Two Plus Two Poker Forums KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets)
 Register FAQ Search Today's Posts Mark Forums Read TwoPlusTwo.com

 Probability Discussions of probability theory

 03-09-2020, 05:19 PM #1 K9Papa newbie   Join Date: Feb 2020 Posts: 39 KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) The math question is: What % of the time will KK flop at least 1 K & AA will not flop an ace? My math is as follows: You have to flop a K: 2(C,1) Then you have a total of: 45 unseen cards that can complete the last 2 cards on the Flop: [52 - 2 As - 3 Ks, as 1 K is on the Flop] unseen cards - the 2 As that can't come on the Flop. 45(C,2) So, it's [2(C,1)] * [45(C,2)] = 2 * 990 = 1980 possible Flops that bring at least one K & no As. We have 48 unseen cards pre, so total possible Flops: C(48,3) = 17,296 1980/17,296 = .11447 or 11.48% chance. I've seen on at least 3 poker sites that it's in the low 20% arena, i.e., 20.17%, 21.67% & I think 22.06% are the numbers I've seen. Or maybe what I've seen it quoted as is 20.17% 21.xx% & one other [that I can't remember but think is was 22.06%] on various sites. What's up with my poor math skillz?
 03-09-2020, 05:22 PM #2 4cardfish newbie   Join Date: Nov 2019 Posts: 37 re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) That’s for flopping a K only. You can also win by making a straight/flush.
03-09-2020, 05:36 PM   #3
K9Papa
newbie

Join Date: Feb 2020
Posts: 39
re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets)

Quote:
 Originally Posted by 4cardfish That’s for flopping a K only. You can also win by making a straight/flush.
This is thru the Flop only.

 03-09-2020, 06:37 PM #4 whosnext Carpal \'Tunnel     Join Date: Mar 2009 Location: California Posts: 6,370 re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) Your approach is correct but your math is not quite right. You should probably split it into two cases. Tally each then add the tallies. Case 1: Exactly 1 King and 0 Aces on Flop Case 2: Exactly 2 Kings and 0 Aces on Flop
 03-10-2020, 11:37 AM #5 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,967 re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) Whatever sites you're looking at are most likely giving a 5-card probability. The answer is nowhere near 20%, in fact your answer was slightly high because you unknowingly double-counted quad kings. Your C(45,2) allows for a 2nd king. There is only one way (2C2) to get both kings, not C(2,1).
 03-11-2020, 01:30 AM #6 K9Papa newbie   Join Date: Feb 2020 Posts: 39 re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) Thank you gentlemen!
 03-11-2020, 04:18 AM #7 K9Papa newbie   Join Date: Feb 2020 Posts: 39 re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) After thinking about it, I'm still a bit confused.... The C(2,1) in my formula is to flop 1 of the 2 Ks available. The C(45,2) are the remaining cards in the deck [available to complete the flop, after 1 K flops & the 45 includes the other K still in the deck, but not the 2 As, so quad Ks are a possibility in my equation....or so I thought....
 03-11-2020, 07:27 AM #8 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,967 re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) Yes they are a possibility. The problem is there are only 44 combos of quad kings with no A, but you unintentionally counted 88 combos. It's rather hidden, and if you're curious to see it I'll try to explain. But splitting the calculation into two separate tallies (as whosnext suggested) would be the easy way to avoid that. When dealing with overlapping sets, you either need to use inclusion-exclusion or avoid overlap altogether. And come to think of it, when you make quads, you probably don't care if an ace comes (in fact you prefer it), so maybe you want the probability to reflect that.
03-11-2020, 01:48 PM   #9
whosnext
Carpal \'Tunnel

Join Date: Mar 2009
Location: California
Posts: 6,370
re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets)

Quote:
 Originally Posted by heehaww Yes they are a possibility. The problem is there are only 44 combos of quad kings with no A, but you unintentionally counted 88 combos. It's rather hidden, and if you're curious to see it I'll try to explain. But splitting the calculation into two separate tallies (as whosnext suggested) would be the easy way to avoid that. When dealing with overlapping sets, you either need to use inclusion-exclusion or avoid overlap altogether.
Since this "overlapping" phenomenon is so widespread (and hidden), I think it would be great if you explained why OP's original formula was not quite correct in this case.

