More Bad Beat Odds Questions

Hi, I'm not sure if you've covered this before exactly, but I checked your bad beat probability tables and I'm not sure exactly how to convert the numbers to x to 1 odds.

So here's my question Mr. Wizard: In Atlantic City, the bad beat jackpots are quads or better must lose with both hole cards playing in both hands and the quads must have a pocket pair. So what are the odds if you have a full 10-handed game and EVERY player plays EVERY hand and sees all 5 board cards EVERY time. What are the odds (like 100,00 to 1 for instance) of it happening?

Also, how about if you have a 9-handed game in the same scenario where everyone sees every flop and all 5 board cards.

One last one. How about if it's folded around to the blinds. Say before either one looks at their hole cards. What are the odds that they would hit the bad beat just the two of them if they played out the hand EVERY time and saw every board card, not knowing in advance if they even had bad beat qualifying hands?

Thanks in advance for answering these questions or referring me to where they've already been answered. Also, how do you convert probabilities like this: 0.00001086 to x to 1 odds?

Greg

Well, doing 1/decimal should give you x to 1 odds, I think (i.e. your probability is .05- 1/.05 = 20 because .05 = 1/20

So you're basically just taking the reciprocal. Hope that helps... hope it's right. X to Y always confuses me, like in blackjack and roulette XD