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

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3 handed at the end of a 30+3 SNG on PacificI pick up JJ on the button and raise to 3x BBFlop comes KcJcJh to nail me with quads.I check, SB bets, BB raises, I call, SB calls.Turn comes Ac, SB goes all in, BB calls, I call.SB shows 10c / 7c, BB shows K - ragRiver comes Qc to beat my quads with a royal.Now thats the first royal I've encountered - not all too surprising at 1 in 649,740 hands.Quads, however, are 1 in 4,165.So is losing quads to a royal really (649,740)(4165) = 1 in 2,706,167,100??

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So is losing quads to a royal really (649,740)(4165) = 1 in 2,706,167,100??
no
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i don't think so, but i've never been much of a math guy.

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since online poker is rigged, conventional math does not apply.

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I don't know the exact answer to this question, but I know your basic numbers aren't right. The odds you are using are for 5-card hands (e.g. 5-card stud), not 7-card hands as in Hold-em. The chances of getting quads or a straght flush are thus MUCH higher than you estimate.In about 25,000 hands I have on poker tracker, I have ended up with quads on 0.14% of hands where the river has been dealt (regardless of whether I stayed it). Thus suggest to me that the chance of hitting quads is better than 1/1000. I know the chance of FLOPPING quads if you have a pocket pair already is 1/400.

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I don't care about the math. You played it horribly though.

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I don't care about the math. You played it horribly though.
No sw here, how would you have played it? I don't mind the slowplay since after the flop he is 99.6% to win and after the turn he's 97.6% to win. If there was a ever a time for a slow play, this is it. Just a really crappy beat. I guess he could have possibly come over the top on the flop, but why? He doesn't want to get isolate so what is the point? He want's to let the draw hit.
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Yeah, I don't see any problem with the way he played it. You have to slowplay the flop, because you want people to draw to their straights and flushes and what not...especially since other people were betting for him. The one thing I don't understand is that he said he checked on the flop, and then the SB bet. If he was on the button, why was he acting first on the flop?

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Yeah, I don't see any problem with the way he played it. You have to slowplay the flop, because you want people to draw to their straights and flushes and what not...especially since other people were betting for him. The one thing I don't understand is that he said he checked on the flop, and then the SB bet. If he was on the button, why was he acting first on the flop?
I noticed that too, but forgot to mention it in my post.
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i don´t know Pacific but on Stars u would have been at least a 2-1 dog to win the hand

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Ask Phil Gordon. He's the math wizard.

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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.

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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.
This looks rather impressive to me, so I'm quite ready to just say that you're right.
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oh no, you are completely right in your math semaj, but i dont think that helps the question of how often does any quads lose to a royal flush...
Well it does tell you how often the situation the OP encoutered will happen. I guess I'll have to go back and calculate the odds of any quads losing to any royal now. I'll return with an answer shortly.Edit: Well at 11:50 this have proved to be just too daunting a task for my tired mind. If someone hasn't calculated it by tomorrow morning I might try it again.
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without really looking at it the number will be too high because its figured out for the exact cards and suits. in the real problem it doesnt matter if the suit is clubs heart spades or diamonds. it also doesnt matter the the other players card is as long as he has the 10 of the suit that the royal will be made with.

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Well, upon further consideration the odds of quads losing to a royal depend on the quads, i.e. it is more likely that your opponent has a royal when you have quads queens than when you have quad 4's. There's just too much math involved for me to do tonight.

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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.
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Ask Phil Gordon. He's the math wizard.
I think you mean David Skalansky or Matt Matros
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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.
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