"hi guys this is one of my first posts on here, i actually cant believe whats happend. i was cruising in the open managed to get a good stack together, as the tournys moving im finding myself losing chips from blinds as im playing good solid poker. 69 players left, i have around 24 BB's blinds are at 200/400. i pick up QQ in early position i raise to 1600 the chip leader at the table has around 6k more than me he calls from the BB, flop comes down 458 he bets 2,500 at me i shove for my remaining 9k of chips he snap calls me with 87 the turn comes down an 8 then the river comes down another 8 lol i cant believe it such a sick beat and he hits the 2 remaining 8's what are the actual odd's of him doing this hitting running 8's and the remaining case 8?"
The odds of making quad 8's are 2/45 * 1/44. This equates to 989 to 1.
Of course, there are other ways for the 87 to win. Any 8, 7 or 6 gives him the win which is 9 nine outs. Using the "rule of 4", we would say that this is approximately a 36% chance to win.
Individually, there are re-draws to most of these outs:
If one of two 8's hits, one of the two remaining Q's will still win it for ginge999. If one of the three 7's hits, one of the two remaining Q's or any of the six remaining 4's or 5's will win it for ginge999. If one of the four 6's hits, one of the remaining 3 7's will split it for ginge999 (and vice versa).
(the final two in the calculations is because the two cards can come in either order)
Therefore we can see that the "rule of 4" is remarkably accurate in this case.
In general the "rule of 4" works pretty well because the true figure for each "out" is 4.5% but discounting it to 4% gives a margin for the re-draws that the opponent has. As you can see from other calculations above, it isn't always so accurate but it is a very reasonable approximation in most cases.
Here's a hand from a recent 7pm deepstack (a tournament that I heartily recommend).
p1: QhTc p2: 9c9d
f=KdJs9h
Well, these are two big hands. p1 has flopped the nuts and p2 has flopped the bottom set.
p1 is winning but what are the odds of p2 winning or it being a tie?
p2 will win if he makes a full house or quads. There are 7 cards for p2 to improve on the turn (in which case the river card is immaterial) and another 10 cards to improve on the river if he doesn't on the turn.
Therefore the odds are 7/45 + (38/45*10/44) = 34.75%
The pot will be split if they both make the same straight. This will happen if a Q or T comes on the turn and the other card comes on the river, this makes 6 cards on the flop and 3 cards on the river.
Here's a question that was asked on a recent show. What are the maximum number of outs that you can have on the river? (if you're not actually winning already!).
Well, the maximum outs would be if you had an opened ended straight flush draw with 4 overcards to the pair on the board (two of which are a pair) against an underpair.
An example is:
p1: 2c2s p2: KhQh
board=ThJhTs8s
In this case, p2 wins in the following cases:
any of eight hearts to make a flush (i.e. not the 2h) any of three (non-heart) K's to make a better two pair any of three (non-heart) Q's to make a better two pair any of three (non-heart) A's to make a straight any of three (non-heart) 9's to make a straight any of three J's to make a counterfeit two pairs either of two (non-heart) 8's to make a counterfeit two pairs
This is 25 outs and the odds are 25/44 = 57% (not too shabby, really!)
Here are two seemingly similar but entirely different hands.
The first is from another 7pm deepstack.
p1: KhKs p2: 6c6d
f=5d4d3d
There had been some pre-flop raising and re-raising.
Anyway, we can see that p2 has flopped an open-ended straight flush draw. If you add in the further two outs for making a set of 6's, this means that he has 17 outs after the flop (nine diamonds, three more 7's, three more 2's and two 6's). The rule of 4 would say that he is 68% to win the hand. In fact, due to the number of re-draws for some of those outs, he is actually about 60% favourite with another 1.2% chance of a split. (Detailed maths available on request!).
The two players got it ai on the flop and the K's held up.
Later that same evening in a Bounty Hunter, I held 6c9s in an unraised big blind.
p1: 6c9s p2: Kc2c f=5c7c8c
In this case, I hold the nut straight and an open-ended straight flush draw. Sadly my opponent already has a flush which makes my straight redundant and most of my flush draws redundant too. In fact, only the straight flush will win the hand for me, so I have only 2 outs.
