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flytrap
12-03-2004, 11:42 AM
I was playing 10/20 hold'em at Greektown Casino in Detroit last night when a simple probability question came up. I flopped top set, and lost to a flopped straight. I grumbled to my friend about how I never make my 2-1 dogs one out of three times. The guy on the other side of me said, 'You're a way bigger dog than that, you're at least 3.1 to 1 to make a full house or better by the river.' Clearly this is absurd, and I wagered \$1000 with this man that I was better than a 2.8 to 1 dog, and he agreed that HEPFAP would be the source used for the correct answer. I consulted the book today, and although I won the bet, I want to know where my math went wrong.
On page 310 the book states that if you flop trips, you will make a full house or better 33% of the time. My math came up with 36.632%. Here's how I got that answer, and can someone please show me where I erred, as I would trust Sklansky and Malmuth over myself.

My hand:AA
Flop A23
47 unseen cards, 7 give us a FH or better. 7/47=14.893%
Turn A237, 46 unseen cards, 10 give us a FH or better 10/46=21.739%.
14.893%+21.739%=36.632%
That's a 3.632% difference from HEPFAP. Is my math right, and the book worded wrong, or is my math wrong? Thanks

fnord_too
12-03-2004, 01:44 PM
Easiest way to do this is to compute the chance you won't make the fh and subtract that from 1.

There is a 40/47 chance you miss on the turn followed by a 36/46 chance you miss on the river. So there is almost exactly a 2/3 chance you fail (~.666) so a 1/3 chance you succeed.

Your specific error is that you double count the times both the turn and river will give you a fh or better (say the final board is A2332 or A232A).

Mike Haven
12-03-2004, 02:26 PM
[ QUOTE ]
My hand:AA
Flop A23
47 unseen cards, 7 give us a FH or better. 7/47=14.893%
Turn A237, 46 unseen cards, 10 give us a FH or better 10/46=21.739%.
14.893%+21.739%=36.632%
That's a 3.632% difference from HEPFAP. Is my math right, and the book worded wrong, or is my math wrong? Thanks

[/ QUOTE ]

the way you have done it, you have forgotten that you have "used up", so to speak, 14.893% of the 100% of time - you are only interested what happens on the turn draw for the remainder of time

therefore you should be calculating 21.739% of (100 - 14.893)% for the second part, or 21.739% of 85.107%, which is 18.501%

DiceyPlay
12-03-2004, 03:03 PM
You added the probabilities of you getting it on the turn and then getting it on the river (given you didn't get it on the turn).

In reality you can get it on the turn or the river or both and to calculate it that way you would have to add these probabilities

p(get it on turn) * p(don't get it on river)
+
p(don't get it on turn) * p(get it on river)
+
p(get it on turn) * p(get it on river)

But that's the hard way. The easy way is to use the fact that the probability of you getting it + the probability of you not getting it is equal to 1.

So p(making fh or better) = 1 - p(not making fh or better)

And p(not making fh or better) is an easy calc.

p(not making fh or better) = 40/47 * 36/46 ~ .666

So p(making fh or better) ~ .334

-DP

MickeyHoldem
12-03-2004, 04:31 PM
Please forgive me for nit-pickin' but all of the answers that have been given do not properly address the situation.

Given that you have flopped the top set... and you need to improve in order to win the hand (FH or better) then the calculation should be as follows:

1 - (38/45 * (6/32 * 35/44 + 26/32 * 34/44)) = ~.3439

a) your opp. obviously has 2 cards that you do not need, therefore you should use 45 unknown cards instead of 47
b) this takes into account the chance that the turn may be a card that your opp. holds, thereby giving you only 9 outs on the river instead of 10.

Mike Haven
12-03-2004, 04:48 PM
whilst i understand your suggestion, unless you actually know the specific cards in your opponent's hand it's conventional to include them in the unknown cards in the full deck for the purposes of these types of calculations

DiceyPlay
12-03-2004, 06:01 PM
The cards you need could be any of the unseen cards. They could be the cards to come. They could be the cards in your opponentns hand. They could be in the deck never to see the light of day during the hand. Each unseen card has the same chance of being any card that you do not know (you only know the cards on the board and in your hand). All the unseend cards have a 1/47 chance of being any of the unseen cards on the turn and a 1/46 change on the river. The fact that your opponent holds two cards does not change this. Your opponent might hold cards of the same rank as the ones you need. If you knew that, that would change things. Otherwise, nothing changes and I stand by my calcs.

MortalWombatDotCom
12-03-2004, 08:23 PM
At least two different questions have been addressed so far.

1) what is the probability that a flopped set improves to a full house or better by the river.

2) what is the probability that a flopped set improves to a full house or better by the river given my opponent has flopped a straight.

these are not the same question, and their solutions are different.

that having been said, i can't figure out why

1 - (38/45 * (6/32 * 35/44 + 26/32 * 34/44))

would be correct for version 2... i get

1 - (38/45 * (6/38 * 35/44 + 32/38 * 34/44))

to miss your FH or better, you need to miss on the turn (which happens 38 ways out of 45 available cards) and then miss again on the river... of the 38 ways that you might have missed the first time, 6 leave you with 9 outs, the remaining 32 leave you with 10 outs.