Any aces vs aces is a 95.65% chance of being a split pot, so the odds of either hand winning outright (which can only be achieved with a flush), is 2.17%.
Each player has 2.17% chance of making a flush. That's the same as a 45/1 shot.
In the long run this will be a chop 22 out of every 23 times, one or other of the players will scoop the pot the other 1 out of 23.
To put this in perspective rolling tow sixes from a pair of dices will only happen 1 in every 36 attempts, so aces beating aces in poker is no big deal really.
A highly unlikely scenario is a chop with both players playing a straight flush on the board.
Sometimes a royal flush will beat the other players king-high straight flush on the board. An extremely unlikely event as it requires the run-out to be one of four combos of 9TJQK suited, but I bet it's happened somewhere.
In Response to Re: unlucky or bad beat : Okay, that's puzzling me. Do you mean that the board comes down all the same suit, so both players make a flush, but one of them will have the nut flush ? Posted by Fortunatus
yes.
you know that it nothing more than being a little bit careful about the question being asked. the OP question sounds as though the answer must be something astrnomical but we are given aces for both players.
Each player has 2.17% chance of making a flush. That's the same as a 45/1 shot. In the long run this will be a chop 22 out of every 23 times, one or other of the players will scoop the pot the other 1 out of 23. To put this in perspective rolling tow sixes from a pair of dices will only happen 1 in every 36 attempts, so aces beating aces in poker is no big deal really. Posted by GaryQQQ
So effectively losing a hand AA vs AA all in pre is the same (give or take) as getting it all in on the turn and oppo hits their 'miracle' 1-outer?
Wouldn't the odds of a flush board with 2 players holding aces be 12/48 * 11/47 * 10/46 * 9/45 * 8/46, then multiplied by 4 to cover any suit. Which is 540/1, or am I competely overthinking this ? Posted by Fortunatus
whatever the first card suit, the odds that the next card is the same suit is roughly 1 in 4. and so on. four cards at i in 4 is 1 in 256 or 255/1
you get a tick for your answer being quite close but lose points for your workings. whereas i get good marks for my workings with a note written in red biro saying that i must pay more attention to detail.
yes. 493/1. you get a tick for your answer being quite close but lose points for your workings. whereas i get good marks for my workings with a note written in red biro saying that i must pay more attention to detail. Posted by aussie09
Comments
less than 15/1... edit ... lower than 31/1... second edit 22/1
also, the chance that both players hit a flush is 255/1
In the long run this will be a chop 22 out of every 23 times, one or other of the players will scoop the pot the other 1 out of 23.
To put this in perspective rolling tow sixes from a pair of dices will only happen 1 in every 36 attempts, so aces beating aces in poker is no big deal really.
A highly unlikely scenario is a chop with both players playing a straight flush on the board.
Sometimes a royal flush will beat the other players king-high straight flush on the board. An extremely unlikely event as it requires the run-out to be one of four combos of 9TJQK suited, but I bet it's happened somewhere.
yes.
you know that it nothing more than being a little bit careful about the question being asked. the OP question sounds as though the answer must be something astrnomical but we are given aces for both players.
Good stat of the day, posters.
whatever the first card suit, the odds that the next card is the same suit is roughly 1 in 4. and so on. four cards at i in 4 is 1 in 256 or 255/1
more accurately,..
11/46 x 10/45 x 9/44 x 8/43
yes. 493/1.
you get a tick for your answer being quite close but lose points for your workings. whereas i get good marks for my workings with a note written in red biro saying that i must pay more attention to detail.