bad beat what are the odds

[barker 0]

Hello Everyone,Great site Daniel, here's a bad beat I thought I'd share.$200 nl party poker, I,m on the button with 7-7. Under the gun brings it in for $20 4 players call to me. Under the gun has $900 and has been playing very tight a raise here has been a-a,k-k in the past couple of hours.I have $650 that I've built up over the last 3 hours. I call of course and see the flop 7 handed. Flop 7c 7d 3cunder the gun bets$200 everyone folds, I raise $200 and ug puts me all in I say thank you I have quads, the turn and river bring 5c and 6c under the gun has 4-4 with the club for a straight flush? they say 5555555555555THANK YOU? What are the odds of that happening and whatds with the 5555555555555555? Anyway I thought that was a bad beat based upon the action and previous play of the ug player.Take Care everyone and hopefully you don't suffer any of these beats in the near future.Todd

[FloppyNuts 0]

Whoops, I screwed up. I'm re-editing my post. I previously posted "1,080 to 1." After the flop, with 45 unknown cards out, the odds of catching 2 and only 2 perfect running cards is 989 to 1.

[DeNuts1 0]

Yep nuts is right 1000 to 1....have had it happen to myself on a couple of occasions...blows. Playing in Vegas 10/20 limit pocket JJ raise late pos. 1 caller flop of K, J, 7. Guy leads out I raise...he reraises?....i call. Turn is another K gives me jacks full him 3 kings. We cap it. River another mf'n King quit betting but it was too late...lost my entire buy in on that hand.

[Guest]

The 'Bad Beats' forum is probably more appropriate place for this.

[justblaze 0]

Yep nuts is right 1000 to 1....have had it happen to myself on a couple of occasions...blows. Playing in Vegas 10/20 limit pocket JJ raise late pos. 1 caller flop of K, J, 7. Guy leads out I raise...he reraises?....i call. Turn is another K gives me jacks full him 3 kings. We cap it. River another mf'n King quit betting but it was too late...lost my entire buy in on that hand.

He has more than one out on the river, genius.

[MrConceit 0]

Yep nuts is right 1000 to 1....have had it happen to myself on a couple of occasions...blows. Playing in Vegas 10/20 limit pocket JJ raise late pos. 1 caller flop of K, J, 7. Guy leads out I raise...he reraises?....i call. Turn is another K gives me jacks full him 3 kings. We cap it. River another mf'n King quit betting but it was too late...lost my entire buy in on that hand.

He has more than one out on the river, genius.

I don't think he was claiming his flopped second set was quite as bad a beat as a runner runner gutshot SF that the first poster had. He was just illustrating a runner runner beat. Top pair vs second set will lose 96.77 percent of the time. Yeah, he picked up more outs on his perfect turn and now will only be losing 84 percent of the time. Wow, you're right, this most definitely wasn't a bad beat!Sorry justblaze, but come now. Just because it wasn't a 1 outer on the river... At least idiot 44 was allin in the first example, the other guy participated in a capped turn when he was 84 percent chance to lose.

[Chip_and_a_Chair 0]

Hello Everyone,Great site Daniel, here's a bad beat I thought I'd share.$200 nl party poker, I,m on the button with 7-7. Under the gun brings it in for $20 4 players call to me. Under the gun has $900 and has been playing very tight a raise here has been a-a,k-k in the past couple of hours.I have $650 that I've built up over the last 3 hours. I call of course and see the flop 7 handed. Flop 7c 7d 3cunder the gun bets$200 everyone folds, I raise $200 and ug puts me all in I say thank you I have quads, the turn and river bring 5c and 6c under the gun has 4-4 with the club for a straight flush? they say 5555555555555THANK YOU? What are the odds of that happening and whatds with the 5555555555555555? Anyway I thought that was a bad beat based upon the action and previous play of the ug player.Take Care everyone and hopefully you don't suffer any of these beats in the near future.Todd

Todd,By my calculation, the odds of your opponent winning after that flop are 1979:1 (roughly .05 percent), because there are only two particular cards in the deck that will help him, and he needs them both (1/45 x 1/44). I faced the same situation at Foxwoods when I held A 3 in the BB and the flop was A A 3. My opponent tried to make a ballsy move and pushed me all-in with 8 8. He caught running 8s, the only two cards that could beat me after I flopped the full house.The good news is you will NEVER take a worse beat than this in your life, because the only way an opponent can have worse than 1979:1 odds two-handed after a flop is if he's drawing dead.P.S. The "555555555555555" isn't some smack-talking code. "5" is located above "T" on the keyboard, and in his excitement he was probably just mashing the keyboard a little too frantically.

