Specific Flop Odds with Pocket Pairs
01-24-2017
, 09:19 AM
first post: correct
second post: when you're using this way of counting, the problem is that you need to count all the permutations. that can lead to confusions that don't appear if you use combinations.
if you have 1 overcard and no paired board: first you can choose 1 overcard (28), then 2 unpaired undercards (24 and 20). but what if you want to choose the undercards first and then overcard? you could write it like this:
28/50 * 20/49 * 16/48 + 20/50 * 28/49 * 16/48 + 20/50 * 16/49 * 28/48, which is just 3 times the same thing. you can choose the overcard first and then 2 undercards, 1 undercard + 1 overcard + 1 undercard or 2 undercards + 1 overcard.
second case (2 overcards) is similar, you can have undercard first, in the middle or in the end.
third case, which confuses you, you're choosing 3 overcards from group of overcards, so it needs to be 28*24*20 and that's the only way. maybe now you see why i told you in the first post why combinations are preferred for this
though that doesn't mean you can't use this "simple" counting, in fact for some situations it's easier to do it that way. that is usually if the order of things matters. for example, suppose flop is "abc" and you want the board after the river to be unpaired ("abcde"). in this case, simple counting is the first thing that should come to mind for solving this because the cards arrive in sequence one after another.
so, recap: in general, when sequencing matters use simple counting and if it doesn't use combinations.
second post: when you're using this way of counting, the problem is that you need to count all the permutations. that can lead to confusions that don't appear if you use combinations.
if you have 1 overcard and no paired board: first you can choose 1 overcard (28), then 2 unpaired undercards (24 and 20). but what if you want to choose the undercards first and then overcard? you could write it like this:
28/50 * 20/49 * 16/48 + 20/50 * 28/49 * 16/48 + 20/50 * 16/49 * 28/48, which is just 3 times the same thing. you can choose the overcard first and then 2 undercards, 1 undercard + 1 overcard + 1 undercard or 2 undercards + 1 overcard.
second case (2 overcards) is similar, you can have undercard first, in the middle or in the end.
third case, which confuses you, you're choosing 3 overcards from group of overcards, so it needs to be 28*24*20 and that's the only way. maybe now you see why i told you in the first post why combinations are preferred for this
though that doesn't mean you can't use this "simple" counting, in fact for some situations it's easier to do it that way. that is usually if the order of things matters. for example, suppose flop is "abc" and you want the board after the river to be unpaired ("abcde"). in this case, simple counting is the first thing that should come to mind for solving this because the cards arrive in sequence one after another. so, recap: in general, when sequencing matters use simple counting and if it doesn't use combinations.
01-24-2017
, 01:40 PM
Quote:
Odds Flop has at least 1 Over card and does not pair
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Example Hand 77:
(28/50)*(44/49)*(40/48) = 41.9%
I thought I might need to multiply the result by 3 as in the calculations provided for quads down to two pair but doing so results in some hands coming out to over 100% which really confuses me.
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Example Hand 77:
(28/50)*(44/49)*(40/48) = 41.9%
I thought I might need to multiply the result by 3 as in the calculations provided for quads down to two pair but doing so results in some hands coming out to over 100% which really confuses me.
You have an arrangement Oxx and you also wish to count xOx and xxO.
But the (44/49) and (40/48) in your expression already allow the x's to be O's. Multiplying by 3 counts the specific arrangement OOU twice because Oxx and xOx each counted it once (because the first x in each was allowed to be an O). Similarly, it double-counts OUO and UOO. It triple-counts OOO.
The over-counting is why you get a result greater than 100% (but over-counting doesn't always break the 100% barrier, and that's when it can dangerously go unnoticed). To fix this, you can use the inclusion-exclusion technique (the principle of inclusion-exclusion, or PIE). The following would have worked:
3*(28/50)(44/49)(40/48) - 3*(28/50)(24/49)(40/48) + (28/50)(24/49)(20/48) = 24/35 =~ 68.57%
1) We take your original expression and multiply by 3 to count the 3 arrangements of Oxx.
2) By doing that, each of the 3 OOx arrangements is counted twice, so we subtract each OOx arrangement once.
3) The first step triple-counted the OOO possibility, because OOO is counted by each Oxx. The 2nd step subtracted it 3 times, because each subtraction of an OOx also subtracted OOO. Thus, by now, OOO isn't counted at all, so we need to add it back in.
PIE can come into play whenever there are overlapping possibilities. The 1st card being an over is not mutually exclusive of the 2nd card being an over, so adding the two possibilities double-counts the possibility of both happening. On the other hand, if you were talking about a specific card like the A
, the possibilities would be mutually exclusive, so you could multiply or add without a care in the world.PIE is a good approach to this problem, and is sometimes a far superior approach, but here, what you did in your latest post (adding non-overlapping cases like md did) was equally good. However, we can improve the execution of both approaches. As md alluded to, your strings of fractions are the same as using permuations. The above is equivalent to writing:
[3(28*44*40) - 3(28*24*40) + (28*24*20)] / (50 P 3)
Another example of fractions vs perms: the probability of an all-diamond flop is (13/52)(12/51)(11/50), but that's the same as (13 P 3) / (52 P 3)
Not only is the latter is easier to type, but calculators have the permute operation. Like you, I began with multiplying fractions. Once I realized they were permutations, the transition to combinations was painless: just replace some P's with C's, and this spares you the extra work of multiplying by 3's and such (when you're not using PIE, that is). When order doesn't matter, you can use perms, but doing so requires you to count all the arrangements in the numerator so that they cancel out with the denominator. With combinations, you don't have to worry about arrangements anywhere. Also notice that nCr = nPr / r!, so combos are just perms with the arrangements "divided out of them".
For a probability problem where order doesn't matter, you'll get the same answer with perms and combos because order either cancels out (perms / perms), or is not counted at all (combos). Normally, it's simpler to use combos in that situation. However, imo this problem is an exception because of how the paired boards aren't allowed.
Here's what the PIE solution looks like with...
permutations: 28[3(44-24)40 + 24*20] / (50 P 3)
combinations: {3*42[28(11C2) - (7C2)40] + (7C3)43} / (50 C 3)
Here's what your latest solution looks like with...
permutations: 28[3(24+16)20 + 24*20] / (50 P 3)
combinations: {42[28(5C2) + (7C2)20] + (7C3)43} / (50 C 3)
Quote:
On a side note I want to thank forum members heehaww and whosnext because I've learned so much from your posts in the last few days.