This means that I have only a 8.65% chance of winning the hand. That's quite a difference to the hand above even though I have flopped a seemingly stronger hand.
These two hands illustrate very well, I think, the importance of putting your opponent on a hand when calculating your odds!
A hand from a recent "Cash Unlimited" show with Tikay and Richard Orford.
P1: AdAc P2: Jh2h
f=6hJc8h
So, what are the odds?
P1 has the best hand so far but P2 has 14 outs. If he hits one of the nine hearts, P1 has no redraws (unless it is the ace of hearts in which case he has 10 re-draws). If P2 hits one of the two jacks, P1 has a re-draw to one of the two remaining aces. If P2 hits his 2, then P1 has eight re-draws to make a better two pair or a set.
The rule of 4 says P2 is 56% favourite. The actual maths say that this is almost exactly 50/50 - another proverbial coin-flip.
I'm going to cover something slightly different to the normal odds questions here.
Last night on the Bounty Hunter coverage, there was a hand where four of the players had pocket pairs - 55, JJ, KK, AA. Trevor and Richard decided that the odds of this were 16 * 16 * 16 * 16/1 - i.e 65,536 to 1.
This would be the correct answer if there were only four players at the table. However, at a 6-seater table, you have to divide these odds by the number of ways of choosing four players from six. There are 15 combinations (to use some mathematical jargon) of 4 from 6. Therefore the odds are actually 4,369 to 1 - still pretty unlikely!
i can see a flaw in your calculations - i may be wrong as i only have an o level in woodwork lol but anonymous stated that u must hit a royal flush card on the 1st 2nd and 3rd flop cards but u dont u could miss them and hit on the turn or the river am i wrong???
The odds of making a royal flush in Hold 'Em (after the river) are smaller.
If you don't specify that you need to use either of your hole cards as part of the royal flush, then you must divide the odds by the number of ways of choosing 5 cards from 7.
There are 21 different ways of choosing 5 cards from 7.
tell anonymous to work out the odds of a royal flush to include the turn and the river and the player getting the royal living in holywood lol Posted by loonytoons
I hope the above post clears up the question of the odds of making a royal flush after the river.
I'm a bit stuck on trying to work out the odds of the player living in Hollywood, If you'd like to provide me with the following information, I'll have a go.
a) The number of poker players in Hollywood b) The number of poker players in the world c) The number of hands that people in Hollywood play d) The number of hands that people in the world play e) Whether Ben Affleck is inherently luckier than the average poker player
In Response to Re: WHAT ARE THE ODDS? : I hope the above post clears up the question of the odds of making a royal flush after the river. I'm a bit stuck on trying to work out the odds of the player living in Hollywood, If you'd like to provide me with the following information, I'll have a go. a) The number of poker players in Hollywood b) The number of poker players in the world c) The number of hands that people in Hollywood play d) The number of hands that people in the world play e) Whether Ben Affleck is inherently luckier than the average poker player :-) Posted by MereNovice
What are the odds of flopping quads with unpaired hole cards?
Well, this is pretty simple to work out, but rather harder to actually do!
The first card can be 1 of any of 6 cards that match either of your pocket cards. The second card must be 1 of the 2 remaining cards that match the first flopped card. The third card must be the remaining (case) card that matches the first two flopped cards.
Therefore, the odds are 6/50 * 2/49 * 1/48.
This equates to 9,799 to 1.
If you were playing a single table for 8 hours a day being dealt 50 hands a hour (and seeing the flop every hand!), you'd expect this to happen about once every 24 or 25 days.
During last night's show, Mr. Orford asked what the odds of hitting "perfect perfect" cards after the flop are. The figure they came up with, off the top of their heads was about 333/1.
If we say that we are playing heads-up and hence 7 of the cards are known after the flop, then the odds are:
2/45 * 1/44 = 0.001 or 989 to 1.