[justblaze 0]

heh his post made it quite clear that he had had been beat by 2 perfect cards, and then provided an example which was not in fact an example. and btw the 55555555 thing IS a smack talking code. Some of these morons think you will only hit your draws if you are holding down the 5 key as the river card comes out. These are the same people wearing tinfoil hats and claiming the sites are all rigged.

[FloppyNuts 0]

There are 45*44, or 1980 total combinations, but 2 of those are the same: 5c on the turn and 6c on the river is the same as 6c on the turn and 5c on the river. You have to divide by 2. Technically, it's: 45!/(43!*2!).The exclamation point is "factorial." "5!," or 5-factorial is 5*4*3*2*1.6-factorial is 6*5*4*3*2*1...Hope I'm right.

[DeNuts1 0]

Yep nuts is right 1000 to 1....have had it happen to myself on a couple of occasions...blows. Playing in Vegas 10/20 limit pocket JJ raise late pos. 1 caller flop of K, J, 7. Guy leads out I raise...he reraises?....i call. Turn is another K gives me jacks full him 3 kings. We cap it. River another mf'n King quit betting but it was too late...lost my entire buy in on that hand.

He has more than one out on the river, genius.

Um...blaze where did I state that he had one out on the river again?? I said I have had it happen and it blows. I then gave a runner/runner beat story myself. Wtf are you talking about?

[DeNuts1 0]

There are 45*44, or 1980 total combinations, but 2 of those are the same: 5c on the turn and 6c on the river is the same as 6c on the turn and 5c on the river. You have to divide by 2. Technically, it's: 45!/(43!)(2!).The exclamation point is "factorial." "5!," or 5-factorial is 5*4*3*2*1.6-factorial is 6*5*4*3*2*1...Hope I'm right.

I vote the word factorial be permanently banned from the poker forum.

[justblaze 0]

Yep nuts is right 1000 to 1....have had it happen to myself on a couple of occasions...blows. Playing in Vegas 10/20 limit pocket JJ raise late pos. 1 caller flop of K, J, 7. Guy leads out I raise...he reraises?....i call. Turn is another K gives me jacks full him 3 kings. We cap it. River another mf'n King quit betting but it was too late...lost my entire buy in on that hand.

He has more than one out on the river, genius.

Um...blaze where did I state that he had one out on the river again?? I said I have had it happen and it blows. I then gave a runner/runner beat story myself. Wtf are you talking about?

you responded to a poster complaining of having someone catch two perfect cards where those were the only two cards in the deck to help the guy and both of them had to fall. you responded by saying you had had this happen to you. If that's not what you meant...

[DeNuts1 0]

blaze....i'm working i don't have time to argue so....your a genius i'm retarded lets just leave it at that.

[justblaze 0]

blaze....i'm working i don't have time to argue so....your a genius i'm retarded lets just leave it at that.

heh im working too, but i make time. your solution sounds fair though 8)

[Hextall27 0]

and btw the 55555555 thing IS a smack talking code. Some of these morons think you will only hit your draws if you are holding down the 5 key as the river card comes out. These are the same people wearing tinfoil hats and claiming the sites are all rigged.

Are you serious? That is, hands down, the dumbest thing I've ever heard in my life.

[justblaze 0]

and btw the 55555555 thing IS a smack talking code. Some of these morons think you will only hit your draws if you are holding down the 5 key as the river card comes out. These are the same people wearing tinfoil hats and claiming the sites are all rigged.

Are you serious? That is, hands down, the dumbest thing I've ever heard in my life.

nah, makes just as much sense as "pokerstars is rigged, i got rivered twice. UB is much better, ive won 12$ there."

[Chip_and_a_Chair 0]

There are 45*44, or 1980 total combinations, but 2 of those are the same:  5c on the turn and 6c on the river is the same as 6c on the turn and 5c on the river.   You have to divide by 2.  Technically, it's:  45!/(43!*2!).The exclamation point is "factorial."  "5!," or 5-factorial is 5*4*3*2*1.6-factorial is 6*5*4*3*2*1...Hope I'm right.

Good point, I hadn't considered the fact that the order of which potential straight flush card arrives is insignificant, leaving two cards on the turn that would keep the opponent "alive" in the hand, so to speak. But without concerning ourselves with factorials, couldn't we get the result we're looking for simply by using 2/45 x 1/44? There are only two possible orders -- 5c then 6c v. 6c then 5c -- so do we really need to mess with permutations? Wouldn't you get your original post of 989:1 odds just by divided 1980 by 2, instead of 1? I'm not the greatest at math, so I'm wondering if factorials are necessary.Since the opponent needs runner-runner to win, the best he can do on the turn is keep from drawing dead -- he can't take the lead. There are two outs that will keep him alive on the turn. If he hits one of those, there is only one card left in the entire deck that would help him -- the other club connector. The order shouldn't matter for the purpose of calculating odds, right?EDIT: I know dividing the factorials in your example gives you the same answer, I'm just wondering if there's a special reason

[justblaze 0]

There are 45*44, or 1980 total combinations, but 2 of those are the same:  5c on the turn and 6c on the river is the same as 6c on the turn and 5c on the river.   You have to divide by 2.  Technically, it's:  45!/(43!*2!).The exclamation point is "factorial."  "5!," or 5-factorial is 5*4*3*2*1.6-factorial is 6*5*4*3*2*1...Hope I'm right.

