*Originally posted by WildBeans*

As far as I know, 6 outs gives you about 24% chance of winning and so thats 24:76 or 1:3,1

edit1: I think you're getting confused, because one street is 12:88 and 2 streets is 24:76, - for odds like this you don't just halve one of the sides... However you do double the percentage chance - e.g. if I have 12% to hit one street, then I have 24% to get it on 2 streets..

edit 2: I would still like to know if you'd always give yourself 6 outs in this situation.. Maybe one should be discounted due to the good possibility of his calling A-rag hands preflop..

No, I understand that you get 2X more outs for the push. But I think with more precise calculations we can get a slightly higher value for the necessary pot odds:

#1>> 6:47 = 12,7% or 25,4% for 2 streets. 25,4:74,5 = 1:3,4

or

#2>> 6outs:41odds = 1:6,83 < I'd need 6,83 pot odds for BEQ in one street

And half of that =

__3,4__ pot odds for 2 streets.

Does it make any sense?