mathematically challenged seeks help

[buzzo]

Can anyone give me a site or the info for the following ? I know with a pair my odds of hitting a set on the flof is 7.5 to 1 against. If I have QQ in a 10 handed game and no other player was dealt a Q what are my odds of flopping a set? If I other player was dealt a Q ?
If I am dealt Ax what are odds of me flopping an A if no other player was dealt an A, 1 other player dealt an A,2 other players dealt an A ?
Appreciate any help!!!!!

Thanks i really have a need for this info

Buzzo

[REDMAN]

Why concern yourself with if others have an A Q or not. Unless you know if they have an A or Q you have to treat though those cards as though they are in the deck. It is possible someone exposed a card or something like that but this the exception and if this happens and you are playing for trips just fold. On the flop you count the number of cards that are in the deck. You have 2 cards so 50 are left in the deck. To count the probability of hitting trips or an A it is easier to count the chance of not hit in an A or tips and then take the inverse.

The math on not hitting trips is on the flop:

(1st card 48/50)*(2nd card 47/49)*(3rd card 46 /48) =

50 cards you do not know to start (you know your 2) and 48 that don’t give you trips.
49 cards on the 2nd card and 47 that do not give you trips.
48 cards on the 2nd card and 46 that do not give you trips.

(48/50)*(47/49)*(46/48)=

.96*.959*.958=.8824

1-.8824= .1176

Or you have 12% chance of hitting trips.

Not hitting an A is:

(47/50)*(46/49)*(45/48)=.827

1-.827=.172

Or 17.2% to hit an A

[buzzo]

I have a reason for the question. Follow you logic until .96*.959*958 =n 88.24 Can understand how you
96, etc, have no idea how those 3 figures equal .8824. If I knew that maybe i could figure probabilities. Thanks for your taking time to try and help.
THE COMMUNITY CARDS MOST LIKELY TO CAUSE ACTION ARE THE CARDS LEAST LIKELY TO APPEAR.

I am an inventor who believes he can change that law without violating the limits of a deck of 52 cards. In other worrds odds against a player getting AA dealt to him remain 1 in 221 e3ven if another player has AK. Sounds impossible but can be done.

Buzzo

[buzzo]

OOPS Think I figured it out. Multip0ly those 3 figures, then subtract result from 1. Will try and answer my own question and post result here for verification. THANKS

[REDMAN]

Here is an easy way to "approximate" the odds. Use the rule of 2. That means for ever card you want to hit times by 2. For example if you have QQ and want to hit a Q on the flop, you have 2 Q’s left in the deck. That means the first card you have 2 x 2 = 4 or 4% chance of hitting with one card. If you are calculating hitting a Q on flop (3 cards) you can add (4% + 4% +4%) = 12%. If you are calculating hitting a Q on the turn you have 4% on change on that card and so on. This is not exact but close.