 03-12-2020, 07:49 PM #10 K9Papa newbie   Join Date: Feb 2020 Posts: 39 re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) Good points. Does the overlap come because I use the 2 Ks in C(2,1) and then use the remaining K again in C(45,2) thus using the card twice & overlapping the K that didn't come in C(2,1)? That would make sense.........but then again I'm a novice at this.
03-12-2020, 10:45 PM   #11
heehaww
Pooh-Bah

Join Date: Aug 2011
Location: Tacooos!!!!
Posts: 4,967
re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets)

Since it comes up a lot, perhaps this should be a sticky, so I went ahead and attempted to write something worrrthy of stickiness. To that end, I went beyond addressing OP's question and wrote an intro to inclusion-exclusion. The specific response to OP is mostly at the beginning of Part 2, but is also partly covered in Part 1.

Part 1: Background

Suppose you flip a coin twice and want the probability of at least one tails. You know it isn't simply 50% + 50%. The big clue is that you know it can't be 100% (nor 150% with 3 flips), but what's the underlying reason probabilities don't add (or multiply) like that? Well, sometimes they do, for instance if you want the A, you have a 1/52 chance with one card and a 2/52 chance with 2 cards. What's the difference between those two scenarios? Overlapping possibilities.

There is no overlap in the ace example. Getting the ace with your 1st card or your 2nd card are mutually exclusive possibilities because the A can only be one of your cards (if any). Each 1/52 is the probability of the A being an exact card, so there is no overlap between the two 1/52 terms. Therefore, when you add them, nothing is counted more than once.

By contrast, when you flip a coin twice, Tails can occur on one or both flips. There is a 50% chance for the 1st flip and a 50% chance for the 2nd flip. Notice, though, that each of those 50%'s allows the other flip to also be tails. There's a 50% chance of Tx and a 50% of xT, where x can be a T or H.

If you add the TX and XT cases, you're adding (TH & TT) + (HT & TT), so you're double-counting the TT case. That's why the chance of at least one T is not P(Tx) + P(xT), but lower. If you remove the excess TT, you get the right answer: P(T) = P(Tx) + P(xT) - P(TT) = 100% - 25%, which is the Principle of Inclusion-Exclusion at work. That can also be illustrated with a Venn diagram, but I prefer to picture the strings of symbols, especially when we extend the concept to N sets.

Let's extend it to 3 flips. We know it's not 150%, but by how much is that too high?

Each Txx possibility counts a TTx possibility, so when any two Txx possibilities are combined, one of the 3 TT possibilities gets double-counted. Here is my mental model of what's happening:

Txx + xTx double-counts TTx
xTx + xxT double-counts xTT
Txx + xxT double-counts TxT

By adding Txx + xTx + xxT, we have double-counted each of the 3 TT possibilities. We've also triple-counted the TTT case. Therefore, it follows that:

P(T with 3 flips) = 150% - P(exactly 2 tails) - 2*P(TTT) = 150% - 3/8 - 2(1/8) = 7/8

If you wanted the chance of exactly one tails, you'd simply adjust the above as follows:
P(exactly one T w/ 3 flips) = 150% - 2*P(exactly 2 tails) - 3*P(TTT) = 3/8

However, for problems where it's worth using this approach to begin with (unlike this coin problem), the exactly-2 case would be just as difficult to calculate as the original thing we're trying to calculate. So rather than subtracting the TTH cases, we'd subtract the TTx cases, which is easier in such problems. In doing so, we'd also be subtracting TTT cases, in fact over-subtracting them. Here's what that would look like.

There are 3 TTx arrangements. Since the x is ignored, or equivalently, has a 100% chance, each TTx arrangement has a 25% chance. So from 150% we subtract 75%. Our original 150% triple-counted the TTT. But for each TTx case we subtracted, we also subtracted the TTT case, so overall it got added 3 times and then subtracted 3 times for a net result of 0. It needs to be added back in.

P(T with 3 flips) = 3(1/2) - 3(1/4) + 1/8 = 7/8

Mind you, the problem can also be solved using only addition, but only if there's no overlap among what you're adding: P(T with 3 flips) = P(exactly one T) + P(exactly 2) + P(TTT)

Part 2: Card Examples

Quote:
 What % of the time will KK flop at least 1 K & AA will not flop an ace?
A common mistake is to make the numerator C(2,1)*C(45,2). That numerator counts the Kxx cases, but since another K is part of the 45, the x's are allowed to be another K, which means the KKx cases are counted twice.