I've seen it happen once (not on SkyPoker) where the player needed to hit two precise cards to win and did. The scenario was:
Comments
"hi guys this is one of my first posts on here, i actually cant believe whats happend. i was cruising in the open managed to get a good stack together, as the tournys moving im finding myself losing chips from blinds as im playing good solid poker. 69 players left, i have around 24 BB's blinds are at 200/400. i pick up QQ in early position i raise to 1600 the chip leader at the table has around 6k more than me he calls from the BB, flop comes down 458 he bets 2,500 at me i shove for my remaining 9k of chips he snap calls me with 87 the turn comes down an 8 then the river comes down another 8 lol i cant believe it such a sick beat and he hits the 2 remaining 8's what are the actual odd's of him doing this hitting running 8's and the remaining case 8?"
The odds of making quad 8's are 2/45 * 1/44. This equates to 989 to 1.
Of course, there are other ways for the 87 to win.
Any 8, 7 or 6 gives him the win which is 9 nine outs.
Using the "rule of 4", we would say that this is approximately a 36% chance to win.
Individually, there are re-draws to most of these outs:
If one of two 8's hits, one of the two remaining Q's will still win it for ginge999.
If one of the three 7's hits, one of the two remaining Q's or any of the six remaining 4's or 5's will win it for ginge999.
If one of the four 6's hits, one of the remaining 3 7's will split it for ginge999 (and vice versa).
Therefore, the true odds are:
For the 87 to win:-
(2/45 * (1 - 2/44)) + (3/45 * (1 - 8/44)) + (4/45 * (1 - 3/44)) * 2 = 35.96%
For the split pot:-
4/45 * 3/44 * 2 = 1.21%
(the final two in the calculations is because the two cards can come in either order)
Therefore we can see that the "rule of 4" is remarkably accurate in this case.
In general the "rule of 4" works pretty well because the true figure for each "out" is 4.5% but discounting it to 4% gives a margin for the re-draws that the opponent has. As you can see from other calculations above, it isn't always so accurate but it is a very reasonable approximation in most cases.
p1: QhTc
p2: 9c9d
f=KdJs9h
Well, these are two big hands. p1 has flopped the nuts and p2 has flopped the bottom set.
p1 is winning but what are the odds of p2 winning or it being a tie?
p2 will win if he makes a full house or quads.
There are 7 cards for p2 to improve on the turn (in which case the river card is immaterial) and another 10 cards to improve on the river if he doesn't on the turn.
Therefore the odds are 7/45 + (38/45*10/44) = 34.75%
The pot will be split if they both make the same straight.
This will happen if a Q or T comes on the turn and the other card comes on the river, this makes 6 cards on the flop and 3 cards on the river.
Therefore the odds are 6/45 * 3/44 = 0.91%
What are the maximum number of outs that you can have on the river? (if you're not actually winning already!).
Well, the maximum outs would be if you had an opened ended straight flush draw with 4 overcards to the pair on the board (two of which are a pair) against an underpair.
An example is:
p1: 2c2s
p2: KhQh
board=ThJhTs8s
In this case, p2 wins in the following cases:
any of eight hearts to make a flush (i.e. not the 2h)
any of three (non-heart) K's to make a better two pair
any of three (non-heart) Q's to make a better two pair
any of three (non-heart) A's to make a straight
any of three (non-heart) 9's to make a straight
any of three J's to make a counterfeit two pairs
either of two (non-heart) 8's to make a counterfeit two pairs
This is 25 outs and the odds are 25/44 = 57% (not too shabby, really!)
The first is from another 7pm deepstack.
p1: KhKs
p2: 6c6d
f=5d4d3d
There had been some pre-flop raising and re-raising.
Anyway, we can see that p2 has flopped an open-ended straight flush draw. If you add in the further two outs for making a set of 6's, this means that he has 17 outs after the flop (nine diamonds, three more 7's, three more 2's and two 6's).
The rule of 4 would say that he is 68% to win the hand. In fact, due to the number of re-draws for some of those outs, he is actually about 60% favourite with another 1.2% chance of a split. (Detailed maths available on request!).
