Probability theory question

[robilaruk]

need help with this. 10 seater stt blinds are substantial to stacks (say 100/200) so its shove or fold with 1500 chips and there are 6 players left you are in the CO and have A6, what is the chance of the button SB or BB having a bigger Ace? you are on the button same question for SB & BB you are in the SB, same question for the BB if there were only 5 players left does it make a substantial difference to the probability? (and if so how much of a difference 4 handed on the bubble?) I have no idea how to work this out, I can play around in pokerstove but that just matches hand vs hand (range) which is not quite the same thing? (or is it?:$) any help most appreciated:ok Cheers Damo

[slapdash]

Re: Probability theory question

if there were only 5 players left does it make a substantial difference to the probability? (and if so how much of a difference 4 handed on the bubble?)

The only way the number of players makes a difference is that if the players in early position have folded, then it makes it less likely that they have an ace (since they're probably more likely to bet with an ace than without one), and this increases the chance that the players in later position have an ace. But it's hard to quantify how much difference it makes, since it depends on what hands they'll fold, which will vary from player to player. It probably doesn't make that much difference, though. If you ignore the information from the early position players folding, it's possible to answer the questions you asked, but it's quite an involved calculation. I don't have time tonight, but I'll see what I can do later ...

[GaF]

Re: Probability theory question Sklansky-Chubukov rankings are very helpful here :ok With blinds of 100/200, cards of A6o, and even if you show your cards to your opponent ...... You should shove from the small blind with a stack of up to 5600 You should shove from the button with up to approx 2800 Of course if you don't show your cards to your opponent, it's possible he'll fold a better hand, so you can do it with bigger stacks..... This is for guaranteed positive ev in cash games, and in tournament chips in tournaments (obviously payout structure may give you slightly different answers...)

[Mr Intensity]

Re: Probability theory question I thought he was just asking how likely it is a player in SB/BB has a better ace? Nothing to do with shoving or where he's ahead.:unsure If I'm too tired to read the question right I'm not going to get the calculator out:lol:lol

[robilaruk]

Re: Probability theory question correct! I am purely interested in this: all hands folded to me in XX postion I have A6 - I have no room to manuever with a normal raise due to my stack vs the blinds so its shove or fold - what is the probabaility that I will run into one of the remaining players holding Ax (bigger than 6) if I shove. I am not interested in any other holdings except Ax I picked A6 as it is an ok hand, but not a great hand for shoving (compared to A10+ where your second card might be good enough to win the pot for you) and how much position makes a difference (if much at all?) - so am trying to figure out whether I should be shoving with it (does that make sense?) Cheers Damo

I thought he was just asking how likely it is a player in SB/BB has a better ace? Nothing to do with shoving or where he's ahead.:unsure If I'm too tired to read the question right I'm not going to get the calculator out:lol:lol

[GaF]

Re: Probability theory question Why are you only worried about Ax? Why not KK? QQ? JJ? 66? 55? etc? Accept I didn't answer the question - but was trying to read between the lines a little, and think you may be asking the wrong question for the wrong reasons :unsure However, to try and answer the question.... The chance of being dealt A6 should be something like.... Dealt an Ace first will be 3/50 followed by a card bigger than a 6 (not an Ace) will be 28/49 which is a combined probability of 84/2450 (about 3.5%) it can be dealt in reverse too, so (28/50) x (3/49) = 84/2450 (naturally, the same) We have a combined possibility therefore of about 7% of one player left to act holding Ax bigger than A6, when you hold A6 in your hand (ignoring the phenomenon Slapdash talked of) - of course sometimes the player may fold to your shove (but lets assume he doesn't and always calls) - your A6 will still beat his Ax about 30% of the time (and if he calls with Ax smaller than A6, you will win 70% of the time) - so the chances, given that you hold A6, of one remaining opponent holding Ax bigger than A6, and beating you in a showdown is about 7% x 70% = 4.9% - about 1 in 20....

[slapdash]

Re: Probability theory question

We have a combined possibility therefore of about 7% of one player left to act holding Ax bigger than A6, when you hold A6 in your hand

That looks right. And if there are two or three players left to act, the chance of one of them having a bigger ace is about two or three times as high (so 14% or 21%). Actually, it will be slightly less, because of the possibility that more than one of them has a bigger ace: roughly 14% or 20%, probably. This is always ignoring any inferences you can draw from the fact that some of the players have already folded.

[robilaruk]

Re: Probability theory question

Accept I didn't answer the question - but was trying to read between the lines a little' date=' and think you may be asking the wrong question for the wrong reasons :unsure[/quote'] I am trying to work out if A6 is the minimum Ace I would shove given the scenario I discussed, hence my question - and how often my Ace will be outkicked why would villian also have the same chance of 3/50 (3/49), surely it should be 2/50 (2/49) as I already have an ace? This is the bit where probabaility gets the better of me, does me holding an Ace make no difference to villians chance of holding the same?:unsure Thanks Damo

[slapdash]

Re: Probability theory question

why would villian also have the same chance of 3/50 (3/49), surely it should be 2/50 (2/49) as I already have an ace? This is the bit where probabaility gets the better of me, does me holding an Ace make no difference to villians chance of holding the same?:unsure

Since one of your two cards is an ace, three of the other 50 cards are aces, so the chance of any particular card in one of your opponents' hands being an ace is 3/50. If you didn't have an ace, it would be 4/50.

[GaF]

Re: Probability theory question

I am trying to work out if A6 is the minimum Ace I would shove given the scenario I discussed, hence my question - and how often my Ace will be outkicked

In that case, I refer you to my first answer ;) In a cash game (I'm choosing a cash game because it isn't complicated by the difference between tournament money and real money - though away from bubble/prize step up situations, it's probably pretty similar in tournaments) if you had A2o in the small blind with blinds of 100/200, and a stack of 2000, everyone folds to you, and BB has a larger stack - it is a clear mistake to fold (even if you show your opponent your cards before he makes a decision) .... further we can say that it is a clear mistake to fold (as opposed to shoving - lets ignore other possibilities for the moment which may be more optimal), even showing your opponent your cards, with any stack size up to about 4600 (the smaller of yours and your opponents stack size)... Keeping the same numbers, 100/200 blinds and your stack (the smaller) is 2000, then you would be wrong to fold (instead of shoving) hands like K3o, Q5s, J8s.... If you fold these hands, you are making a clear error and losing ev - even where your opponent sees your cards (i.e. you're not bluffing - you're playing the strength of the cards you have)..... The situation where you are the SB and everyone folds to you, and there is a relatively short stack is a relatively simple (small) mathmatical problem - actions are limited - shove or fold - and Sklansky - Chubukov have done the maths for you!!! I think by considering "what size Ace do I need to avoid being dominated and therefore happy to shove" considers insufficient information, will leave you playing clearly too tight, and losing expected value.... You can (and should) actually play quite a bit looser that S-C suggest - they consider situations where it is assumed your opponent knows your cards and therefore (with complete information) is able to play optimally against you - in reality your opponents dont know your cards and dont play optimally, so the value of shoving can be expected to be beneficial with a wider range of hands than can be proven. It would definitely not be beneficial to shove (as opposed only to folding) with a narrower range of hands.....

[robilaruk]

Re: Probability theory question

Since one of your two cards is an ace, three of the other 50 cards are aces, so the chance of any particular card in one of your opponents' hands being an ace is 3/50. If you didn't have an ace, it would be 4/50.

thats a very good point - y'see I told you I needed help:$ Damo ps can I blame it on the fact that I am a womble? :lol