You can visualize it the same way I illustrated it with the 3 coinflips when listing the Txx vs TTx cases. There's also a numerical explanation.

If instead of the Kxx cases, you wanted to count the KKx cases, you'd correctly say C(2,2)*44. Notice how the two kings are both picked at the same time, hence the C(2,2) term instead of C(2,1).

Let's make it explicit by rewriting the bad numerator:

C(2,1) * [C(44,2) + C(1,1)*44]

What you've done is counted the KKx cases C(2,1)*C(1,1) times instead of C(2,2).

A warning sign was that you picked from the same set (of two kings) twice separately. As a rule of thumb, any time you find yourself picking from the same set more than once (which I'll call "double-dipping"), especially when you don't intend for the order of selections to matter, you have to watch out for overcounting.

To get the correct answer, you can either subtract the excess KKx combos or, from the start, split the problem into non-overlapping cases. If we let "o" represent non-kings, you can add the K-o-o cases and the KKo cases: C(2,1)*C(44,2) + C(2,2)*44

Quote:
 5 hold'em players go to the flop and it's all diamonds. What's the chance that at least 2 of them flopped a flush?
49 cards remain, 10 of which are diamonds.

Unlike the other examples, inclusion-exclusion is a time-saving approach to this problem, and probably the most efficient one. It makes this otherwise tedious problem quick and easy (potentially).

Start with the FFxxx cases. There are C(5,2) ways to pick 2 players who to get the flushes. The other 3 players are allowed to have anything, so we can greatly simplify the calculation by ignoring them, which looks like:

C(5,2)*C(10,4) / C(49,4)

If we didn't ignore the unchosen players, we'd have a larger numerator and a proportionally larger denominator:

C(10,4)*C(45,6) * 3*(5*3) / [C(49,10) * 9*7*5*3]

Ignoring is clearly better! But notice how the 45 includes the rest of the diamonds, so C(45,6) means they can be picked again. Just like in the previous example, this double-dipping results in the cases of >4 diamonds to be overcounted. With the shorter method, the same thing is happening, except it's visible in the form of the denominator being too small rather than the numerator being too large. What the two expressions have in common is that they're counting FFxxx cases where the x's are all-inclusive. The inclusiveness of the x's results in overlap, and so things get overcounted. This time, however, we're overcounting on purpose (with the plan of correcting later), since the x's make the calculations easier.

Anyway, we started with P(FFxxx). That triple-counts P(FFFxx), so we have to subtract that twice:

- 2*C(5,3)*C(10,6) / C(49,6)

As a rule of thumb, +'s and -'s always alternate, so our next step will be to add something. Next up is P(FFFFx). It was counted C(4,2) times, then subtracted 2*C(4,3) times for a net count of -2. To make the count +1, we'll add it back 3 times:

+ 3*C(5,4)*C(10,8) / C(49,8)

Finally, the case of all 5 players having a flush. Without any math, I already know it needs to be subtracted 4 times, because these error/correction coefficients follow patterns.

- 4 / C(49,10)

In entirety, after simplification: C(5,2)*C(10,4)*[1/C(49,4) - 2/C(49,6)] + 3*5*C(10,2)/C(49,8) - 4/C(49,10)

 03-13-2020, 01:55 PM #12 K9Papa newbie   Join Date: Feb 2020 Posts: 39 Re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) I've read it all, but haven't studied it all. Got caught up with the stock market. It appears from Part I, that I'm correct, in that my C(4,1)*C(45,2) IS inclusive of any one of the 4 aces appearing on the board and if it isn't chosen, it has a 2nd chance. I get that part now......... If the A is what appears otf in C(4,1), then the other 3 suits, which had a chance to be chosen in C(4,1) gets another chance in C(45,2) Thanks for the in-depth article! I hope it gets pinned.
 03-13-2020, 05:54 PM #13 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,967 Re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) It's included, but too much so. You counted the same combos twice. Your numerator should be 1936 rather than 1980, and the correct answer is 11.19% as opposed to 11.45%. Your C(2,1) is selecting one of the remaining two suits, say spade and heart. Multiplying by C(2,1) is equivalent to adding C(1,1)*C(45,2) + C(1,1)*C(45,2). That is, you've added (K)(x) + (K)(x) with each x potentially being the final K. The possibility of getting both kings got counted twice: once by the 1st term and once by the 2nd term added.
 03-13-2020, 06:52 PM #14 whosnext Carpal \'Tunnel     Join Date: Mar 2009 Location: California Posts: 6,370 Re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) heehaww has explained this very well but just so it is crystal crystal clear to any future reader of this thread ... If order of the flop cards does not matter, the following flops are considered the same (count as only one): (K)(K)(9) AND (K)(K)(9) so some care needs to be taken not to count them separately as two different flops.
03-14-2020, 11:14 AM   #15
K9Papa
newbie

