I been playing poker for over 40 years-bought my house on playing stud, no offenses, but i think i have seen more poker hands in my sleep then you have live:)…Good luck my friend.
Nothing you say offends me I don’t take people like you very seriously. You have no clue what you are talking about. Again, thank you for participating in this discussion, you added so many laughs to this discussion.
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lihiue, your Royal Flush analogy has nothing to do with this conversation and is irrelevant. The odds of making pocket aces for instance is 0.45%. The odds of one player getting 9-7 is 0.905%, those odds drop for each of the other players. The odds of three players getting 9-7 five handed is astronomical, I never said it is impossible, people win the lottery for instance. The odds of making a straight in hold’em is less than 5%. So taking the odds of three players in a 5 handed game being dealt a 7 and a 9 is very little and then making the straight is is EXTREMELY LOW. The probability of these improbable hands are quite frequent on this site, this was just one example.
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Hi,
I guess I am a Replay apologist. I work here, so please dont hold that against me.
The point I would like to make is that 9-7 is not the only way to make a straight.
Shouldn’t the question being asked be:
'What are the chances three players will get ANY same hole cards AND THEN make a straight?
If you are insisting the hole cards have to be exactly those two ranks (9 and 7), it obviously makes it much harder.
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I calculated the exact odds:
3 players being dealt a 9 & 7 is 1 in ~3,000,000
3 players being dealt AsTc, KhQh and 6c6s is 1 in ~300,000,000 (I was out by a factor of six before, it’s actually 96 times less likely than 9 & 7)
3 players being dealt the same cards happens about 1 in 40,000 hands
Someone who has played millions of hands, which I would think is quite likely over a 40 year career, would definitely have seen 3 players be dealt the same cards many times.
I wouldn’t describe the odds as astronomical, but low enough that you shouldn’t be seeing it frequently (frequently is different to a lot, depending on how many hands you play). You’ve only given one example though, so I have nothing to go on here.
The royal flush analogy was only supposed to illustrate that any specific hand is extremely unlikely and has the exact same odds as a royal flush. One just seems special to us, but mathematically there’s nothing unique about it.
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Chase the River, the odds would be the same as the three players hole cards must match and then a straight must occur. It could be J-8 for example, once one player has a card, the other two would need that same card eventually and when the first player was dealt the second card the other two would also need to have the 2nd card eventually. The order can be different of course, 7 or 9 first or second. Then of course a straight must be dealt using both those cards. I was asking for the odds of all those events occurring but it seems my question triggered people. Didn’t realize math questions could get so hostile, oh well.
lihiue, where is your math for 3 players being dealt the same cards happen in 1 out of 40,000 hands when 5 handed? Also, you are assuming the player you are defending (who claimed this scenario happens A LOT) is not lying to you and you are assuming that player has played millions of hands. You are also not taking account there were only 5 players in the hand, that player would have had to play “millions” of hands 5 handed. My question was not just for 3 players being dealt the same hand, it also had the component of a straight occurring after 3 players were dealt the same value of cards. That doesn’t happen “a lot”, just saying.
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The number of extra players in the hand does not affect the odds at all. We don’t care or know what cards anyone except the 3 players have, so anyone else’s cards may as well be still in the deck.
The math works out like this:
Player A can be dealt any card first, so that’s 52/52 possibilities. There are 48 possible cards Player A could be dealt for their second card and 51 cards left. (We have to exclude the 3 cards that would pair the first card).
Player B now has 6 possibilities for their first card (3 matching player A’s first card & 3 matching the second) - there’s 50 cards remaining. Whichever card they get for their first, there will be 3/49 options for the second.
Player C then has 4/48 and 2/47 options for their two cards.
(51 * 50 * 49 * 48 * 47) / (48 * 6 * 3 * 4 * 2) = 40782.291666…
I was also assuming that bill8888 was only referring to people being dealt the same hand and possibly including cases where just 2 players got deal the same cards and made an unlikely hand, because why would anyone keep track of exactly the situation you describe? If you really need to know the odds though, just multiply the above by the odds of making a straight with any 2 cards.
To be honest, I don’t really care about the anecdotal side of this, the only thing I was trying to show is that if you think feelmysins was wrong to state:
The odds of that hand happening is exactly the same as any other hand happening. Each and every hand is equally likely to be dealt at any given time. That’s how random odds work.
then you have misunderstood what they were saying. I’m not sure I can explain it much better if you still feel that way, but I would challenge you to provide 3 hands that you think should occur far more often if so.
