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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.
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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.
Playing the 7:00pm Forum Deepstack the other night I got dealt AA 3 times in 7 hands. Mr. Orford who was seated next to me also got dealt AA twice around the same time.
Grimstar30 (if it was someone else, I apologise) asked me what the odds of that happening are so here are my calculations.
We reckoned that between us we had AA 5 times in 30 hands.
The odds of being dealt AA are 4/52 * 3/51. This is 220 to 1 (a figure often quoted on TV).
The odds of being dealt AA 3 times in succession (at a random time) are 4/52 * 3/51 * 4/52 * 3/51 * 4/52 * 3/51 = 10,793,860 to 1.
To calculate the odds of being dealt AA 3 times in 7 hands, we must divide those odds by the number of "combinations" of 3 from 7 which is 35.
Therefore the odds of being dealt AA 3 times in 7 hands is 308,395 to 1.
The odds of two people being dealt AA 5 times in 30 hands on a 10 seater table are as follows:
4/52 * 3/51 * 4/52 * 3/51 * 4/52 * 3/51 * 4/52 * 3/51 * 4/52 * 3/51 * 142,506 * 2 * 45 = 41,103 to 1.
The 142,506 is the number of combinations of 5 hands from 30.
The 2 is to account for the 2 players.
The 45 is the number of combinations of 2 players from 10 seated at the table.
As always, I'm happy to be corrected if anyone sees a flaw in my calculations.
When you flop a set and an opponent flops a bigger set, you're going to lose a lot of chips.
While playing the Tikay Sleepstack last night, RazorKev told me that it had happened to him 5 times in quick succession on two tables (3 times in a 10-seater tournament and 2 times in a 6-seater cash game).
He wondered what the odds of this happening are.
For the purpose of this calculation we will assume that it happened within 100 hands on each of the tables.
We will work it out in stages. There are a lot of calculations here so those who are just interested in the answer can skip to the bottom and look at the number in bold!
The odds of being dealt a pocket pair are 3/51 = 16 to 1.
The odds of two people heads-up being dealt a pocket pair are 3/51 * 3/51 = 288/1.
The odds of one person and an opponent at a 10-seater table being dealt pocket pair are 3/51 * 3/51 * 9 = 31 to 1.
To get the odds for both of these people to flop a set we must multiply the odds by 2/48 * 2/47 * 44/46 * 6.
(i.e. the first flop card must be one of 2 cards that pairs player A, the second flop card must be one of 2 cards that pairs player B and the last card must be one of 44 cards that pairs neither player and the cards could fall in any of 6 ways).
This gives us 3/51 * 3/51 * 9 * 4/48 * 2/47 * 44/46 * 6 = 3,155 to 1.
The odds of this happening three times in succession are obtained by multiplying this number together three times, i.e. 31,424,257,606 to 1.
To get the odds of it happening 3 times in 100 hands, we must divide the odds by the number of combinations of 3 from 100 hands which is 161,700.
Therefore the odds are 194,336 to 1. Finally, we will assume that RazorKev is just as likely to have the over-set as the under-set (I'm saying nothing about his likely range of starting hands) so we will multiply this by 2 to give odds of 388,673 to 1.
If we do the same calculation for it happening 2 times in 100 hands at a 6-seater table we get odds of 13,035 to 1.
If we wish to know the odds for both happening we need to multiply the odds together.
This gives us a final figure of 5,066,773,532 to 1.
As always, I'm happy to be corrected if anyone sees a flaw in my calculations. There may be a horrible mistake in there somewhere, I haven't spent a lot of time checking it.
Now I am a complete spanner, but isn't it 10,648,000 to 1? (ie, 220x220x220)
Just trying to understand
If the odds of something are 220 to 1 then we are saying it happens 1 in 221 times.
(This is the same as saying that if something happens 1 in 3 times, it has odds of 2 to 1).
Therefore, for it to happen three times, it is 1 in (221 * 221 * 221) times which is 1 in 10,793,861 times.
This is the same as saying 10,793,860 to 1.
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Which is obviously 6+5+4+3+2+1 - But how do you work that out? How about 4 cards into 7? or 3? No way am I bothering to write those all out manually!
Interestingly ... that gives odds against of 48,840:1 of being dealt AA in the next two hands, but in Matthew Hilger's 'Odds and Probabilities' book he sets that as a test question (page 28) and reveals the answer to be 49,383:1! Not helpful! lol
I'm not sure where Matthew Hilger gets his number from.
"Combinations" and "permutations" have very specific (and different) meanings for mathematicians.
Combinations are the number of ways of choosing the objects where the order doesn't matter.
Permutations are the number of ways of choosing the objects where the order does matter.
For example, if we're choosing lottery balls, the number of combinations is what we are interested in, i.e. it doesn't matter what order the balls come out.
The number of combinations of 6 from 49 balls is: 13,983,816
The number of permutations of 6 from 49 balls is: 10,068,347,520
(So, it's just as well we don't have to predict the balls in the right order or else there would be a lot of rollover weeks!).
The way that I calculate this is to use the PERMUT and COMBIN functions in an Excel spreadsheet!
If you want to go into the actual equations that are used to calculate these functions, please PM me and I'll be more than happy to go through it.
Alternatively, if anyone else is interested in this, I'll post it on here.
I went out of the 7pm deepstack last night playing like a donkey against an opponent who flopped a straight holding J8.
I was asked what are the odds of this happening (flopping the straight - not me playing like a donkey).
There are two ways to flop a straight holding J8 (or any double gapped cards that are in the range KT to 52).
The flop can be QT9 or T97.
The odds of the flop being QT9 or T97 are 12/50 * 8/49 * 4/48 * 2 = 152 to 1 (approximately).
The odds for flopping a straight with triple gapped cards are double this (i.e. approximately 304 to 1) and the odds of flopping a straight with connected cards (in the range JT to 54) are half this (i.e. approximately 76 to 1).
i feel out of my depth here lol
The actual odds of a 2nd player having a pocket pair are approximately 15.78 to 1.
This is calculated by (48/50 * 3/49) + (2/50 * 1/49)) which is derived as follows:
The first card is one of the 48 cards that don't match the first player's pair and then his second card must be one of the three cards that pair that or his first card is the same as the first players pair and his second card must be the last remaining card in the deck that matches this.
I also didn't include any consideration of multiple players holding pocket pairs and seeing the flop hence increasing the possibility of people flopping a set.
Thanks for pointing this out. I very much welcome any comments like this.
Cards like 67 suited are known as "ace crackers" since they give you the best chance of beating aces.
However, you are still only about 23% to win pre-flop.
If the 67 is unsuited then you are only about 19% to win (it varies a little depending on whether the suits clash).
The best way to find out the odds of hands like this is to use one of the hand odds calculators that are available on the internet. I don't think that I'm allowed to give a link to any specific sites but if you type "poker odds calculator" into a popular search engine it will give you a list of some main ones. It's quite informative to try out various hands and see what the odds are - some might surprise you.
I posted a table of some commonly quoted hands near the top of this thread.
So, to calculate the odds of flopping a royal flush say, the probability is:
20/52 * 4/51 * 3/50 * 2/49 * 1/48.
Generally when working out odds like this we only consider what we have in our own hands since we do not know where any of the other cards are.
It's interesting (to me anyway!) to see how this is implemented on TV programs where they show the odds for the respective hands. Say at a 6-seater table we have 2 players heads up going to the river and one player has 4 outs. Generally this is shown as 9% (4 * 2.3%). However, some programs include all the known cards folded by the other players which is why you will see different percentages quoted sometimes.
If this wasn't what you meant, please let me know.