One for the mathematicians.

[GR1ZZL3R]

I've never really taken much notice of the challenges while I'm playing because they are mostly irrelevant and don't require any action from the player.

 

This one I suppose you could see more flops more in hope than anything else but it got me thinking.

Firstly, I don't see that many flops in the early levels, in nine minutes there may be 15-20 flops of which I might see 3-4. What are the odds of seeing two sevens on the flop, my odds are probably about 4/1 or 5/1 of being in the hand when it happens, so roughly how many games would I have to play to complete this challenge three times for the points. :Wonder: I have an actual figure but just wondered on the odds for a specific pair, the next challenge is a pair on the flop but they must be in 1st and 3rd position. Does anyonyone read these or take any notice of them?

 

[MadAdo]

@GR1ZZL3R  it should be something like this:

Pair a certain hole card is 5:1
Pair either hole card is 2.45:1 preflop

The method Practical Poker math uses is find 2 card combo that do not contain ace multiply by number of remaining unseen aces resulting in number of 3 card combos that contain at least an ace

subtract 2 cards in hole and 3 remaining aces leaving 47 unseen none of which are aces then use comb(47,2) (2 card combinations of 47 cards)
(47*46)
-------
(1*2) = 1081, 2 card combos that contain no ace
3 card flops that will flop an ace is 3 (the remaining aces) * 1081 (the 2 card combos that have no ace)
3*1081 = 3243 - the combos that have an ace

The total 3 card combos is
(50*49*48)
----------
(1*2*3) = 19600

So 19600-3243 = 16357, the willnots
willnots:wills
16357:3243 is
5:1
2 unpaired hole cards >>> pair a specific hole card

[Brocky]

Isn't the first and third card being the same just 3 in 51?

For the 2 x 7s on the flop, I think I've worked out that it's roughly 1.5%. 1 in 13 times the first card will be a 7 of which either of the next 2 will be a 7 about 1 in 8 = 1 in 104. Then you add a 1 in 196 chance that the 2nd and 3rd cards are a 7 when the first one isn't.

Disclaimer - I probably don't know what I'm talking about.

[MadAdo]

---

@Brocky wrote:

Isn't the first and third card being the same just 3 in 51?

 

For the 2 x 7s on the flop, I think I've worked out that it's roughly 1.5%. 1 in 13 times the first card will be a 7 of which either of the next 2 will be a 7 about 1 in 8 = 1 in 104. Then you add a 1 in 196 chance that the 2nd and 3rd cards are a 7 when the first one isn't.

 

Disclaimer - I probably don't know what I'm talking about.

---

Exactly! 3 in 51 seems legit to be first/third cards are the same rank!

FOr the 2x7s on the flop:

first card is 7 (4 out of 50 - assuming that you did not hold any 7 !!!) than 2nd or 3rd are 7 as well: 3/49 + (3/48 just roughly) => 1/101 :D :D :D

first card is no 7 and 2nd and 3rd are both 7s: 4/49 * 3/48 = 1/196 @Brocky  you seems legit again 🆒

 

It seems like whenever you play often you know the basics! :popeye:

[GR1ZZL3R]

We could get really deep into this but that's not my intention, actually not playing poker this week has made my mind wander onto many other things. In the first few levels with 5 players, some of the time some of the sevens will already have gone as pocket cards, so that would skew the maths a bit. I just wondered if anyone knew the actual odds of a specific pair being dealt on the flop. 

If we take @Brocky's thoughts as averaging 1 in 150, and I see only 1 in 5 flops that would suggest 750 flops for me to see one with a pair of sevens. At say 25 flops per first three levels, which could be a bit high, I would have to play 30 games to see this once and 90 games to see it three times, which actually is extremely close to what happened, though only a sample size of one attempt.

Maybe you do know what you're talking about @Brocky, I'm not sure I do, just thinking out loud. 😃

[MadAdo]

@GR1ZZL3R  dear mate, since you wrote that: "Firstly, I don't see that many flops in the early levels, in nine minutes there may be 15-20 flops of which I might see 3-4" and I figure out that you play at least as many as 95 5-man S&G I assume you see somewhat 95*3,5 = 332 flops you COULD for sure complete that pair 77 on flop major mission 3 times!!! => we figured-out that your odds are nearly 1/100 to achieve it once! Glad to serve/answer this one but know I do not provide PRECISE numbers ... what a shame 😠

[GR1ZZL3R]

Nicely summed up @MadAdo I was writing my answer before I saw your last one, seems to be on the right track.😃

[GR1ZZL3R]

The numbers are near enough @MadAdo to confirm what I suspected, that it would take a fairly long time.👍

[MadAdo]

---

@GR1ZZL3R wrote:

We could get really deep into this but that's not my intention, actually not playing poker this week has made my mind wander onto many other things. In the first few levels with 5 players, some of the time some of the sevens will already have gone as pocket cards, so that would skew the maths a bit. I just wondered if anyone knew the actual odds of a specific pair being dealt on the flop. 

If we take @Brocky's thoughts as averaging 1 in 150, and I see only 1 in 5 flops that would suggest 750 flops for me to see one with a pair of sevens. At say 25 flops per first three levels, which could be a bit high, I would have to play 30 games to see this once and 90 games to see it three times, which actually is extremely close to what happened, though only a sample size of one attempt.

Maybe you do know what you're talking about @Brocky, I'm not sure I do, just thinking out loud. 😃

---

@GR1ZZL3R  1 in 150 is roughly still your odds (no matter what you getting dealt ...) - you cannot multiply it by 5 for any reason as it make no sense here !!! If you play 1 flop of 5 - you still see as much as 150 flops to be able to complete challenge at least once (probably)!

