Probablity Question for Advanced Players

[AdvancedPlaya]

I need help, I understand in the book percentages by number of outs you have. But how do you know your outs if you are only hoping for say, you have Ace-Jack-flop comes-King-3-7...and you need both a queen and a ten to make a straight. How would you calculate that. Or if u want to calculate that another Ace and Jack will come as next two cards to make two pair. Thank you. This is for needing the next two to be exactly that, not just one card u r waiting for.

[PygmyHero]

To reiterate, you have AJ and the flop is K 3 7. You did not give any information on suits, so I will ignore these in my answer (though be aware that some of the combinations I give below will lead to flush boards).

First lets look at making the running nut straight. You would first need to catch a queen or a ten on the turn, the odds of which are 8/47 (I am assuming we do not know any our opponents cards though you seem to put him on a king - if this is the case then it would be 8/46). Then we would need to catch a queen or a ten on the river - whichever one we did not catch on the turn. This is 4/46 (again, 4/45 if you put him on the king).

So, (8/47) * (4/46) = 32/2162 = about 1.5% or 66.6:1 against (the satanic implications here may mean you're drawing thin).

If you put him on the king it is no better.
(8/46) * (4/45) = 32/2070 = again 1.5% and 63.7:1 against

Now your other outs...from your question it seems that you're putting him on a king and if that is the case then just getting one ace would be enough...unless he has AK of course. You're not putting him on K7 or K3 are you? It also seems that running aces or running jacks should win the pot for you. If you want this info, I'll do it next time. I'll just address the question you asked, which seemed to be for hitting two pair.

You must catch an ace or jack on the turn. This is 6/47 (again, 6/46 if we take his king out of our hypothetical deck). Then you must catch an ace or jack (again whichever we did NOT get on the turn) on the river. This is 3/46 (3/45 without his king in the deck).

(6/47) * (3/46) = 18/2162 = only 0.8% or 119.1:1 against

With the knowledge of him having a king:

(6/46) * (3/45) = 18/2070 = up to 0.9% and 114:1 against.

So taking your question (again, not necessarily all your outs for two reasons - it seems that running jack or running aces would usually win and you will sometimes even if you catch the nut straight because it will make someone else a flush), it would seem that the answer you're looking for is the sum of

1.5% + 0.8% = 2.3% if you don't know any of his cards, and

1.5% + 0.9% = 2.4% if you put him on a king

Hope that's what you wanted.

[New York Jet]

Figure out your odds to hit each action individually, then multiply them.

Hitting a Queen/Ten combo on Turn and River (8/47)*(4/46) = 32/2162 = 0.0148 = 1.48%
Hitting an Ace and Jack on the Turn and River (6/47) * (3/46) = 18/2162 = 0.0083 = 0.83%
Note that these numbers are for hitting your hand exactly. It does not figure "or better" possibilites (i.e. Running Aces for trips). In the particular hand you mention, assuming you need two pair or better to win and neither hand having a flush possibility, you have just over a 4% chance to win at showdown.
I hope this helps. For more info, try "Getting the Best of It" by David Sklansky (you're at the correct site).

[gamble4pro]

That reverts to calculate combinations (this is the easisest and shortest way in figuring out the odds). In your example: you hold AJ, flop: K37.
The odds for straight (10 Q needed):
the favorable combinations (for turn and river card) are (10 Q) (the order does not count). The number of these is 4*4=16 (number of outs for the first card * number of outs for the second). The number of all possible 2-card combinations is C(47,2)=1081 (the formula of combinations is C(n,k)=n!/(k!(n-k)!), where n!=1*2*3*...*n). So the odds are 16/1081=1.48% . These odds also stand for any other combination of cards with 4 outs each (ex: 89, 56, etc.)
Attention! If you want the odds for AA to come:
The number of favorable combinations is C(3,2)=3 (3 aces left taken each 2)and not 3*3=9 (by doing the "*" operation, you are double-counting some combinations).
And so on.