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paulish
02-04-2005, 07:26 PM
Let's say I'm on the button with [Q/images/graemlins/heart.gif J/images/graemlins/diamond.gif]. Both SB, BB and my self is in a push or fold modus.
SB and BB only plays a pair or an ace
I push with my my QJo

<font color="red"> What is the probability of getting atleast one caller?</font>

----- ----- -----
There are 52x51=) 1.326 different two cards combinations in the deck
There are (6x13=) 78 different combinations of pairs in the deck
There are (10x12=) 120 different combinations of Ax in the deck
Since I hold [Q/images/graemlins/heart.gif J/images/graemlins/diamond.gif], there are 8 less Ace combinations in the deck, and 6 less pairs combinations left in the deck
So my opponent, only has ([78+120]-[8+6])184 playbale hands.
----- ----- -----

I'm not sure what to do with my numbers next. Please give the "working", not just the answer /images/graemlins/laugh.gif

<font color="blue"> paulish </font>
JackWilson
02-04-2005, 08:10 PM
I'm too lazy to do the exact working, but just to alert you of a pretty big flaw in your workings right now:

[ QUOTE ]
Since I hold [Q J], there are 8 less Ace combinations in the deck, and 6 less pairs combinations left in the deck. So my opponent, only has ([78+120]-[8+6])184 playbale hands.

[/ QUOTE ]

That is true, but remember that you need to subtract a LOT of hands from the total number of two-card combinations because of the two cards you hold. It is roundabout 60 or so fewer hands I would guess? Okay, that's a pretty lousy guess but whatever.

So work out exactly how many two-card combinations your QJo eliminates and then:

184 playable hands / total number of combinations left in the deck (which is equal to 1326 minus those eliminated combinations above)

This is the individual chance of each having one.

Now let's call that percentage x. You would then go x/100 multiplied by x/100. This would give you another percentage. Invert this percentage. This is the chance of getting 1 or more callers.

Sigh. Late so my apologies if I made any big mistakes.
binions
02-04-2005, 08:20 PM
[ QUOTE ]
Let's say I'm on the button with [Q/images/graemlins/heart.gif J/images/graemlins/diamond.gif]. Both SB, BB and my self is in a push or fold modus.
SB and BB only plays a pair or an ace
I push with my my QJo

<font color="red"> What is the probability of getting atleast one caller?</font>

----- ----- -----
There are 52x51=) 1.326 different two cards combinations in the deck
There are (6x13=) 78 different combinations of pairs in the deck
There are (10x12=) 120 different combinations of Ax in the deck
Since I hold [Q/images/graemlins/heart.gif J/images/graemlins/diamond.gif], there are 8 less Ace combinations in the deck, and 6 less pairs combinations left in the deck
So my opponent, only has ([78+120]-[8+6])184 playbale hands.
----- ----- -----

I'm not sure what to do with my numbers next. Please give the "working", not just the answer /images/graemlins/laugh.gif

<font color="blue"> paulish </font>

[/ QUOTE ]

Theoretically, the formula for at least one of them having a pair or ace is 1 - (chance that SB won't have A or Pair * chance that BB won't have A or pair given that SB did not have A or pair)

Conceptually, this is similar to the formula for figuring the odds of hitting a flush draw with 2 to come [1-(38/47 * 37/46)].

Since you have QJo, there are 50*49/2 = 1225 possible hands left that the SB could have.

You are correct that there are 72 possible pairs left the SB can have. However, there are 16x12=192 possible Ax hands, and you have 8 blocked, so 192-8 = 184 possible Ax hands.

So, the chance the SB having will not have a pair or A is (1225-256)/1225 = 969/1225

Figuring the odds against the BB having a pair or Ace is more complicated. Note, we now are making assumptions about the SB hand (ie no ace, no pair), so there are only 48*47/2 = 1128 possible hands left that the BB can have.

If the SB does not have an ace or pocket pair, does it have a QJ, Qx, Jx, or xy (where xy are not paired or an A, Q or J)?