[MadAdo]

Calling all you math-graded members: does anybody knows PRECISE numbers ? e.g. no roughly calculations just give us pure formula-e ? 👼

+ as well for 1st/3rd card of the same rank -> 3/51 is not as precise as you can figure-out (2nd card could be also the same as the first/third one) ...

[GR1ZZL3R]

Sorry to disagree @MadAdo but if something only happens once in 150 flops and I'm only seeing 30 of those, surely the odds are that I won't  see that one specific flop, therefore I have to see five times as many as normal to see it happen once! :wonder:

Edit: I don't mean see as be in the hand, only as an observer.

[FeelsBadMan]

idk why you need the precise number since it really doesn't matter, you can't do much to increase your probability other than playing. Seeing more flops increases it but it's not really worth it, just play your game and it will come. 

The exact math is ((4/52)*(3/51)*(48/50)*3) + ((4/52)*(3/51)*(2/50)) = 0.013212669683258 or 1.3212669683258%  or 1 in 75.684931506849 flops per pair of 7 on the board. There might be a more elegant way to do this but who cares. Didn't remove your holecards because we're not searching for something relative to them so all cards are relevant.

Ran a quick simulation to check my result and out of 132600 possible simulated flops I found 1752 flops that have at least 2 7's , which leads to the exact numbers above. 

[GR1ZZL3R]

Thanks @FeelsBadMan  I never needed precise numbers, as I said I don't usually take any notice as they are just random flops which I can't affect the game play of, just wanted an idea of the odds. Now I know the exact odds I don't think it will affect my play in any way, but thanks again.😃

[FeelsBadMan]

@GR1ZZL3R Sorry, my reply was confusing cause I kinda merged everyone in the thread and gave a general response, which in conversation doesn't really work that well :laugh: . @MadAdo asked for the precise numbers, not you 😃 

[MadAdo]

I like MATH never mind on elementary school nearly never managed to got the best rating (I like doing small mistakes as well) ... :puking:

So I do my homework and here is what I think the exact calculation should looks like:

quest: Probability for 'at least' pair of 77s on the flop:

(4/52)*(3/51)*(48/50) = 0.00434389140271493212669683257919 // first 7, second 7, third any other than 7 (X)

(4/52)*(48/51)*(3/50) = 0.00434389140271493212669683257919 // 7X7

(48/52)*(4/51)*(3/50) = 0.00434389140271493212669683257919 // X77

(4/52)*(3/51)*(2/50) = 1.8099547511312217194570135746606e-4 // 777

Sum it up:

4	3	48
52	51	50
0.0769230769	0.0588235294	0.96	0.00434389140271493
4	48	3
52	51	50
0.0769230769	0.9411764706	0.06	0.00434389140271493
48	4	3
52	51	50
0.9230769231	0.0784313725	0.06	0.00434389140271493
4	3	2
52	51	50
0.0769230769	0.0588235294	0.04	0.00018099547511312
			0.01321266968325790

Prove that Mr. @FeelsBadMan  do it correctly :Ok: (with that assumption that we don't care if we were dealt any 7 as our holecard)!!!

...

I would like to 'understand another approach' ... (could anyone of you correct it = I don't see what is wrong with it :Sad: ?!?)

Definition: Combinations are using ® items from (n) possible choices without repetition and order is not important!

comb(r,n)=n! / (r!*(n-r)!)

Let's think that we are interested in all combinations of at least pair of 7-ens on the flop ... so those are combinations 2 from 4 => 6 possible combinations * any rest card from the deck (so 50) => 6*50=300 possible flops which contains at least pair of 77s

Total possible flop combinations (we don't care our holecards): Comb(3, 52) = 52! / (3! * 49!) => 22100

Just devide numbers we get 300/22100 = 0.01357466063348416289592760180995

-> but this is not exactly the same number as above and I cannot see WHY ? :Annoyed: Does anybody of you ? ...

[FeelsBadMan]

The problem with grouping combinations like that is that some of them will be doubles (permutations of the same board) 

if you do C(4,2) *C(48,47) + C(4/3) = 6*48 +4 = 292

22100/292 = 75.684931506849 flops so the same result as above.

 

[MadAdo]

@FeelsBadMan  Many thank you - you are right! so surprisingly ... many thank you man, I can sleep good now! 😀

[MadAdo]

---

@FeelsBadMan wrote:

The problem with grouping combinations like that is that some of them will be doubles (permutations of the same board) 

if you do C(4,2) *C(48,47) + C(4/3) = 6*48 +4 = 292

22100/292 = 75.684931506849 flops so the same result as above.

 

---

@FeelsBadMan  yep man! I think that we are talking about permutations of 777 flop this time and those could be really exactly 8.

Original combinations:

1) 7h7s7d

2) 7h7s7c

3) 7h7c7d

4) 7s7c7d

Possible (double) permutations:

1A) 7h7d7s or 7d7h7s (same "pair" of first 2 not matter here)

1B) 7s7d7h or 7d7s7h

 

2A) 7h7c7s or 7c7h7s

2B) 7c7s7h or 7s7c7h

 

3A) 7h7d7c or 7d7h7c

3B) 7d7c7h or 7c7d7h

 

4A) 7s7d7c or 7d7s7c

4B) 7c7d7s or 7d7c7s

... or something like that :Angel:

[GR1ZZL3R]

This thread may have gone slightly deeper than I intended but nevertheless still very interesting, thanks for your input. My original guess ( purely in my head, no calculator ) was about 90 games to do this challenge, which was actually done in 87, wether that was by good maths or good guessing I really don't know. Looking at the figures I'm guessing the guess was slightly off. 😃

The original point was that I don't think these challenges affect the game play at all except in some minor way, "fold a hand preflop containing a 2", so maybe they ought to be renamed. Any thoughts? :wonder: