Gotta See This Pls Omg

[F1man]

Originally Posted by Masterplan

if the above aituation did occur it would most probably be the most unlikely event that has ever happened short of the big bang.
considering that except for the queens it could ONLY occur with those starting hands so you can start off with:
1/52*1/51*1/50*1/49*1/48*1/47*1/46*2/45 ish (i'm not great at math or probabilities if i'm wrong i'm wrong)
then the board cards
1/44*1/43*1/42*1/41*2/40
then add in personal probabilities like the chance of the 8/9 calling whatever raises had come previously, possibly none if slow played so you might choose to ignore it.
anyone got a calc to do it?
i work it out as 15/3954242643911239680000

I think you have gone way over on your calc. First of all, when dealt the cards, it does not matter which person gets dealt which cards, or in what order. The prob you have calculated in for a person to be dealt the cards in that order. Therefore to get the same kind of result in screen shot, there are plenty more ways to do it. (I could have done the calc one day)

The second part of the calc goes wrong, as the cards can be dealt in any order. For ex, the chance of the first card on the flop being dealt which matches the cards on the table is infact 6/44, then 5/44 and so on. (Although a bit more complicated as the two on one card makes it a bit tricky).

In short this is one mother of a calc. But remeber every hand you play ie everyone get dealt 2 random cards, and then the cards are in the middle played, as as much chance of happening as this one on this thread.

[Masterplan]

i think its slightly more tha 60bn to 1, because if you have a king or ace of a suit for example in a purely heads up situation.
its 8 in 52, but then to have that matching partner is then 1 in 51. which if you then consider the 8/9 musdt be the same suit, you now have 2 in 50 then 1 in 49, then you have the QJT of that suit coming down, 3/48, 2/47, 1/46 with 1 per board card chance of it being chosen. i remember vaguely from years ago when i did statistics it should be something like 45 permute 5 * 45 permute 2 * 45 permute 1. or something like that.
remember that 8 9 T J Q K A all of the same suit have to be selected from the deck in specific orders in order to be dealt.
this alone is probably more than 60bn to 1

[slapdash]

There are 5040 ways of choosing which of the 10 players gets the royal
flush, which gets the other straight flush, etc. Have you taken that into
account?

Let's do this carefully:

There are 52*51*50*49*48/5*4*3*2*1 = 2598960 possible boards.

To get QJ10 suited, there are 4 possible suits. To get one of those cards
paired, there are 9 possibilities. To get another card, nine or lower, say,
but not the 8 or 9 of the QJ10 suit, there are 30 possibilities.

So there are 4*9*30=1080 boards that will work. So the probability of
getting a good board is one in 2598960/1080 = 2406.44444...

Now look at how likely it is that particular players get the royal, etc.:

Given the board, there are 47*46/2=1081 hands, only one gives the royal,
so there's a 1 in 1081 chance of player A getting this.

Given the board and first hand, there are 45*44/2=990 possible second
hands, so there's a 1 in 990 chance of player B getting the straight flush.

Given the board and first two hands, there are 43*42/2=903 possible
third hands, so there's a 1 in 903 chance of player C getting quads.

Given the board and first three hands, there are 41*40/2=820 possible
hands for player D. There are 6 ways for him to get a high full house,
giving a 1 in 136.66666 chance.

In total that gives a 1 in

2406.44444 * 1081 * 990 * 903 * 136.66666 = 317824286579800

chance.

But there are 10*9*8*7 =5040 ways to pick players A,B,C,D, giving a total
chance of 1 in

317824286579800/5040 = 63060374321.3888.

[Penelopeys]

He he...

Let me introduce Slapdash... The PL Master Brain

[rednutt]

Originally Posted by slapdash

There are 5040 ways of choosing which of the 10 players gets the royal
flush, which gets the other straight flush, etc. Have you taken that into
account?

Let's do this carefully:

There are 52*51*50*49*48/5*4*3*2*1 = 2598960 possible boards.

To get QJ10 suited, there are 4 possible suits. To get one of those cards
paired, there are 9 possibilities. To get another card, nine or lower, say,
but not the 8 or 9 of the QJ10 suit, there are 30 possibilities.

So there are 4*9*30=1080 boards that will work. So the probability of
getting a good board is one in 2598960/1080 = 2406.44444...

Now look at how likely it is that particular players get the royal, etc.:

Given the board, there are 47*46/2=1081 hands, only one gives the royal,
so there's a 1 in 1081 chance of player A getting this.

