View Full Version : Tough question???

11-25-2003, 08:42 PM
Okay I actually have two questions, one which is very easy.

The first question is: is the probability that someone will hit a pocket pair simply (4/52)*(2/2)=.0769? This should be basic intersection probability right?

The next question I am very interested in and I am not sure if it is even possible to calculate but here it is anyways.
suppose the flop has put a pair on the board: AA4, what is the probability that someone has one other A in their hand to make trips? is it:
which if I calculated correctly would be .003673

any help with this would be great

11-25-2003, 09:41 PM
Is the first question, what are the chances of flopping a
set or having a pair dealt in your hand? The first
(including quads) is 1-C(48,3)/C(50,3) and the last is 1/17
(pick first card; second can only be 3 of 51 of the
remaining). By the way, the first number is about 0.117551
and C(n,r) = the number of combinations of n objects chosen
r at a time (or the number of possible sets of size r
chosen from a set of n>=r objects). It is just
n!/((n-r)! x r!) where k! = k x (k-1) x...x 1.

If the board reads XXY then for a two card hand to have one
of the two remaining Xs (or more), the combinations are 1
(XX) + 2 x 47 (Y on flop excluded) = 95 out of C(49,2) =
1176 hands. I suppose you also should note that there are
C(3,2) = 3 hands that flopped a full house without an X.
Altogether, the chances of a specific hand being trips or
better is 98/1176 = 1/12 = 0.083333, a very simple number
to remember. On the other hand, obviously if a specific
action (or lack of action) has occurred, these odds don't
necessarily help you in the play of the hand. Also, hands
containing an ace tend to get played.

11-25-2003, 10:36 PM
I realize that it most cases an A probably will be played, so basically what you are saying is if a pair is on the board, the probability that someone has trips or better is 8.33%?
Also can I get you to possibly explain the formula you used. dumb it down a little for me please. I am currently in a Stats class right now but am a bit confused by your formula, probability has been the most confusing part for me.
Thanks in advance

11-26-2003, 12:02 AM
Say AA4 is on board. Now there are only 52-3 = 49 cards
left in the deck. For a two card holding there are now
C(49,2) = 49 x 48 / 2 combinations. The holding of 7s 2c
is exactly the same as 2c 7s so the combinations just
count the possible hands exactly once. Each of these
C(49,2) = 1176 combinations are equally likely.

The number of combinations for trips or better is:

Quads: 1
An ace+ other: 2 x 47 = 94
(two choices of which remaining ace x 47 choices of
the remaining non-ace: one 4 is gone on the flop)
pocket 44: 3

so that makes 98/1176 = 1/12.

There is another way of determining this instead of
enumerating the above possibilities (there may be several
ways of getting the correct answer): consider the
chance of not getting an ace in two cards. For a hand
not to have an ace, the probability is 47/49 x 46/48 or
the probability of not having an ace 190/(49x48) and the
same result follows (add 3x2/(49x48) for 44).

11-26-2003, 02:26 AM
Probability you will be dealt a pair is 1*3/51 = 1/17.

With a flop of AAX, and you don't have an A, the probability that at least one of your N opponents has an A assuming random hands is 1 minus the probability that there are no aces out. This is 1 - C(45,2N)/C(47,2N). That is, there are 47 cards remaining, and 45 of them are not an A, so there are C(45,2N) ways to choose 2N cards without an A, and C(47,2N) total ways to choose 2N cards.

For example, if you have 4 opponents, 1 - C(45,8)/C(47,8) = 1 - (45*44*43*..*38/8!) / (47*46*45*...*40/8!) = 1 - 39*38/(47*46) = 31.5%.

Here are the solutions for different numbers of opponents.

<font class="small">Code:</font><hr /><pre>
opponents P(A out)

1 8.4%
2 16.5%
3 24.1%
4 31.5%
5 38.4%
6 45.0%
7 51.2%
8 57.0%
9 62.4%
10 67.5%</pre><hr />

11-26-2003, 05:33 PM
And dont make the mistake of thinking that if you have 9 opponents and 4 fold that the probability of an A being out is only the 5 player probability of 38.4%. You are probably much closer to the 9 player probability of 62.4%.

(8% times the number of players is close enough to not bother memorizing the table).

11-27-2003, 12:36 AM
if that is what the probability would be at the start how would you calculate it if people fold?