random number.
So what's the probability of full house there?
I got my answer with 10C1*9C1/10K

But I just choosed 2 digits here. How is this correct? What about remaining 3 digits?

 

[pbuk]

hmm i think now I get it.

I hope so. The key is that we don't calculate probabilities at each stage we count frequencies. It would be possible to do it by calculating probabilities but this would be really difficult because each card is unique and so at each stage after the first you have to consider the conditional probability of drawing a card that satisfies a certain condition given that you have already drawn cards satisfying certain other conditions.0

I'm trying to find full house for 5 digits(not poker card now, but different name same thing tbh).

Not the same at all:

• Digits don't have suits, only values.
• Without imposing further criteria digits are independent so there is no issue of conditional probability.
• Because digits are independent the probabilities of combinations are not equal (how many numbers between 00 and 99 have 2 ones? how many have 1 one and 1 two?)

We've pool of 10K random numbers and we've 5 digits of random number.

If you include leading zeros there are 100,000 5 digit numbers { 00000...99999 }, but this is the number of permutations. How many combinations are there? Where does the "pool of 10K random numbers" come in?

So what's the probability of full house there?

I got my answer with 10C1*9C1/10K

I assume you include 01010 as a full house? In that case there are indeed 90 different full houses, and dividing by 10,000 does give the right answer, but why?

Spoiler

This is only the right answer by coincidence: the workings for the right answer should lead to $\frac{(2 + 4 + 4)90}{100,000}$; can you see why?

But I just choosed 2 digits here. How is this correct? What about remaining 3 digits?

Explain why you used $\binom{10}{1} \text{ and } \binom{9}{1}$; that should answer your own question.

 

[PeroK]

It can be done step by step. The probability of all of the following can be fairly easily calculated.

After three cards we need either:

1) All 3 cards the same.

Or:

2) Two cards the same, and the third different.

Case 1) is easy to finish off. The last two cards must be the same.

Case 2 splits into two cases after 4 cards:

2a) two pairs

2b) a triple and a single.

These can easily be finished off by calculating what you need for the last card.

It's worth doing it this way and checking that you get the same answer as using the counting method.

 

[PeroK]

Another, quicker, method is to calculate the probability of being dealt case 1) above. I.e. XXXYY. Then use the fact that you are equally likely to get this hand with the X's and Y's in any order. So, the probability is $\binom 5 2$ times this.