Given the board and first hand, there are 45*44/2=990 possible second
hands, so there's a 1 in 990 chance of player B getting the straight flush.

Given the board and first two hands, there are 43*42/2=903 possible
third hands, so there's a 1 in 903 chance of player C getting quads.

Given the board and first three hands, there are 41*40/2=820 possible
hands for player D. There are 6 ways for him to get a high full house,
giving a 1 in 136.66666 chance.

In total that gives a 1 in

2406.44444 * 1081 * 990 * 903 * 136.66666 = 317824286579800

chance.

But there are 10*9*8*7 =5040 ways to pick players A,B,C,D, giving a total
chance of 1 in

317824286579800/5040 = 63060374321.3888.

[jeffers]

Originally Posted by slapdash

There are 5040 ways of choosing which of the 10 players gets the royal
flush, which gets the other straight flush, etc. Have you taken that into
account?

Let's do this carefully:

There are 52*51*50*49*48/5*4*3*2*1 = 2598960 possible boards.

To get QJ10 suited, there are 4 possible suits. To get one of those cards
paired, there are 9 possibilities. To get another card, nine or lower, say,
but not the 8 or 9 of the QJ10 suit, there are 30 possibilities.

So there are 4*9*30=1080 boards that will work. So the probability of
getting a good board is one in 2598960/1080 = 2406.44444...

Now look at how likely it is that particular players get the royal, etc.:

Given the board, there are 47*46/2=1081 hands, only one gives the royal,
so there's a 1 in 1081 chance of player A getting this.

Given the board and first hand, there are 45*44/2=990 possible second
hands, so there's a 1 in 990 chance of player B getting the straight flush.

Given the board and first two hands, there are 43*42/2=903 possible
third hands, so there's a 1 in 903 chance of player C getting quads.

Given the board and first three hands, there are 41*40/2=820 possible
hands for player D. There are 6 ways for him to get a high full house,
giving a 1 in 136.66666 chance.

In total that gives a 1 in

2406.44444 * 1081 * 990 * 903 * 136.66666 = 317824286579800

chance.

But there are 10*9*8*7 =5040 ways to pick players A,B,C,D, giving a total
chance of 1 in

317824286579800/5040 = 63060374321.3888.

I think we'll take your word for it.

[slapdash]

I'll have to remember this for future reference. Just put a few big numbers in
a post, and they'll all stop arguing!

[Penelopeys]

Originally Posted by slapdash

I'll have to remember this for future reference. Just put a few big numbers in
a post, and they'll all stop arguing!

[guesswest]

Party, who account for 70% of the market, and used to account for more, have dealt 4.2 billion hands since they started. So I'd guesstimate 6 billion hands dealt online total. That'd make it 1 in 10. But also a huge proportion of hands will have been dealt on heads up and six seater tables and in other kinds of poker games. Even ignoring all that though, certainly much more likely than not that this hand has never happened.

[lemonjon]

I can see what ur all saying, but

if we multiple the interflux generator gore readings by 12 and deduct the gamma ray particle inducer then clearly the answer is: warp 6

[slapdash]

Originally Posted by lemonjon

I can see what ur all saying, but

if we multiple the interflux generator gore readings by 12 and deduct the gamma ray particle inducer then clearly the answer is: warp 6

I was hoping nobody would spot that.

[Masterplan]

Originally Posted by lemonjon

I can see what ur all saying, but

if we multiple the interflux generator gore readings by 12 and deduct the gamma ray particle inducer then clearly the answer is: warp 6

i thought the answer was 42?

[lemonjon]

Well masterplan, the answer would have been 42 had I wrote core instead of gore, this little detail was enough to transcube the particle switch which makes the answer warp 6

[moFace]

i like cheese

[Penelopeys]

Originally Posted by moFace

i like cheese

That made my sunday

[Valiant23]

@moface!

Originally Posted by lemonjon

Well masterplan, the answer would have been 42 had I wrote core instead of gore, this little detail was enough to transcube the particle switch which makes the answer warp 6

But when you times that by 9 you get 42.

Obvious really..

[slapdash]

Originally Posted by Valiant23

But when you times that by 9 you get 42.

Hmm, I can see I have a long way to go here before my work is done.

[lemonjon]

but

if u had 4 and 2 together u get 14, divide that by the number of gamma data particles in a triangle and u get warp 9.8765432098764567738902, this gets rounded to the nearest hyper point which is 6


she cannnnnnie take much more of this captain

[Penelopeys]

And... 3 + 3 is 6....

[lemonjon]

r u sure

can u prove it