what are the percentages or odds of this

[Mikey]

Flop is Q 4 A.

You hold KQ.

Your opponent bets and you call.

Turn is an A.

He bets again.

Disregard the action.

What is the likelyhood of your opponnent actually holding trip Aces now?

Can you show the calculation too..

Thanks in Advance.

[dabluebery]

I guess I'd do it like this. There are 46 cards left in the deck. There are 3 aces. The chances of him having two aces in his hand are;

3/46 * 2/45 = 6/2070 which is 1/345.

[Ghazban]

Sounds right to me IF he is equally likely to be holding any two cards. Given the betting and whatnot, I think that the pure calculation will be lower than what you'd see in practice. From a purely mathematical standpoint, I get 1/345, too (same calculation).

[David K.]

You guys missed the turn....

There are 46 cards left, two of them being aces.

The number of combinations of two cards including at least one ace is 45 (one ace times 45 other cards, including the other ace).

The total number of combination of any two cards from the remaining 46 is C(46,2)

The probability of a set of two cards containing at least one ace is 45/C(46,2)=45/[46!/(46-2)!2!]
=45.2.44!/46!
=2/46=1/23=.0435==4.35% , or 1:22 odds.

[dabluebery]

I didn't miss the turn. I missed the Ace on the flop. I'm stupid. Definitely this is the way to do it;

2/46 * 1/45 = 1/1035. Sorry.

[David K.]

Now you're giving him the probability that he has quad aces /images/graemlins/wink.gif

[Ghazban]

Oops; I missed the one on the flop, too. Still, is it really useful to make this calculation as if all hands are equally likely? The fact that the guy is betting makes it more likely that he has a piece of the flop than that he's holding garbage. Clearly, the effect of this cannot be quantified with any certainty but you won't ever be using this information in a vacuum.

[David K.]

[ QUOTE ]
The number of combinations of two cards including at least one ace is 45 (one ace times 45 other cards, including the other ace).

[/ QUOTE ]

I'm sorry, I was wrong. That number is actually 2x44+1 (2 ways to have an ace and 44 non-ace cards, and 1 way to have two aces).

The correct result is then (2x44+1)/C(46,2)=.08599 or 8.599%

[daryn]

time for the correct answer.

he could have 946 hands without an ace, 1 hand with both aces, and 88 hands that contain 1 ace.

thus the chances he has trip aces are just 947:88 or 10.76:1 against.

[David K.]

We find the same result using a different method, except I was counting the hand with two aces as part of the probability. Under your system, the odds would be 946:89 or 10.629:1 which give us a probability of 1/(10.629+1)=0.08599 or 8.599%

odds that the opponent holds EXACTLY trips and not quad aces are, you are right, 947:88 or 10.761:1 which give us a probability of 1/(10.761+1)=0.08503 or 8.503%

Same difference. /images/graemlins/wink.gif

[daryn]

sorry it just bugs me when i see a question about probability, and then 10 guys answer, all wrong. i know people make mistakes, blah blah blah...

[David K.]

Yeah I wish I could have edited my first post but the forum won't let me. I'll be more patient next time before answering. But at least I saw my mistake and corrected it.

[daryn]

true but if you want to be technical, and i often do, your second post was wrong too /images/graemlins/smile.gif..

i know you know what you did, and i know it's nitpicking, so we'll leave it at that /images/graemlins/grin.gif

[David K.]

No wait, I want to understand... wrong because I included the AA probability? It's always hard to know when someone say "what's the probability that I'll flop trips" if that means exactly trips or "at least" trips, don't you agree?

Or was I wrong somewhere else?

[daryn]

yeah that's what i meant. i guess i went back and read his question before answering, and it said "what's the probability he has trips"... and 4 of a kind is not trips.

like i said, just a nitpick. i know BruceZ always subtracts the chances of flopping 4 of a kind when he talks about the odds of flopping a set.

[David K.]

Oh alright, just wanted to make sure.