View Full Version : # of limpers needed for 33

10-21-2003, 07:02 PM
So Im taking this probability course, and Id like to tackle some holdem problems for exercize. These exercizes are mostly combinitorical, and I can elaborate upon request.

Should I limp with 33?

Lets say for the sake of argument the answer would be 'yes' povided you will flop a set. The probablility of flopping a set (according to my calculations) is:

(2,1)(49,2)/(50,3) =6/50 ...pronounced "2 choose 1, times 49 choose 2, devided by 50 choose 3"

6/50 = 1/8.33333 so pot odds require us to be playing against at least 8 other players on the first betting round. This doesnt take into account implied odds however. If we expect to be able to take 8*(cost to limp) out of the game on that hand, then limping is profitable. For example: 2 limpers to me on the button with 33. I limp, SB completes, BB checks. Thats 4*(cost of limping) I flop my set, first limper bets out, second limper calls, I raise, and the limpers call behind me. Thats the other 4*(cost of limping), regarless of what the blinds do. This of course assumes that my set of 3s is good.

So if i believe my small pairs are going to be played >4 handed, its likely Ill be able to get enough out of the hand to make limping profitable. /images/graemlins/crazy.gif

10-21-2003, 08:11 PM
The probablility of flopping a set (according to my calculations) is:

(2,1)(49,2)/(50,3) =6/50

This has a subtle and common error. By saying (49,2) you are apparently trying to also include quads and full houses (or you didn't really think about that at all), but you are not including them the right way. If you are going to include cases of the board pairing, these must be handled explicitly in separate terms, or you will be counting them more than once. This is because it is not clear if the (49,2) will include your rank, or if it will be a pair itself, and this will always cause a problem when you are multiplying combinations this way.

If you want the probability of a set, full house, or quads, here is the correct expression. I will use the notation C(n,k) instead of (n,k) as you have used.

[ C(2,1)*48*44/2 + C(2,1)*48*3/2 + C(2,2)*48 ] / C(50,3) = 11.8% or 1 in 8.5 or 7.5-1.

The first term is for a set, the second is for a full house, and the last is for quads. Of course if you just wanted a set without a full house or quads, you would just use the first term, and this would be 10.8%, 1 in 9.3, or 8.3-1.

The rest of what you say looks fine.

10-22-2003, 01:07 AM
I had intended to include cases where one flops quads or a full house. My calculation should have given the frequency that at least one '3' comes on the flop. In the probability section of HEFAP, it lists the probability of floping a set or better as ~11.8% just like you stated... where I have 12% exactly. So Im pretty sure youre right, but Im failing to see my error.

I guess what youre saying is im counting flops like:
3 /images/graemlins/spade.gif 5 /images/graemlins/club.gif 5 /images/graemlins/heart.gif
3 /images/graemlins/spade.gif 5 /images/graemlins/heart.gif 5 /images/graemlins/club.gif
as different, but I dont think I am.

In my calculation:

C(2,1) = 2 different ways to draw a 3 in one spot. And C(49,2) = 49*48/2*1 different ways to draw any two remaining cards.

Could you name two flops that arent different that I counted as discrete? Thanks for any help. /images/graemlins/confused.gif

10-22-2003, 02:12 AM
Could you name two flops that arent different that I counted as discrete?

3 /images/graemlins/spade.gif 3 /images/graemlins/club.gif 5 /images/graemlins/heart.gif
3 /images/graemlins/club.gif 3 /images/graemlins/spade.gif 5 /images/graemlins/heart.gif

Actually, the only ones you counted twice were the quads. You counted full houses correctly. You counted 2352 total flops, and I counted 2304 total flops. The difference is exactly the 48 quads.

10-22-2003, 09:43 AM
Another thing to remember when playing small pairs is how likely your opponent are to pay you off. If your table is full of calling stations that will call you down with any pair, then you can get by with less limpers. You get 6-1 implied odds heads-up against one person who will call you to the river should you flop your set with out ever getting in a raise. If there are a couple of these people in your game, you can generally play you small pairs from any position as long as you don't anticipate a raise.

10-23-2003, 01:45 AM
Thanks for setting me straight... that makes perfect sense. A more appropriate expression might be:

[C(2,1)(49,2) - C(1,1)C(48,1)]/C(50,3) ...in order to subtract the quads i counted twice.

Nottam... if theyre calling me down with any pair, i ought to hold a medium pair like 88 or 99 to rely on those extra implied odds. Im not really good enough at this point to be comfortable ramming 33 into calling stations, but maybe ill get there some day.

10-23-2003, 01:47 AM
Sorry, I totally misread your post. I agree 100%.