2* Royal Flush whilst 6 tabling

[mostsmooth]

[ QUOTE ]
what are the odds of getting A2, 34, 56, 78, 9T, JQ, and AK simultaneously in that order while 7 tabling, where the leftmost table on your screen receives the A2, the next receives 34, etc etc, all of the same suit and every other hand flops a straight flush while the other hands flop 4 of a kind?

[/ QUOTE ]
the probability is 100%

[drbst]

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

[Luzion]

alphatmw: I PM'ed you the answer

[p-i]

[ QUOTE ]
This is a poorly worded question I think. His question is asking the expected amount of hours he needs to play before you flop two royal flushes at different tables at the same time. You can flop a royal flush by playing any two suited broadway cards; AKs, JTs, QTs, KQs, etc etc.

Probability to flop a royal flush while holding two suited broadway cards

3/50)(2/49)(1/48) = 0.00005102 = 0.005% or basically around 1 in 19600 flops.

This is a bit confusing to calculate the probability of getting a royal because first you to calculate the probability of getting suited broadway cards.

AKs, AQs, AJs, ATs, KQs, KJs, KTs, QJs, JTs.

9 x 4 different suits = 36 ways to get dealt broadway suited cards

36/1326 = 0.027149321 * 0.00005102 = 0.000001385 or 0.0001385% to flop a royal with random pocket cards.

Probability not to flop a royal on one dealt cycle of cards on any of your 6 tables...

(1 - 0.000001385)^6 = 0.99999169

Probability of flopping exactly one royal during one dealt set of cards on your 6 tables.

6 * (0.000001385) * 0.99999169^5 = 0.00000831

Chance of flopping at least two royals at the same time
= 1 - (0.00000831 +.99999169) = 0 [img]/images/graemlins/frown.gif[/img]

My calculator only has so many decimal points and can't figure it out for you.

[/ QUOTE ]

Just following your math in bc I get:

tworoyals = 3.16578 x 10^-10
tworoyals^100 = 1.11694 x 10^-955

[TomBrooks]

I know why the OP asks the question. He got two royal flushes at the same exact time while six tabling. And they were both in spades. I don't recall if he held the same hole cards with both.

I think the way he worded his question was a little ambiguous, but I think it would be of some interest, and a little simpler, to know the probability of getting two royal flushes at the exact same time while playing 6 tables or x tables.

If one then wanted to figure out what the odds would be of getting them both in spades, I suppose one could divide by 16 ( 4x4 = 16 ).

[p-i]

Recalculating but using 9/1326 for a single suit gives:

tworoyals = 1.79878 x 10^-12
tworoyals^100 = 3.14535 x 10^-1193

I think...