Probability of getting a decent hand in the next orbit

[HoldingFolding]

If I am in danger of being blinded out what is the probability of gettting a decent hand in the next x hands. Decent being (AA-TT,AK,AQs,AJs)? And how do you do the calculation.

Thanks in anticpation.

[Shoe]

If you are in danger of being blinded out in the next orbit, you shouldn't worry about if your AK, AQ, or AJ is suited or not.

[AngryCola]

[ QUOTE ]
If you are in danger of being blinded out in the next orbit, you shouldn't worry about if your AK, AQ, or AJ is suited or not.

[/ QUOTE ]

That may be true, but it wasn't the question. He was just bringing up an example. It's kind of snide of you to tell him that his example sucks so therefore it doesn't deserve an answer. This is the probablity forum, right?

How about this:
It isn't a tournament, and you are in no danger of going broke. Also, let us define "good hands" as:

AA-99, AKs, AK, AQs, AQ, AJs, AJ, KQs
/images/graemlins/spade.gif

[Precision1C]

There are 1326 combinations of hold em hole cards. AA-TT represent 6 each *5=30 AK=16 AQs/AJs 4 each=8 so a total of 54 of 1326 are good hands. Assuming that you have 10 hands in an orbit the odds of getting a good hand are 1-[1272/1326]^10=34%.

[uuDevil]

Here is my effort:

There are 6 ways to get each of AA/KK/QQ/JJ/TT, 30 total
There are 16 ways to get AK
There are 4 ways each to get AQs/AJs, or 8 total
There are C(52,2)=1326 total ways to deal 2 cards from 52

(30+16+8)/1326=.04072

So those hands represent about 4% of the hands you can be dealt. The probability of being dealt one of these hands at least once in x opportunities is

1-P(not being dealt selected hand)= 1-(1-.04072)^x

In 10 opportunities, this would be .340 or odds of 1.94:1.

Edit: Precision1C beat me to it. I'm slow. /images/graemlins/frown.gif

[AngryCola]

uudevil got to it already. /images/graemlins/grin.gif

[HoldingFolding]

Excellent stuff. Yes, it was really the formula I was looking for.

Would I be correct then in saying - all other things being equal - that I should go all in with any hand where:

1-(x/1326)^10=50%

In other words, if postion wasn't relevant (!) & you were guaranteed to get called and not simply steal the blinds, you should push with any of the top 89 hands for your first hand.

Ignoring suit:
AA-88 & AK-AJ for 90 hands?


Then adjust for the following hands: 1-(x/1326)^9=50% etc.

Hands Hands
Remaining to play
10 89
9 98
8 110
7 125
6 145
5 172
4 211
3 274
2 388
1 663
0 1326

Or am I missing something?

[HoldingFolding]

Can some bright spark confirm the formula if not the numbers, thanks.

[gaming_mouse]

[ QUOTE ]
Can some bright spark confirm the formula if not the numbers, thanks.

[/ QUOTE ]

Precision's formula is correct.

The strategy you suggest does not necessarily follow. It depends on what your goal is. To maximize expected profit? Maximize the chance you win the hand you play? This part of the question may not be simple.

gm

[Neil Stevens]

[ QUOTE ]
If you are in danger of being blinded out in the next orbit, you shouldn't worry about if your AK, AQ, or AJ is suited or not.

[/ QUOTE ]
The truth of that statement is exactly what the poster's question would determine.