Geothe where are you!?

Poker_Fail

Robbie Barrs Lol..Odds of this happening me kq him a 10...all in on flop of q105..

turn 10 river 10 for quad 10's GTFO!!!

Need something settled pls.

 

I think the odds of runner runner 10s is 2/45 * 1/44 = 1/990

DJ says it's 2/37 * 1/36 or 1/666

unknown_username

I'm very flattered that someone's put my handle into a subject line . . . I'm sure there's some on the board here who'd like me to stick my abacus right up (yeah you guessed...).

So ending up with quads with two odd cards after going all-in pre-flop? Let's take it by flop, turn and river:

Flop - 50 unseen cards, three tens in the deck somewhere with three chances to pull one:

3/50 x 3/1 = 9/50,

Turn - 2 tens left in the deck from 47 unseen cards:

2/47

River - 1 ten left in the deck from 46 unseen cards:

1/46

The coincidence of all three events occuring = 9/50 x 2/47 x 1/46  =  18/108,100  = 1/6,006

or 6,005-1

Pretty long odds although not quite as long as flopping a quad from one of two odd cards (which has happened to me once). I would say that there are variations to the calculation in this resulting from two or three of the tens showing on the flop instead of just one. Also, as someone has pointed out before, it isn't quite bang on the 3/50 x 3/1 as this bit is also subject to variation - ie if the 10 card shows first, second or third on the flop. But I reckon its good enough for doing simple sums in your head.


==

As to the question following as to whether you count all of the unseen cards or just those left to deal when doing the sums? The answer is to take account of all of the unseens, regardless of whether they've been dealt out, mucked or are left in the deck.

Take an example of a game with 23 players - 46 hole cards would be dealt out, with some hands folded, leaving just 6 in the undealt deck. Assuming you only accounted for those 6, the odds of pairing one of two odd cards would be calculated as:

(3x2)/6 (three of each of two odd cards left in the deck)  = 6/6 = 1/1.

1/1, or 100%, is a dead certainty. Of course pairing one of your two cards wouldn't be a dead cert (on balance of probability some if not all of those 6 outs will have been dealt to the other 22 players) so the basis of the calculation must be wrong.

==

Learn a few numbers and impress your friends! Just remember that 62% of statistics are made up on the spot and are usally bu11s#it . . .

Good cards.

Poker_Fail

In Response to Re: Geothe where are you!?:

I'm very flattered that someone's put my handle into a subject line . . . I'm sure there's some on the board here who'd like me to stick my abacus right up (yeah you guessed...). So ending up with quads with two odd cards after going all-in pre-flop? Let's take it by flop, turn and river: Flop - 50 unseen cards, three tens in the deck somewhere with three chances to pull one: 3/50 x 3/1 = 9/50, Turn - 2 tens left in the deck from 47 unseen cards: 2/47 River - 1 ten left in the deck from 46 unseen cards: 1/46 The coincidence of all three events occuring = 9/50 x 2/47 x 1/46  =  18/108,100  = 1/6,006 or 6,005-1 Pretty long odds although not quite as long as flopping a quad from one of two odd cards (which has happened to me once). I would say that there are variations to the calculation in this resulting from two or three of the tens showing on the flop instead of just one. Also, as someone has pointed out before, it isn't quite bang on the 3/50 x 3/1 as this bit is also subject to variation - ie if the 10 card shows first, second or third on the flop. But I reckon its good enough for doing simple sums in your head. == As to the question following as to whether you count all of the unseen cards or just those left to deal when doing the sums? The answer is to take account of all of the unseens, regardless of whether they've been dealt out, mucked or are left in the deck. Take an example of a game with 23 players - 46 hole cards would be dealt out, with some hands folded, leaving just 6 in the undealt deck. Assuming you only accounted for those 6, the odds of pairing one of two odd cards would be calculated as: (3x2)/6 (three of each of two odd cards left in the deck)  = 6/6 = 1/1. 1/1, or 100%, is a dead certainty. Of course pairing one of your two cards wouldn't be a dead cert (on balance of probability some if not all of those 6 outs will have been dealt to the other 22 players) so the basis of the calculation must be wrong. == Learn a few numbers and impress your friends! Just remember that 62% of statistics are made up on the spot and are usally bu11s#it . . . Good cards.
Posted by Goethe

What I specifically meant was after the flop has came, just the odds ofthe 10 coming on the turn and river.

The bit I bolded is basically the point I was trying to make in my side of the arguement, but I couldn't explain it very well at all lol

 thanks : )

unknown_username

Well. . . 2/2,162 (2/47 x 1/46) = 1,080-1

2,162 is a figure worth remembering - it's the number on the bottom for runner-runner draws calculations.

So making two pair on the turn/river would be (6*3)/2,162. Divide by ten and then by two (so overall by 20 and not 18) and you get 1/108 or greater than 107-1 (if you actually divide it by 18 it comes to 120, so the actual answer is 119-1) Rounding up or down to the nearest 10 means you can do it in your head at the table, and not have to call up Pokerthingy or whatever it's called. Where odds are greater than 100-1, I think the actual figure becomes largely academic - most people will play hands where the odds are more than 99-1 in their favour?