# is this math right? (this hand one in 2.7 billion??)

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Well, let's see here. The first thing to calculate is the probability of the starting hands being dealt:let x = event that a player is dealt JJ without the J of clubslet y = event that a player is dealt TcXclet Xc = any club 9 or lesslet z = event that a player is dealt Kx where neither card is a clublet S = event that x,y, and z occur simultaneouslyP(x) = 3/52 * 2/51P(x) = 0.05769231 * 0.03921569P(x) = 0.00226244P(y|x) = 1/50 * 8/49P(y|x) = 0.02 * 0.16326531P(y|x) = 0.00326531P(z|x&y) = 3/48 * 36/47P(z|x&y) = 0.0625 * 0.76595745P(z|x&y) = 0.04787234P(S) = P(x) * P(y|x) * P(z|x&y)P(S) = 0.00000035So the odds are about 2,857,141:1 against those three specific starting hands being dealt on a given deal. Now, given that, we only need to know the probability of the specific board.let B = event that the board reads KcJpJqAcQclet Jp = a jack of either suit not dealt in xlet Jq = the remaining jackP(B|S) = 1/46 * 2/45 * 1/44 * 1/43 * 1/42P(B|S) = 0.02173913 * 0.04444444 * 0.02272727 * 0.02325581 * 0.02380952P(B|S) = 0.00000001P(S&B) = 0.00000035 * 0.00000001P(S&B) = 0.0000000000000035So, if my first year stats still serves me the odds against the specific hand occuring are about  285,714,285,714,285:1I'm sure I've done something wrong so if someone would correct me, I'd appreciate it.
I see Skalansky has posted :shock: :shock: Now if he could post in english, it would be very helpful.
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I just realized what I posted contains a mistake...on the odds of having the quad and flush hands, where I use 2/(48*47), it should be 2/(47*46), and where I use 2/(46*45), it should be 2/(45*44)...changes the final result to 1 in 160 million, not 1 in 175 million.

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Well, let's see here. The first thing to calculate is the probability of the starting hands being dealt:let x = event that a player is dealt JJ without the J of clubslet y = event that a player is dealt TcXclet Xc = any club 9 or lesslet z = event that a player is dealt Kx where neither card is a clublet S = event that x,y, and z occur simultaneouslyP(x) = 3/52 * 2/51P(x) = 0.05769231 * 0.03921569P(x) = 0.00226244P(y|x) = 1/50 * 8/49P(y|x) = 0.02 * 0.16326531P(y|x) = 0.00326531P(z|x&y) = 3/48 * 36/47P(z|x&y) = 0.0625 * 0.76595745P(z|x&y) = 0.04787234P(S) = P(x) * P(y|x) * P(z|x&y)P(S) = 0.00000035So the odds are about 2,857,141:1 against those three specific starting hands being dealt on a given deal. Now, given that, we only need to know the probability of the specific board.let B = event that the board reads KcJpJqAcQclet Jp = a jack of either suit not dealt in xlet Jq = the remaining jackP(B|S) = 1/46 * 2/45 * 1/44 * 1/43 * 1/42P(B|S) = 0.02173913 * 0.04444444 * 0.02272727 * 0.02325581 * 0.02380952P(B|S) = 0.00000001P(S&B) = 0.00000035 * 0.00000001P(S&B) = 0.0000000000000035So, if my first year stats still serves me the odds against the specific hand occuring are about  285,714,285,714,285:1I'm sure I've done something wrong so if someone would correct me, I'd appreciate it.
oh i get it now
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Now if he could post in english, it would be very helpful.
He is posting in English. You just failed 9th grade math class, and then English 101, 201, 302, etc. So sorry. Try babelfish translations from "slightly above average intelligence" to "idiot troll post pad scavenger".
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I guess you are assuming two players. 5 community cards. 2 hole cards.Of the five community, 3 must be broadway and of one suit. Total combinations? C(5,3)*4 = 40. [the 4 was for each suit].Several ways to go after this on the board:a.pairs below tens(99-22). Combinations? 8*[C(4,2)] = 48b.pair of broadway cards including the royal card. 2*[C(3,1)] = 6.c.pairing one of the already existing broadways, and any non royal card: 3*3*38=342.d. pairing royal card, and another royal card: 3*3*2=18.e.double pairing the royal cards: 9*6= 54.f.putting a set of one of the royal cards on board= 3*3=9.hole cards:a. royal holes * quad holes = 1 * 1 = 1b. royal * quad = c[44,1]*1 = 44.c. royal * quad = 1*1=1d. 44*1=144e. 1*2=2f. 1 * 44 = 44.combos of [letter]:a. 1*48=48b.6*44=264c. 342*1=342d.18*44 = 792e. 54*2=108f.9*44=396sum total = 1950total combos of community cards = C(52,5) = 2,598,960total combos of hole cards = C(47,2) * C(45,2) = 1070190probability of quads and royal = (combos of 3 royals)*(sum of each letter combo)/combos of community cards/ combos of hole cards= (40)*(1950)/(2,598,960)/(1070190) = 2.8044E-8 = 2.8044 out of 10^8 = = one out of 35.66 millionSo for 2 players, 2 hole cards, 5 board cards, it's one out of 35.66 mil. Is everything copamalecetic with the calculations? I think this is right...

