Hello! Let's assume that U are...

[SittingBull]

observing 2 stud tables. Table 1 has 2 different player each having an "A" as an upcard on 3rd and 5 other players with different cards. Table 2 has one "A" showing on 3rd. The other 5 players have different cards.Which table is more likely to have a PR of A's on 3rd?
HappyPokering,
SittingBull [img]/images/graemlins/smile.gif[/img]

[AdamK]

Im too lazy to do the arithmetic - and that sort of thing gives me a headache nowadays [img]/images/graemlins/confused.gif[/img]
But im gonna guess Table 2 ..
what do i win ?

[SittingBull]

respect from me when u sit at my table!
SittingBull [img]/images/graemlins/smile.gif[/img]

[wolle]

Table 1 has 17.98% probability for a pair of aces
Table 2 13.19% for a pair of aces.
Hope it's correct calculated.

[Hauser_III]

It's Table 1, with the two different players having an "A" upcard on 3rd. There's an analysis proving it in the stud/8 section of SuperSystem 2 written by Brunson's son.

[Roland]

[ QUOTE ]
Table 1 has 17.98% probability for a pair of aces
Table 2 13.19% for a pair of aces.
Hope it's correct calculated.

[/ QUOTE ]

Does this mean that it's 8.99% for each ace to have another ace in the hole? Or am I getting this completely wrong.

[OrangeKing]

[ QUOTE ]
[ QUOTE ]
Table 1 has 17.98% probability for a pair of aces
Table 2 13.19% for a pair of aces.
Hope it's correct calculated.

[/ QUOTE ]

Does this mean that it's 8.99% for each ace to have another ace in the hole? Or am I getting this completely wrong.

[/ QUOTE ]

Off the top of my head, I'm going to say slightly less - you have to account for the possibility of another player having wired aces.

[SittingBull]

those players who passed the test. U have won a prise: my respect! [img]/images/graemlins/cool.gif[/img]

[Andy B]

They have your respect because they paid $30 for a book?

[img]/images/graemlins/grin.gif[/img]

[wolle]

Does this mean that it's 8.99% for each ace to have another ace in the hole?

yes