If the SB has QJ, then there are 68 pocket pairs left for the BB. If the SB has Qx or Jx, then there are 67 pairs left for the BB. If the SB has xy, then there are 66 pairs left for the BB.

9 possible QJ hands left,
240 possible Qx and Jx hands where x not an A, Q or J
720 possible xy hands where xy not paired nor A, Q or J

(720*66 + 240*67 + 9*68)/969) = 66.266 possible pairs left for the BB to have.

Add to that 184-8=176 possible Ax hands, and the chance the BB will not have a pair or Ax hand is (1128-176-66.266)/1128 or
885.734/1128

1 - (969/1225 * 885.734/1128) = 37.89% chance that SB and/or BB will have pair or Ace, or odds of 1.64:1 against.

By the way, if you are on the button and don't have a pair, it's 11.45% that at least one of the blinds will have a pair, or 7.73:1 against.
paulish
02-07-2005, 06:46 AM
Thx binions!
The reason I posted this question, Is that I'm trying to figure out when a steal from the button is mathematically correct.

But my math ability is too poor /images/graemlins/frown.gif
I found these two;
ranking hands against "push hands" preflop (http://forumserver.twoplustwo.com/showflat.php?Cat=&amp;Number=592219&amp;page=&amp;view=&amp;sb=5&amp;o =&amp;fpart=all&amp;vc=1)
Theorem of Blind Stealing - by MJ (http://teamfu.freeshell.org/tournament/theorem_blind_stealing.html)
...but I'm not sure what to do with these numbers next.

Do you have any thoughts binions?

<font color="blue"> paulish </font>
binions
02-07-2005, 11:55 AM
[ QUOTE ]
Thx binions!
The reason I posted this question, Is that I'm trying to figure out when a steal from the button is mathematically correct.

But my math ability is too poor /images/graemlins/frown.gif
I found these two;
ranking hands against "push hands" preflop (http://forumserver.twoplustwo.com/showflat.php?Cat=&amp;Number=592219&amp;page=&amp;view=&amp;sb=5&amp;o =&amp;fpart=all&amp;vc=1)
Theorem of Blind Stealing - by MJ (http://teamfu.freeshell.org/tournament/theorem_blind_stealing.html)
...but I'm not sure what to do with these numbers next.

Do you have any thoughts binions?

<font color="blue"> paulish </font>

[/ QUOTE ]

Factors to consider:

1. Do either of the blinds have extremely large or small stacks? If so, they are more likely to call with less than premium hands.

2. Some players defend their blinds with mediocre holdings, others don't. Some are more apt to defend against a "suspicious" button raise. Some aren't. How do the blinds play in your specific game?

3. How close are you to the money? The "gap concept" (see Sklansky's Tournament Poker book, or do a search here) works better the closer you are to the bubble, or the closer you are to a significant payout increase.

If the blinds will defend with any ace or pair, mathematically you will win the blinds (and antes) 62.1% of the time without a fight. The other 37.9% of the time, you know from Eastbay's list that you will win 27.1% - 86.9% against that range of hands (aces and pairs), depending on whether you have 32 offsuit (27.1%) or AA (86.9%).

So, let's say you have 32 offsuit. The blinds are 100-200. You each have 1000 chips. You decide that the blinds will call all in only with any ace or pair. Should you go all in?

Well, you will win T300 62.1% of the time when they fold. So your "fold equity" is T186

What happens when you are called? You win a pot of T2100 or 2200 (depending on which blind calls you) 27.1%, and it costs you T1000 to play, so your "caught equity" is (T569 or T596) - T1000 or -431/-404 (this assumes only one blind will call you. Sometimes both will).

Now, caught equity -431/-404 * 37.9% caught percentage = -163/-153 caught EV

Fold equity plus caught EV:
186 + -163/-153 = +23/+33

So, pushing with 32 has positive EV under these exact circumstances.