Maths help please

[pokerpops]

I witnessed a hand the other evening and it got me wondering about the maths of the situation on the river...

No comment required on any part of this other than the river please

.50/1 live

UTG+1 raises to £6
3 callers including BB
Flop -    ah

Original raiser shoves for £23 (I know, I know, but this is live poker!)

Two callers -
MP, playing c£300 bought in for £250 I think. Always thought him to be a good reg, has a few decent scores to his name and is a tricky winning player
BB, playing c£350. Works offshore but is in regularly when he"s home. Has had a good evening so far and is well up on his £100 BI. His call preflop closed the action.

Turn -  
check/check

River  ::
BB bets £50

So - if you"re MP and have an Ace, how often does BB have to have the other 4 in order to make a fold profitable?

Could you fold on the basis of "Best I can do is split the pot"

Am I over-analysing this?

[Marty719]

Def a case of over-analysing.  We do not want to be folding full of aces in this spot when our action closes the betting, esp when the turn goes c/c.

[GarethC]

Def a case of over-analysing.  We do not want to be folding full of aces in this spot when our action closes the betting, esp when the turn goes c/c.

[ ] Answered the question.

To make a call of £50 to win £120.50 you need to be winning 29% (50/(120.50+50)*100) of the time to break even. Therefore if you think he has the 4 (or aces full or a straight flush) >71% of the time you should fold.

[WYoung83]

cant fold, The ammount of time he has the same hand as you.

[Swinebag]

Def a case of over-analysing.  We do not want to be folding full of aces in this spot when our action closes the betting, esp when the turn goes c/c.

[ ] Answered the question.

To make a call of £50 to win £120.50 you need to be winning 29% (50/(120.50+50)*100) of the time to break even. Therefore if you think he has the 4 (or aces full or a straight flush) >71% of the time you should fold.

Pretty sure there were 2 callers on the flop making it £50 to win £143.50 which means you need to be winning even less of the time.

Can"t find a fold here, especially after the lack of turn action

[pokerpops]

Def a case of over-analysing.  We do not want to be folding full of aces in this spot when our action closes the betting, esp when the turn goes c/c.

[ ] Answered the question.

To make a call of £50 to win £120.50 you need to be winning 29% (50/(120.50+50)*100) of the time to break even. Therefore if you think he has the 4 (or aces full or a straight flush) >71% of the time you should fold.

Pretty sure there were 2 callers on the flop making it £50 to win £143.50 which means you need to be winning even less of the time.

Can"t find a fold here, especially after the lack of turn action

4 x £6 pre (.50 from SB fold) = £24.50
3 x £23 0n flop = £69 + £24.50 = £93.50 - which is what is available to split

In a vacuum I"m a caller here every time, but... there"s some maths behind it and I was curious about how often we would have to be wrong (ie BB has the 4 or AA - discard the straight flush he just never has that)
Also - how does the maths vary if the first pot has to split three ways with the all-in player also having an A?

[GarethC]

Wow, good question. The equation I posted assumes you win the full pot when you win, which obviously isn"t going to happen very often. So it"s completely useless. As well as being wrong because I calculated the pot wrongly, thanks Swinebag!

Assuming you can only ever chop or lose, the £50 bet can never be won as part of the pot, you either lose an extra £50, or profit 50% or 33.3% of the pot, depending on whether it"s chopped 2 or 3 ways. So if the pot was £93.50 before the river action you stand to profit either £46.75 or £31.17 when it"s a chop. Or, if the all-in player has you beat, you get your £50 back.

This is where I got stuck when thinking about this situation in bed last night as I was trying to get to sleep. I guess it"s just a case of risk/reward? It probably comes down to a ratio between those numbers and the chance of either player having you beat. Seems simple with two players but the third all-in guy is complicating things a bit in my head. I"ll have to think about it a bit more.