The two players got it ai on the flop and the K's held up.
Later that same evening in a Bounty Hunter, I held 6c9s in an unraised big blind.
p1: 6c9s
p2: Kc2c
f=5c7c8c
In this case, I hold the nut straight and an open-ended straight flush draw.
Sadly my opponent already has a flush which makes my straight redundant and most of my flush draws redundant too.
In fact, only the straight flush will win the hand for me, so I have only 2 outs.
This means that I have only a 8.65% chance of winning the hand.
That's quite a difference to the hand above even though I have flopped a seemingly stronger hand.
These two hands illustrate very well, I think, the importance of putting your opponent on a hand when calculating your odds!
P1: AdAc
P2: Jh2h
f=6hJc8h
So, what are the odds?
P1 has the best hand so far but P2 has 14 outs. If he hits one of the nine hearts, P1 has no redraws (unless it is the ace of hearts in which case he has 10 re-draws). If P2 hits one of the two jacks, P1 has a re-draw to one of the two remaining aces. If P2 hits his 2, then P1 has eight re-draws to make a better two pair or a set.
The rule of 4 says P2 is 56% favourite. The actual maths say that this is almost exactly 50/50 - another proverbial coin-flip.
I'm going to cover something slightly different to the normal odds questions here.
Last night on the Bounty Hunter coverage, there was a hand where four of the players had pocket pairs - 55, JJ, KK, AA.
Trevor and Richard decided that the odds of this were 16 * 16 * 16 * 16/1 - i.e 65,536 to 1.
This would be the correct answer if there were only four players at the table.
However, at a 6-seater table, you have to divide these odds by the number of ways of choosing four players from six.
There are 15 combinations (to use some mathematical jargon) of 4 from 6.
Therefore the odds are actually 4,369 to 1 - still pretty unlikely!
You need to hit both of your cards in the first three cards dealt.
Therefore, the odds are 2/50 * 1/49 * 3 = 0.00245 or 407/1.
The final three in the calculation is because there are 3 ways of choosing 2 cards from the 3 dealt cards.
So, this isn't quite as rare as you might think.
HAHAHAH U LOT NEED TO GET OUT MORE... ANYONE GOT A POKER ODDS CALCULATOR......MMMM ;-))
If you don't specify that you need to use either of your hole cards as part of the royal flush, then you must divide the odds by the number of ways of choosing 5 cards from 7.
There are 21 different ways of choosing 5 cards from 7.
Therefore the odds are 30,940 to 1.
I hope the above post clears up the question of the odds of making a royal flush after the river.
I'm a bit stuck on trying to work out the odds of the player living in Hollywood,
If you'd like to provide me with the following information, I'll have a go.
a) The number of poker players in Hollywood
b) The number of poker players in the world
c) The number of hands that people in Hollywood play
d) The number of hands that people in the world play
e) Whether Ben Affleck is inherently luckier than the average poker player
:-)
Well, this is pretty simple to work out, but rather harder to actually do!
The first card can be 1 of any of 6 cards that match either of your pocket cards.
The second card must be 1 of the 2 remaining cards that match the first flopped card.
The third card must be the remaining (case) card that matches the first two flopped cards.
Therefore, the odds are 6/50 * 2/49 * 1/48.
This equates to 9,799 to 1.
If you were playing a single table for 8 hours a day being dealt 50 hands a hour (and seeing the flop every hand!), you'd expect this to happen about once every 24 or 25 days.
-----------------
During last night's show, Mr. Orford asked what the odds of hitting "perfect perfect" cards after the flop are.
The figure they came up with, off the top of their heads was about 333/1.
If we say that we are playing heads-up and hence 7 of the cards are known after the flop, then the odds are:
2/45 * 1/44 = 0.001 or 989 to 1.
I've seen it happen once (not on SkyPoker) where the player needed to hit two precise cards to win and did.
The scenario was:
AA v 44
f=44J
All the money went in.
t=A
r=A
Now, that's a bad beat.