AKo in UTG+1 (originally posted in microlimits)

[easypete]

I originally posted this in the Micro-Limits forum. It was suggested that I post it here. My question for the Probability forum is what are the odds with 6 others seeing the flop, with 2 spades, that someone has 2 spades? Should I have bet the river here?

Party 1/2 10 Handed.

Dealt to hero A /images/graemlins/spade.gif K /images/graemlins/heart.gif
UTG calls, hero raises, UTG+2 folds, MP1 calls, MP2 calls, MP3 calls,CO calls, Button folds, SB folds, BB calls, UTG calls

Flop (14SB) Q /images/graemlins/spade.gif, 3 /images/graemlins/spade.gif, T /images/graemlins/diamond.gif <font color="blue"> (7 players) </font>
BB checks, UTG checks, hero bets, MP1 calls, MP2 calls, MP3 folds, CO calls, BB calls, UTG calls

Turn: (10 BB) J /images/graemlins/heart.gif <font color="blue">(6 Players) </font>
BB checks, UTG checks, hero bets, MP1 calls, MP2 calls, CO folds, BB calls, UTG calls.

River: (15 BB) 6 /images/graemlins/spade.gif
BB checks, UTG checks, hero checks, MP1 checks, MP2 checks

Pot (15 BB) between BB, UTG, hero, MP1 and MP2.

Summary in white
<font color="white">
UTG lost $5 [ 5d Qd ] [ a pair of queens -- Qd,Qs,Jh,Td,6s ]
hero lost $5 [ As Kh ] [ a straight, ten to ace]
MP1 lost $5 [ Kd 9d ] [ a straight, nine to king]
MP2 lost $5 [ 8d Ad ] [ high card ace ]
BB bet $5, collected $29.50 [ 8s 2s ] [ a flush, queen high]
</font>

[sthief09]

C(9,2)/C(47,2) = .0333 is the probability someone has a flush

so P(doesn't have a flush) = .9667

.9667^6 = .8161

I think I might've had to do a binomial distribution for this, but this seems right

so roughly 18% of the time someone has a flush. that's only if they play any 2 spades. you have the A /images/graemlins/spade.gif making it much less likely someone has a flush. However, these people are chasing you, so it's more likely someone has it, so it isn't as easy as that. But just know that the chances aren't as overwhelming as they may seem, especially when people are calling down with hands like A8 unimproved. This is an easy river bet.

[easypete]

It's been only since my freshman year in college (oh my.... 16 years.... i'm getting old) since I had a class in statistics.

If I get 2 random cards, the chances of me getting 2 spades is 9/52*8/51 right?

That's 2.7% right there... right?

Now I have a spade. So it's actually more like 8/52*7/51. So 2.11% chance of a random hand having 2 spades. There are 6 random hands facing me on the flop. Shouldn't it be closer to 12.6% chance of a spade draw on the flop?

[easypete]

OK... I'm an idiot.... I've got flush draws on my mind... There are 13 spades in a deck... not 9.

So assuming I have one spade w/ 51 cards remaining...

12/51 * 11/50 = 5.18% chance of getting 2 spades.

6 players see flop 6 * 5.18 = 31.06% chance?

I see that you took the percent not having the 2 spades to the 6th power... If that's the case then 0.9482^6=72.68% chance that one player doesn't have it. that's 27.3% chance they have it.

Place in the human factor... they called (all but one) to my flop bet... I think the river bet is a bad idea.

[sthief09]

there are 3 spades on the board and one in your hand. your opponents can't have those cards. I don't know how to do it without combinations.

C(9,2)/C(47,2) is the probability that a random player has 2 spades. that I know for a fact.

EDIT: actually you can do it the way you were doing it, but you have to consider exposed cards. there are 3 spades on the board, and 2 non-spades that your opponents CAN'T have. there is one spade in your hand and one non-spade that they also can't have.

so I think 9/47*8/46 should work out right too

you also can't just multiply 5% by 6 to somehow get 30%. mathematically that makes no sense.

there's a way to do it but I don't know how. P(random player doesn't have 2 spades)^(#players) should give a decent result. 18% is what I got.

[easypete]

[ QUOTE ]
C(9,2)/C(47,2) is the probability that a random player has 2 spades. that I know for a fact.

[/ QUOTE ]

It somehow bothers me that I can't remember how to do combinations. I tried to find my statistics book at home... guess I didn't like it enough to keep it.

I was playing this through my head on the way into work this morning. I do need to take into consideration all the spades i have information on. One in my hand, and three on the board. That leaves 8 spades available w/ 45 unknown.

8*45 / 7*44 = 2.8%

Bear with me for a little while I try to piece this together.

If I'm reading this right[ C(9,2)/C(47,2) ]:

C(9,2) = the number of possible combinations of 2 card starting hands containing the possible remaining 9 spades?

C(47,2) = the number of possible combinations of 2 card starting hands out of the remaining cards in the deck? Should this be C(45,2) after the river? Not that it makes much of a difference (1081 vs 990).

So on the flop, I only know of 3 spades.

C(10,2) / C(47,2) = 4.2%

P(not) = 95.8% ---&gt; 95.8%^6 = 77.3% --&gt; 22.7% chance?

After the river:

C(9,2) / C(45,2) = 3.6%

P(not) = 96.4%^4 = 86.6% ---&gt; 13.6%

I need to take into consideration that any 2 spades will probably stay in to the river from the flop.

P(not) = 96.4^6 = 80.3% ---&gt; 19.7% chance I'm beat.

I have 15 BB in the pot w/ 4 opponents. What do I gain w/ a bet on the river? What if BB is c/r here? I went into check-call mode here in case someone downstream bets. I would only have to be right 1 out of 20 times for this to be right if all call the bet. If I bet then what do I gain? I have to be right 1 out of 4 times for this if all call?

So sthief, you get about 18% chance, i get about the same (slightly higher). How do you adjust this for the information that you get when you bet?

I bet the flop and the turn, I got callers... The information I was getting was that my opponenets either wanted to see the showdown cheaply w/ weaker hands, or someone was on a draw from the flop. There were straight draws and a flush draw on the flop. I think (or at least hope) that this exercise gave me a little better understanding of probability, but I still don't think this is a good place to bet the river. (IMHO)

[sthief09]

yes, 45 not 47. my mistake.

the answer to your question is that none of this is relavent when you're at the table. I find the only real use of obscure stats like this is reassurance. you'd think the chances that someone has a flush on the river is like 75%, but mathematically it's only 20%. technically this isn't right, because there are a lot of hands that they would've played differently. either they'd fold something like 3 /images/graemlins/spade.gif2 /images/graemlins/spade.gif or something like 72 with no spades, so it's probably significantly higher, since there are so many hands that thye can't possibly have. regardless, you have to bet.

[easypete]

Thanks sthief...

I think I have a better understanding of how to calculate the odds using combination now.

I agree, there is no way to really deal with this in the moment. I deal with it here so that in the future, in a similar situation, I have an idea of what to do.