Join Date: Feb 2020
Posts: 39
Re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets)

Quote:
 Originally Posted by whosnext heehaww has explained this very well but just so it is crystal crystal clear to any future reader of this thread ... If order of the flop cards does not matter, the following flops are considered the same (count as only one): (K)(K)(9) AND (K)(K)(9) so some care needs to be taken not to count them separately as two different flops.
Which is called permutations. You don't care if flop comes KQJ or QJK

So when you determine total possible flops when you only know your cards, it's not 50*49*48, it's [50*49*48]/[3*2*1] you start with 3 because that's how many objects [cards] there are. If you wanted total possible 5 card boards, it's' 5*4*3*2*1....the 1 really isn't necessary.

The easy way to write the above possible flops for 50 unseen cards is: C(50,3)

 03-14-2020, 12:31 PM #16 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,967 Re: KK vs AA Why is My Math Wrong? (Inclusion-Exclusion and Overlapping Sets) He didn't become mod of this forum without knowing that He's saying you counted the K permutations when you meant to count their combinations, which is another way of looking at it. A permutation is a special case of multinomial coefficients, which in turn can be expressed as products of binomial coefficients (combinations). For instance the homework problem everyone on the internet seems to have had at one point: how many ways to arrange MISSISSIPPI ? One solution is 11*C(10,2)*C(8,4). The same reasoning works when there are no repeating letters, for instance the arrangements of CARD: C(4,1)*C(3,1)*C(2,1) = 4! There are C(2,2) combos of KK, but you counted C(2,1)*C(1,1), ie 2 factorial, so rather than counting their combos, you counted their perms (of which there are twice as many). Whichever way you slice it, the combos got double-counted.

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Poker News & Discussion     News, Views, and Gossip     Poker Blogs and Goals     Poker Beats, Brags, and Variance     YouTube Podcasts & Twitch Streams     General Poker Discussion Online Poker Sites & Marketplaces     Online Poker Sites         Discussion of Poker Sites         Global Poker         BetOnline.ag     Coaches & Schools         Seeking Coaching         Study Groups         General Coaching & Schools Discussion     Staking         Offering Stakes         Seeking Stakes         Selling Shares - Live         Selling Shares - Online         Staking Rails     Poker Software         General Software Discussion     General Marketplace     Transaction Feedback & Disputes Live Poker     Las Vegas Lifestyle     Venues & Communities     Tournament Events         WPT.com     Home Poker     Casino & Cardroom Poker Poker Strategy     Live No-Limit Hold’em Cash     Online No-Limit Hold’em Cash     No Limit Tournaments         Heads Up SNG and Spin and Gos     Mid-High Stakes MTT     Omaha         Omaha/8     Other Poker Games         Mid-High Stakes Limit         Micro-Small Stakes Limit         Stud     Psychology     Books and Publications     Poker Theory & GTO     Beginners and General Questions 2+2 Communities     Other Other Topics         OOTV     The Lounge: Discussion+Review     BBV4Life         omg omg omg     House of Blogs Sports and Games     Sporting Events         Single-Team Season Threads         Fantasy Sports     Sports Betting     Fantasy Sports         Sporting Events     Wrestling     Golf     Chess and Other Board Games         Backgammon Forum hosted by Bill Robertie.     Video Games         League of Legends         Hearthstone     Puzzles and Other Games Other Topics     Politics and Society     Business, Finance, and Investing     History     Health and Fitness     Travel     Science, Math, and Philosophy     Religion, God, and Theology     Laughs or Links!     Probability     Other Gambling Games     Computer and Technical Help Two Plus Two     About the Forums

All times are GMT -4. The time now is 09:25 PM.

 Contact Us - Two Plus Two Publishing LLC - Privacy Statement - Top