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I guess you are assuming two players. 5 community cards. 2 hole cards.Of the five community, 3 must be broadway and of one suit. Total combinations? C(5,3)*4 = 40. [the 4 was for each suit].Several ways to go after this on the board:a.pairs below tens(99-22). Combinations? 8*[C(4,2)] = 48b.pair of broadway cards including the royal card. 2*[C(3,1)] = 6.c.pairing one of the already existing broadways, and any non royal card: 3*3*38=342.d. pairing royal card, and another royal card: 3*3*2=18.e.double pairing the royal cards: 9*6= 54.f.putting a set of one of the royal cards on board= 3*3=9.hole cards:a. royal holes * quad holes = 1 * 1 = 1b. royal * quad = c[44,1]*1 = 44.c. royal * quad = 1*1=1d. 44*1=144e. 1*2=2f. 1 * 44 = 44.combos of [letter]:a. 1*48=48b.6*44=264c. 342*1=342d.18*44 = 792e. 54*2=108f.9*44=396sum total = 1950total combos of community cards = C(52,5) = 2,598,960total combos of hole cards = C(47,2) * C(45,2) = 1070190probability of quads and royal = (combos of 3 royals)*(sum of each letter combo)/combos of community cards/ combos of hole cards= (40)*(1950)/(2,598,960)/(1070190) = 2.8044E-8 = 2.8044 out of 10^8 = = one out of 35.66 millionSo for 2 players, 2 hole cards, 5 board cards, it's one out of 35.66 mil. Is everything copamalecetic with the calculations? I think this is right...

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when this happened, I laughed hysterically.... then quit poker for a week. I'd say I handled it pretty well!At the time I figured it to be 1-24million though I dont have my work anymore, but Im thinking its higher than that.Maybe it would be easier to find the odds of pocket pair drawn to quads dying to a straight flush, then the odds of a straight flush being a royal flush?

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Now if he could post in english, it would be very helpful.
He is posting in English. You just failed 9th grade math class, and then English 101, 201, 302, etc. So sorry. Try babelfish translations from "slightly above average intelligence" to "idiot troll post pad scavenger".
Hah... let the truth be told.
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im a poker math guy.the odds are low.its as irrelevant as possible.

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I'm in your elite company OP. I too had quads taken down by a royal, though I turned my quads vs. flopping them. And I was holding a non-paired pocket (QJ) vs. your pocket (JJ). The odds are the stuff that could give MIT grads a headache. What I found extraordinary in my case was the fact that this hand occured when we were playing 3-handed at the end of a tournament. The hilarious part was the reaction of the 5 people left in the room. When my opponent went all-in on the river, which gave him the royal, I simply said in a rather dry manner as I called, "I have the queen." He flipped over his K J on the board of Q Q T Q A . I was stunned, as was everybody else, but not less than 10 seconds later we were playing the next hand, and most of the commotion was over. Everybody was talking about the royal, but nobody seemed to care about the quads losing. I still find it one of the quirky moments in my life that seems surreal.

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I think everyone has had this happen to them.so its much lower how many hands have you ever played.its not even 1 in 50 million.think about itits not even 1 in a million.its not even 1 in 100k.we have all seen it happen to others and ourselves, so the number is so far off its just hilarious.my friend beat quads with a runner runner royal flush like 3-4 weeks ago. :cry: ##### Share on other sites
since online poker is rigged, conventional math does not apply.
LMAO did you math freaks factor in PayforUSC's "online poker is rigged" constant. If you did, I bet the odds would be a lot less from 1 in eleventy brazillian to something like 1 in 5.
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Debug error. Double posted.

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Edit: double post

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Now if he could post in english, it would be very helpful.
He is posting in English. You just failed 9th grade math class, and then English 101, 201, 302, etc. So sorry. Try babelfish translations from "slightly above average intelligence" to "idiot troll post pad scavenger".
Mr Jeepster Sir,Why is it that your posts seem to turn every topic into an intelligence contest? What is your deal? I cruise through the forum infrequently, yet I always seem to see a post from you flaming away at someone. The guy was making a light hearted comment... and , as usual, it somehow ends up being about how "smart" you are, and how stupid someone else is.Try happiness. You might like it. Yeah, I know... I'm an idiot too...At least I'm not an insecure prick.
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This thread made me so dizzy, I didn't even read it all.However, with online poker being rigged factored into the math, I think the odds of this happening is about 1 in 3.To the OP, did the suckout artist see your quad jacks before the flop???

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Isn't it easier just to round to aprox zero? It's never going to happen to you again in your lifetime so zero is a good estimation.george

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Now if he could post in english, it would be very helpful.
He is posting in English. You just failed 9th grade math class, and then English 101, 201, 302, etc. So sorry. Try babelfish translations from "slightly above average intelligence" to "idiot troll post pad scavenger".
Mr Jeepster Sir,Why is it that your posts seem to turn every topic into an intelligence contest? What is your deal? I cruise through the forum infrequently, yet I always seem to see a post from you flaming away at someone. The guy was making a light hearted comment... and , as usual, it somehow ends up being about how "smart" you are, and how stupid someone else is.Try happiness. You might like it. Yeah, I know... I'm an idiot too...At least I'm not an insecure prick.
Jeepster is good stuff.
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Now if he could post in english, it would be very helpful.
He is posting in English. You just failed 9th grade math class, and then English 101, 201, 302, etc. So sorry. Try babelfish translations from "slightly above average intelligence" to "idiot troll post pad scavenger".
Mr Jeepster Sir,Why is it that your posts seem to turn every topic into an intelligence contest? What is your deal? I cruise through the forum infrequently, yet I always seem to see a post from you flaming away at someone. The guy was making a light hearted comment... and , as usual, it somehow ends up being about how "smart" you are, and how stupid someone else is.Try happiness. You might like it. Yeah, I know... I'm an idiot too...At least I'm not an insecure prick.
Jeepster is good stuff.
Maybe so, but more often than not, his posts are centered around ridicule.