If your position now is just that 3 players being dealt the same cards and making a straight is rare, then I fully agree.
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lihiue, The correct way to calculate the odds of 3 players having the same two hole card values would be (8/52)(7/51)(6/50)(5/49)(4/48)(3/47) which would give you approximately 1 in 727,100. Then take the odds of straight happening after that event (less than 1 out of 20 times) and you have astronomical numbers. The same as when you calculate getting pocket Aces (4/52)(3/51) which is 1 in 221. In fact, two players being dealt pocket Aces is 3 times more likely than 3 players being dealt the same hole card values.
If you were assuming hands where two people have the same hole cards that is a complete different discussion than what is being discussed here.
You are making assumptions of what the other players are saying based on your own admission. So saying I “misunderstood” when you are basing your answers on assumptions leaves you on shaky ground in your responses. Have to keep it real.
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Loved playing cash games against poster and odds guy on here ( i remember in the casino where a guy had a pencil and paper and wrote EVERY hand down) lol, Fun to see you guys here 2:) G L at the tables, do not forget your calculator before sitting down here guys:).
P.S. I was playing at Rivers casino last week-the casino here in Pittsburgh P.A. in a very small game (1-3) and the PPL. on the table next to me ALL won a lot of money on a bad beat. A guy lost with 4 Aces to a royal, He got 350k, winner of hand got 270k and the other ppl. at the table got 45k each, hands happen, look it up if you think i am joking on this:).
I think I remember seeing that in my news feed. It was a record breaking jackpot, or something like that?
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My math is for three players having the same value cards, LITERALLY the point of the entire post. My math is not wrong, your assumptions and your math are incorrect and I’ll explain why. Your math is assuming each player gets two cards at a time when they don’t, that is why your 8/52 and 4/51 is wrong. The first player is 8/52, the second player is 7/51, the third player is 6/50, then the figures change after that depending on what each player got. You did not account for that, that is why your math assumptions and math are incorrect. In fact if you calculate the incorrect numbers you posted in your response (8/52)(4/51)(6/50)(3/49)(4/48)(2/47), you will see the odds are WAY more than 40k like you claimed before. You should have checked your numbers before claiming your math is correct, you lost credibility by not doing so. Do the math even with your numbers and I hope you see why your 40k number is dead wrong.
If you want to be specific about this hand, this is how it was dealt. Assuming 7&9’s as this is what the hand was dealt: Player one got a 7 (8/52); Player two got a 9 (7/51); Player three got a 7 (6/50). So two 7’s and one 9 has been dealt leaving two 7’s and three 9’s: Player one got a 9 (3/49); Player two got a 7 (2/48); Player three got a 9 (2/47). So (8/52)(7/51)(6/50)(3/49)(2/48)(2/47). This increases the odds of getting this hand by the way, it doesn’t decrease it. The odds end up in the millions. Also, don’t forget to calculate the odds of the straight on top. Remember the player who claimed to see this hand “lots of times”? See why what he said is nonsense? If you don’t understand, then do the math and then you will understand.
Obviously, there literally are 44 out of 52 cards that can never be included in three identical hands. In the 7-9 example there are the 44 cards that are not 7 or 9.
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Cool story bro, thanks for participating. Sorry, no participation trophies to hand out though. 
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It doesn’t matter what order players get the cards in, it just makes it much harder to calculate the odds if you do it your way. My numbers are correct, you should have double checked yours. In your first post, you give 5/49 for the odds of Player A’s second card, but as they must already have been dealt either a 7 or 9, there can’t possibly be more than 4 options for their second card.
Also, when you calculate the odds for hand dealt in the specific order, which I agree is correct, you get increased odds, but this is impossible. You can’t restrict the number of combinations by forcing the cards to come out in a certain order and end up with increased odds, which proves at least one of your calculations is wrong (it’s the first one).
I pointed out initially that the odds of specifically 9 & 7 are way lower than 1 in 40,000 - they’re one in ~3,000,000 (3181018.75 to be exact). The 40,000 comes in because if you’re talking about any two cards, you have to add the probabilities of all possible unpaired 2 card combos. There’s 13 * 12 / 2 = 78 of those.
(8/52)(4/51)(6/50)(3/49)(4/48)(2/47) * 78 = (51 * 50 * 49 * 48 * 47) / (48 * 6 * 3 * 4 * 2) = 40782.291666…
Your last statement clearly shows a misunderstanding of what I’ve been trying to tell you. There are 44 cards that won’t be included in three identical hands, but that’s very different from 44 cards in the deck that can never be part of any three identical hands.
