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bigchips
11-09-2004, 01:07 PM
Let's assume the following, and that my hole cards don't contain any of the cards in question. (please check my calculations)

1) After the flop, there is a 3 card flush on the board. What is the probability that one other player holds two flush cards?

1 - (38/47)*(38/46) = 33.2%

2) After the flop there is a pair on the board. What is the probability someone is holding at least Trips?

2/47 + 2/46 = 8.6%

3) After the flop there is a 3 card connected straight on the board (like 5 - 6 - 7). What is the probability that someone is holding a straight?

Under the above assumptions, it is possible to hold a straight by holding any of the following combinations:
8 and 9 or
8 and 4 or
4 and 3

For 8 and 9, &amp; 4 and 3 (the answer seems too high to me):
1 - (43/47) * (42/46) = 16.5% x 2 = 32.9% chance

My own thought is that the answer should really be:
(4/47) * (4/46) = .74% x 2 = 1.5% (but this seems too low)

I'm not sure how to solve for 8 and 4 because some of the numbers are already used in the above equation.

4) After the flop there is no pair on the board. What is the probability that someone flopped two pair? The flop comes (A Q 7)

If someone is holding:
A and Q or
A and 7 or
Q and 7
then they have two pair.

Any corrections / help on any of the above would be greatly welecome.

Thanks,

scandal
11-10-2004, 04:51 PM
[ QUOTE ]
Let's assume the following, and that my hole cards don't contain any of the cards in question. (please check my calculations)

1) After the flop, there is a 3 card flush on the board. What is the probability that one other player holds two flush cards?

1 - (38/47)*(38/46) = 33.2%

[/ QUOTE ]

What you were trying to calculate is the probability that the opponent has at least one of the remaining flush cards, not that he holds two (There are 10 flush cards left, so its
1 - (37/47)*(36/46) =~ 38%)

To calculate the probability that he holds two of the flush cards in the hole:

C(10,2)=45 posible starting hands with 10 remaining flush cards

C(47,2)=1081 possible starting hands when you've seen five cards.

45/1081= 4.2%

[ QUOTE ]

2) After the flop there is a pair on the board. What is the probability someone is holding at least Trips?

2/47 + 2/46 = 8.6%

[/ QUOTE ]

You can't add the probabilities this way because they are not independent events. To calculate that the opponent has at least one of the two remaining cards to make trips or better:

1-(45/47)*(44/46)= 8.4%

[ QUOTE ]

3) After the flop there is a 3 card connected straight on the board (like 5 - 6 - 7). What is the probability that someone is holding a straight?

Under the above assumptions, it is possible to hold a straight by holding any of the following combinations:
8 and 9 or
8 and 4 or
4 and 3

For 8 and 9, &amp; 4 and 3 (the answer seems too high to me):
1 - (43/47) * (42/46) = 16.5% x 2 = 32.9% chance

My own thought is that the answer should really be:
(4/47) * (4/46) = .74% x 2 = 1.5% (but this seems too low)

I'm not sure how to solve for 8 and 4 because some of the numbers are already used in the above equation.

[/ QUOTE ]

Better to use combinations to figure this one out:

There are 4 cards of each rank. So there are 16 possible combinations for each pair of 9-8, 8-4 and 4-3. That makes 3*16=48 possible starting hands making a straight on the flop. As above, there are 1081 starting hands possible after the flop, so the probability is 48/1081=4.4%

[ QUOTE ]

4) After the flop there is no pair on the board. What is the probability that someone flopped two pair? The flop comes (A Q 7)

If someone is holding:
A and Q or
A and 7 or
Q and 7
then they have two pair.

Any corrections / help on any of the above would be greatly welecome.

Thanks,

[/ QUOTE ]

Calculate this just like the last example. The probability that your opponnent holds cards matching two the flop cards (3 different pairs) is 4.4%

BruceZ
11-11-2004, 10:03 AM
[ QUOTE ]
[ QUOTE ]

2) After the flop there is a pair on the board. What is the probability someone is holding at least Trips?

2/47 + 2/46 = 8.6%

[/ QUOTE ]

You can't add the probabilities this way because they are not independent events.

[/ QUOTE ]

You can't add the probabilities this way because they are not mutually exclusive events, meaning that they can both occur at the same time. Independent events would not necessarily allow us to add their probabilities, and being able to add their probabilities does not imply that the events are independent. Many people confuse these two terms, and they are completely different things.

Independence means that one event occurring does not affect the probability of the other event occurring. Independent events would allow us to multiply their probabilities to get the probability of both events occurring simultaneously. Mutually exclusive events are events that cannot both happen at the same time. Only if the events are mutually exclusive would we be able to add their probabilities to get the probability of either or both of these events occurring.

[ QUOTE ]
[ QUOTE ]

4) After the flop there is no pair on the board. What is the probability that someone flopped two pair? The flop comes (A Q 7)

If someone is holding:
A and Q or
A and 7 or
Q and 7
then they have two pair.

Any corrections / help on any of the above would be greatly welecome.

Thanks,

[/ QUOTE ]

Calculate this just like the last example. The probability that your opponent holds cards matching two the flop cards (3 different pairs) is 4.4%

[/ QUOTE ]

The answer for 2-pair is different than in the last example for a straight because 2 of the cards are already on the board. Since there are 3 cards of each rank remaining, the correct probability that a particular opponent holds 2-pair is 3*3*3/1081 = 2.5%.