Good point, I hadn't considered the fact that the order of which potential straight flush card arrives is insignificant, leaving two cards on the turn that would keep the opponent "alive" in the hand, so to speak. But without concerning ourselves with factorials, couldn't we get the result we're looking for simply by using 2/45 x 1/44? There are only two possible orders -- 5c then 6c v. 6c then 5c -- so do we really need to mess with permutations? Wouldn't you get your original post of 989:1 odds just by divided 1980 by 2, instead of 1? I'm not the greatest at math, so I'm wondering if factorials are necessary.Since the opponent needs runner-runner to win, the best he can do on the turn is keep from drawing dead -- he can't take the lead. There are two outs that will keep him alive on the turn. If he hits one of those, there is only one card left in the entire deck that would help him -- the other club connector. The order shouldn't matter for the purpose of calculating odds, right?

works in this instance, but gets messy if he has 4 cards to keep him alive on the turn, say, and then 1 on the river which changes depending on the turn. or 3 on the turn, 2 on the river, etc etc. basically you need the algebra if you want to understand WHY this is the answer, not just that it is in fact the answer.

[Chip_and_a_Chair 0]

There are 45*44, or 1980 total combinations, but 2 of those are the same:  5c on the turn and 6c on the river is the same as 6c on the turn and 5c on the river.   You have to divide by 2.  Technically, it's:  45!/(43!*2!).The exclamation point is "factorial."  "5!," or 5-factorial is 5*4*3*2*1.6-factorial is 6*5*4*3*2*1...Hope I'm right.

Good point, I hadn't considered the fact that the order of which potential straight flush card arrives is insignificant, leaving two cards on the turn that would keep the opponent "alive" in the hand, so to speak. But without concerning ourselves with factorials, couldn't we get the result we're looking for simply by using 2/45 x 1/44? There are only two possible orders -- 5c then 6c v. 6c then 5c -- so do we really need to mess with permutations? Wouldn't you get your original post of 989:1 odds just by divided 1980 by 2, instead of 1? I'm not the greatest at math, so I'm wondering if factorials are necessary.Since the opponent needs runner-runner to win, the best he can do on the turn is keep from drawing dead -- he can't take the lead. There are two outs that will keep him alive on the turn. If he hits one of those, there is only one card left in the entire deck that would help him -- the other club connector. The order shouldn't matter for the purpose of calculating odds, right?

works in this instance, but gets messy if he has 4 cards to keep him alive on the turn, say, and then 1 on the river which changes depending on the turn. or 3 on the turn, 2 on the river, etc etc. basically you need the algebra if you want to understand WHY this is the answer, not just that it is in fact the answer.

Great answer... that's where I was confused. Thanks!

[FloppyNuts 0]

No, the factorials aren't important. It was just a way of writing it down. We're basically saying the same thing.

[justblaze 0]

No, the factorials aren't important.  It was just a way of writing it down.  We're basically saying the same thing.

uhh, see above, they are important.

[FloppyNuts 0]

It's just mathematical "language." If the problem is: Figure out the total number of 5-card combinations out of a 52 card deck, you'd multiply 52*51*50*49*48.But you can write it more simply, like this: 52!/47!I shouldn't have brought it up, but I was simply plugging in terms into a probability formula I was familiar with.(To solve the above problem, you'd need to divide result by 5-factorial, because the order of the cards doesn't matter.)

[justblaze 0]

It's just mathematical "language."  If the problem is:  Figure out the total number of 5-card combinations out of a 52 card deck, you'd multiply 52*51*50*49*48.But you can write it more simply, like this:  52!/47!I shouldn't have brought it up, but I was simply plugging in terms into a probability formula I was familiar with.

ah but the two expressions above are the same mathematical function, just written in different notation. the two solutions to the original problem are totally different approaches to solving the problem. one works on a theoretical level, but the other does not.

[Johnny Socko 0]

I think he was hitting the "5555555555555555" because he was hoping for that card. Like yelling out "5" during a live game. Are you sure the 5 wasnt the card that hit on the river? Either way, its one hell of a bad beat.