Of course it matters what order the cards come in for the particular hand I shared, it makes a HUGE difference. You literally contradicted yourself when you argued against the way I calculated the first time vs, the second time when you say order doesn’t matter. Do you understand how you argued against yourself? You are talking out of both sides of your mouth with that comment as well as with your 40k comment and your 3 million comment. Good grief.
Your (8/52)(4/51)(6/50)(3/49)(4/48)(2/47) * 78 = (51 * 50 * 49 * 48 * 47) / (48 * 6 * 3 * 4 * 2) = 40782.291666… is DEEPLY FLAWED. That assumes player one was dealt two cards first (8/52 then 4/51). That is not how dealing in holdem works. Your math with your equation is DEEPLY flawed. Sorry my accurate formula which is correct made it “harder to calculate” for you. Nothing hard about it by the way, I laid it out really easily for you as well.
My last statement shows no misunderstanding of what you are “trying to tell me.” Try to tighten up your argument so you don’t keep going down a rabbit hole by making strawman arguments.
I love how you wrote “I pointed out initially that the odds of specifically 9 & 7 are way lower than 1 in 40,000 - they’re one in ~3,000,000 (3181018.75 to be exact).” You literally typed that the odds are lower than 1 in 40,000 then wrote they are 1 in 3,181,018.75 to be exact. NEWS FLASH 1 in 3,181,018.75 is not lower than 1 in 40,000 
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You will probably invent some ridiculous flaw in this logic too, but I’m going to try and explain why the order the cards are dealt doesn’t matter anyway, so here goes…
We’re assuming the deck is in random order, so any card can be anywhere in the deck right? In that case, dealing ABCABC can produce identical result to dealing AABBCC if the deck starts in a different order. In fact, for any order you choose to deal the cards, there’s an order the deck can be in that will give you identical results as any other order. As all deck configurations are equally likely, the order the cards are dealt cannot possibly affect the probability of getting any hand.
Your equations are wrong because while there are 8 cards player A can get first up, 7 for player B and 6 for player C, how many cards player A can get second is either 4, 3 or 2, depending on if AB&C get all the same first card, B&C get different cards, or B&C get the same card but it’s different to A’s. It just gets more complicated from there, but if you work out the entire tree though you will end up with exactly the same answer I got.
BTW, odds of 1 in 3,000,000 are lower than the odds of 1 in 40,000. It’s equivalent to 0.00003% vs 0.0025%
I don’t need to invent any ridiculous flaw, your logic has been ridiculous enough. Yes, we are assuming a well shuffled deck, which is random. I went through the odds of the exact hand…which somehow you found flawed or complicated…too funny. Yet those numbers I gave provided the 1 in about 3,000,000 number which you are now using, imagine that.
You saying my equation is wrong then making the argument I made in the my first equation (which you called flawed) is TOO FUNNY! Of course the order the cards are dealt changes the odds, if a 7 is dealt first then only three 7’s are left. Not a hard concept to grasp unless you just want to be argumentative for the sake of being argumentative like you have.
I misread your comment about the odds being lower. Your 40,000 number was completely flawed and irrelevant based on your flawed math which you shared. BTW, which you initially claimed was the odds of the hand with your flawed logic.
Now that we agree on the 3 million number, which my math clearly showed, then there is the fact that other cards need to come to make the straight which make the odds what? ASTRONOMICAL! Thanks for playing!
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If a 7 comes first, there’s only three sevens left, but if a 9 comes first, there’s only three 9’s left, hence the odds don’t change.
Anyway, if we agree the odds of getting one combo of 3 people having the same hand is 1 in 3,000,000, the odds of getting any one of the 78 possible combos has to be 1 in (3,000,000 / 78) = 1 in ~40,0000 right?
I think I’m being trolled though? Calculating odds is complicated, but you figured out where my equations came from, so it’s getting hard to believe you can’t grasp far more simple concepts. If so, I applaud you, you got me good.
If not, then I can’t help you. Doing the math was kind of fun, pointlessly arguing on the internet, not so much.
Your 1 in 40,000 number is DEEPLY FLAWED. You are being trolled? How, by coming to my thread to discuss? I didn’t invite you here, you came of your own accord. You should really look up the definition of trolling. I believe the only troll here is you now. Anyway, like I said, Thanks for Playing!
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