A counting problem!

[lillumultipass]

Hi!

today, I was getting bored at a conference, so I tried to make a few poker calculations! But since the last time I has to solve counting problems was a few years back in undergraduate, I am rather unsure of my results (above all as it was made on a tiny sheet of paper, and so mostly in my head!)!Anyway, my goal was to calculate the probability of getting a set of 10's (with 2 in your hand in Hold'em) in the flop, but without any Q, K or Ace.
Okn the first step was to count the total number of flops: 50 cards remain, so there are C(3,50) possiblities, that is 19600 different flops.
Then, there are 2xC(2,49) flops with at least a ten, that is 2352 flops.
Finally, the hardest part: finding the number of flops containing one ten, and Q,K or A...
So, my calcu;lations went as follow: let's say the 10 of heart flop and the Q od heart after that, 48 cards remain. SO its 96 flops for the 2 10, that is 384 for the four queens. Doing the same calculations for the K and the A, I found 376 and 368, for a total of 1128 flops containing at least a ten and a Q, K or A. Thus, the proba that I was looking for is 1224/19600 that is around 6% or 15.5 to 1.
That's it!
I am really unsure of that, so if somebody could correct me if I am wrong, it would be great!
Thanks! /images/graemlins/smile.gif

[PygmyHero]

Here's how I did the problem:

To reiterate: you have TT, you want to know the probability of flopping a set with NO Q, K, or A on the flop.

C(2,1) * C(36,2) = 1260

There are 2 tens remaining in the deck, and we are choosing one. 36 represents the universe of remaining cards that meet your requirements (52 in deck - 2 hole cards - 2 other tens that are in first part of equation - 12 because that's how many Q, K, and A there are in the deck), and we are choosing 2 of them.

You are correct on the number of flops. C(50,3) = 19,600.

So your answer is 1260/19600 = 6.4% In odds this is 14.6:1 against.

For the sake of being thorough, the odds of flopping quads with NO Q, K, or A:

C(2,2) * C(36,1) = 36
36/19600 = 0.18% or 543.4:1 against

But really you would not care if the last card was a Q, K, or A because if you flop quads you will almost never lose that hand. If you don't care what the last card is then your odds are:

C(2,2) * C(48,1) = 48
48/19600 = 0.24% or 407.3:1 against

By the way, when reading this I wasn't sure how this would be useful. I love flopping a set with an ace on board since it virtually guarantees me action, and I am a huge favorite. Nonetheless, I hope I helped you out.

[lillumultipass]

Thx pygmyhero!
yes, it is simpler that way! I really don't know why I have first tried to find all the flops with a 10...ah, yes, I guess it was to find the probability of flopping the set! BTW, is this calculation correct???
It is indeed true that this stat is basically useless...I was really bored you know /images/graemlins/smile.gif

[PygmyHero]

Happy to help. I actually try to do these problems just so I can stay sharp on my own stuff.

I just looked over your post and you seem to have gotten to about the same number as me. I didn't actually check your calculations though. I'm not a good enough mathematician to do that. But I can say I'm fairly confident in my calculation. No doubt someone will correct it if I am wrong.

Also, I try not to look at other people's methods too much for fear that I'll end up just copying their work over instead of actually solving the problem. Once I knew what your question was and saw that your method was different than mine...I was off to reply.

[BruceZ]

Not bad, you basically just made the one fundamental type of mistake that almost everyone makes. First I'll show how I would do the problem, and then I'll correct the errors in your method and hopefully get the same thing.

First of all, most everyone here uses C(n,k) for choosing k things out of n, where n > k, so I will use that format.

There is no need to separately count the flops containing A,K,Q. If you want to exclude these, simply start with a deck with 38 cards without these denominations (and without your two hole card Ts), and count the flops with at least one T that you can make with this deck. Then divide by C(50,3) for the total flops that can be made with the original deck. Done! One thing that you MUST do that you did not do is to separate the case of two Ts from the case of just one T. We'll see why in a minute. Here is the whole answer:

2*C(36,2) + 36 = 1296 flops.

1296/C(50,3) = 6.6% or 1 in 15.1, but ODDS of 14.1-to-1.

The first term is for exactly one ten and two of the 36 non-tens, and the second term is for the 36 flops with two tens and one of the 36 non-tens. Remember to subtract 1 to convert probability (1 in) to odds (losers-to-winners).

That is very close to your answer, so the errors you made are not significant numerically, but they are significant conceptually, so we'll go through them.

[ QUOTE ]
Then, there are 2xC(2,49) flops with at least a ten, that is 2352 flops.

[/ QUOTE ]

You must separate the cases of one ten from two tens in separate terms or you will be counting the cases of two tens twice. You may say, "but the 49 can be a ten or a non-ten". It doesn't work that way for a reason which may seem subtle. Take the case of Ts x Tc, where x is any card. You counted the Ts as one of the 2 tens, and then the Tc as one of the 49 cards. But wait, you also counted the Tc as one of the 2 tens, and the Ts as one of the 49. This is the same flop, and you counted it twice. The correct way to count this is:

2*C(48,2) + 48 = 2304.

The first term counts flops with exactly one ten and two of the 48 non-tens, and the second term is for the 48 flops of two tens and one of the 48 non-tens. Notice that this number differs from yours by 48, since you double counted the 48 flops with two tens. This problem is epidemic, even among advanced folks, so if you can learn to avoid it by this simple example, you will be ahead of the game. Just remember that whenever you think you can get away with combining cases like this, you will invariably be wrong.


[ QUOTE ]
Finally, the hardest part: finding the number of flops containing one ten, and Q,K or A...

[/ QUOTE ]

We already showed that there need not be anything to this at all, but we will continue with your method...


[ QUOTE ]
So, my calcu;lations went as follow: let's say the 10 of heart flop and the Q od heart after that, 48 cards remain. SO its 96 flops for the 2 10, that is 384 for the four queens.

[/ QUOTE ]

In this case you can have 1 T + 1 Q, 2 Ts + 1 Q, or 1 T + 2 Qs. You hoped to take care of all of this with the "48", but as we learned above, it doesn't work this way. We need 3 terms. Let x mean neither a T or a Q.

TQx: 2*4*44 = 352
TTQ: 1*4 = 4
TQQ: 2*C(4,2) = 12
-------------------
total Q flops: 368

Your count of 384 is too high by 16 since you double counted the TTQ and the TQQ cases.

Now if we do the same thing for the K and the A, we will get 368 for both of these too, BUT, we will be counting TQK, TQA, and TKA in two of the cases. Since there are 2*4*4 = 32 of each of these, 96 will be double counted, so the final count will be 368*3 - 96 = 1008. Subtracting this from the total of 2304 found above gives 2304 - 1008 = 1296, which agrees with my original method, yay!

[lillumultipass]

wow, what a great explanation Bruce!!! Thanks a lot!
I know I should have paid more attention in Statistics classes lol.
Fortunately for me, it was not my major!
As to the ODDS thing...yes it was 1 in 15 that I meant...another big mistake!!!

[Orpheus]

I was taught a different notation (long ago), but here's how I did it, using grade school math.

1) How many dealt flops won't have a K, Q or A? (50-12)*(49-12)*(48-12) = 50,616

2) How many dealt won't have a K, Q, A or T? (50-14)*(49-14)*(48-14) = 42,840

3) Therefore, 50,616-42,840 (= 7776) lack a A/K/Q , but have a T.

4) I didn't bother with order permutations. It cancels out anyway: 7776/117,600 = 6.61% or 14.12 :1