IA (HL) Combinations and Permutations help please

[God]

For my IA i want to do about the probabilities of poker. I didn't do the topic in class about combinations and permutations, but i am slowly learning it myself. 

 

Just stuck on something: Lets say 3 people are playing poker, and they all get 2 cards each at the start of the game. So there are 46 cards left, 5 of which will come in the middle. 

 

knowing that i have a pair in my hand ( a pair of 2 for example), how can i find the probability of the five cards in the middle to have another card of the same rank (e.g. another 2), to make a triple, or even to make 4-of-a-kind.

 

 

So i would start with 46C5 to find the number of combinations possible for the 5 cards in the middle. Now i need to find out how many of those combinations contain either 1 of the same ranked card, or 2 of the same ranked cards that are in my hand.

 

Do you guys understand what im talking about? I hope you can help

 

-God (doesn't know everything)

[God]

I tried to answer my question, and i want to know if i am correct , or why i am wrong.

 

 

 

so we first choose the rank, which there is only 1 you can have out of 13.  Then we choose the suit, which there are only 2 left in the pile for that specific rank (the other 2 are in your hand). Then for the remaining 4 cards: 4 ranks out of 12 are chosen, and  for each rank a specific suit is chosen.

 

 

(1C1) x (2C1) x (12C4) x (4C1)^4  = 253440

 

This should be the number of the frequency of 5 card combinations that include one of the ranks that are in your hand already (the pair in your hand), when the other 4 cards have completely different ranks.  so for example, i have a pair of 2s in my hand, this is the frequency hands if the 5 cards in the middle consists of one 2 card, and 4 other cards of different ranks.

 

 

 

 

Then if i change the amount of possible ranks that the last 4 cards can have:

 

(1C1) x (2C1) x (12C3) x (4C1)^4 = 112640    (This one shows the frequency that one card is a 2 and the other 4 cards are consists of 3 other ranks, so there has to be another pair. e.g.: 2, 5,5,8,9)

 

(1C1) x (2C1) x (12C2) x (4C1)^4 = 33792 (eg: 2, 5,5,5,9)

 

(1C1) x (2C1) x (12C1) x (4C1)^4 = 6144 (eg: 2,5,5,5,5)

 

 

Then add all of them up:

253440 + 112640+ 33792 + 6144 = 406016

This number shows all the possibilities of getting a 2 in the middle 5 cards, if you have a pair of 2s in your hand. ( or any other rank card)

 

 406016/1370754 = 0.296 = 29.6%

 

1370754 is (46C5), the amount of possible combinations with 5 cards with 46 possible cards.

 

Sorry for explaining really badly, but is this correct?

 

I would really appreciate the help

Edited July 9, 2014 by God

[ctrls]

How much probability have you done? I'm just curious, because I would presume combinations and permutations are typically taught before it. I'm going to assume you've covered it though, since it's somewhat required to properly explain my point.

 

First of all, I'm getting a different result compared to yours. My final result was 551/2695 = 20.4% (3sf), which is the probability that one or two cards of the same rank as the pair you have is in the middle 5. Bear in mind however that this may be wrong (this kind of probability isn't my strongest area), so I would appreciate it if someone else can check it.

 

It seems like you made at least two errors, but I'm not sure what the second is - I'm having trouble following your solution. I don't think you are using the term 'frequency' correctly in this context and there doesn't seem to be much justification in some areas (eg: when you calculate the number of ways things can occur).

 

One mistake however, is that you haven't accounted for the possibility that one of the other players also draws a card of the same rank - there's still one card of that rank left in the deck, so you can still have one in the middle 5. I'm guessing there's a second because you obtained a higher probability than I did, which is odd considering you missed a case. I'm not particularly sure how you got 4 terms either.

 

Instead of trying to do everything at once however, perhaps it may help to break the problem down into smaller, manageable cases. This way its also easier to justify your result and for the examiner to understand your working. Here's how I approached the problem:

 

I'm going to call the rank of your pair x. The event that none of the other players has a card of rank x will be denoted A, the event of one of the other players having one card of rank x will be denoted B and the event of one or more cards in the middle 5 will be denoted C. For the sake of notation, I'll also let D denote the event that you draw a pair at the start.

 

Since A and B are mutually exclusive events, we can consider the two cases separately and add the probabilities (using a result from the formula booklet. As a result, the required probability is,

 

P(C | D) = P( (A or B) and C | D)     (given P(A or B)=1)

               = P( A and C | D) + P(B and C | D)

               = P(A | D)*P( C | A and D) + P(B | D)*P(C | B and D)

 

Hope that makes sense. From there, you work out the individual probabilities which can be done with similar methods you've considered above - though you still need to be careful with your justification.

 

It may be useful to look up the hypergoenetric distribution, which can be used to find the probabilities of C for the various cases.

 

This post is a bit jumbled up and messy, so sorry about that. Good luck with your exploration!

Edited July 9, 2014 by ctrls

[God]

 

I'm going to call the rank of your pair x. The event that none of the other players has a card of rank x will be denoted A, the event of one of the other players having one card of rank x will be denoted B and the event of one or more cards in the middle 5 will be denoted C. For the sake of notation, I'll also let D denote the event that you draw a pair at the start.

 

Since A and B are mutually exclusive events, we can consider the two cases separately and add the probabilities (using a result from the formula booklet. As a result, the required probability is,

 

P(C | D) = P( (A or B) and C | D)     (given P(A or B)=1)

               = P( A and C | D) + P(B and C | D)

               = P(A | D)*P( C | A and D) + P(B | D)*P(C | B and D)

 

Hope that makes sense. From there, you work out the individual probabilities which can be done with similar methods you've considered above - though you still need to be careful with your justification.

 

It may be useful to look up the hypergoenetric distribution, which can be used to find the probabilities of C for the various cases.

 

This post is a bit jumbled up and messy, so sorry about that. Good luck with your exploration!

 

Thanks bro, that actually helped me. That hypergeometric distribution thing is pretty cool too. i did get a slightly different answer though : 20.7%. However, I used hypergeometric distribution for working out all 4 of the probabilities  P(A | D)*P( C | A and D) + P(B | D)*P(C | B and D). And it is assuming that i get dealt 2 cards first, and then the other 2 people are dealt 2. I would really like to know how you worked out P(A | D) and P(B | D). 

Also, can you please tell me how you got from P( A and C | D) to P(A | D)*P( C | A and D). I don't see it in the formula booklet.

 

I would really appreciate your help again, or anyone else's!

 

Thanks!

 

-God

[ctrls]

The approach using the hypergeometric distribution for each casr should be valid, I just checked and it seems be fine. One subtlety I didn't mention however is that for P(C | A and D), you have consider the cases of one card being in the middle 5 and the case of both cards getting there, so you end with with 2 probabilities you have to sum.

 

My solution may be wrong though, for some reason I'm getting different values when I try again. Not too sure why, but it's possible you may be correct.

 

As for how I made that manipulation, it's using the idea that P(A and B)=P(A | B)P(B), which is included in a rearranged form in the formula booklet.