Calculating the true odds of winning

dingle

Playing live last night I pushed on a flushdraw (I was shortstacked) and got called by a bigger flushdraw and two pair. I said something like "F@*!" but some other random guy said that I still had a chance.

My question is "what chance?"

Flop
As 5s Jd

My hand: Js 7s
Other hands involved: Ks 6s and Ac 5c

I know I have a 8 or 9% chance of hitting a Jack but as there are 11 other cards ( AdAh5d5h + 7 remaining Spades) which could appear which will beat me even if I do hit the Jack, what is my chance of winning the pot.

So, to all those math heads out there. How do I calculate my true chance of actually winning the pot.

Please note I'm not that interested in the actual number generated by the example above but in the method used.

LuckyLloyd

This post has been deleted.

Mellor

I think of situations like this its best to total possibilities and then the ones were you win.

43 cards left in the deck.
You need to hit one of 2 jacks, or running sevens
and avoid 2 aces, 2 fives, and 7 spades.

No of possible deals = (43*42)/2 = 903

no of deals were you win = (2*31)-1 = 6........ Jacks,
______________________= (3*2)/2 = 3........running sevens

so chances of winning are 65/903 = 7.1%

5starpool

Running sevens?

phantom_lord

equity 	win 	tie 	      pots won 	pots tied	
Hand 0: 	07.530%  	07.53% 	00.00% 	            68 	        0.00   { Js7s }
Hand 1: 	28.793%  	28.79% 	00.00% 	           260 	        0.00   { Ks6s }
Hand 2: 	63.677%  	63.68% 	00.00% 	           575 	        0.00   { Ac5c }

Mellor

5starpool wrote:

Running sevens?

forgot them, changed now, thanks

Bandana boy

Not 100% accurate but pretty close approximation

cards %win
Js 7s 7.53
Ks 6s 28.79
Ac 5c 63.68



http://twodimes.net/poker/

Hitman Actual

Using combs:

43 cards left, so 43C2 combs of turn/river = [(43x42)/(2x1)] = 903.

We win with JJ, 77, J7, and Jx, x not a spade, Ace or 5:
JJ: 2C1 = 1 combination
77: 3C2 = 3
J7: 5C2 - 1 - 3 = 6. (i.e. subtract the JJ and 77 combs).
Jx: x cant be a spade, 5 or Ace or 7 (we've already calculated the J7 combs). That's 14 cards, so we win with 43-14 = 29 cards. There are two jacks left, so we have 58 Jx combs.

This gives a total of 68 combs where we win:

68/903 = 7.53% = 12.3/1

QED

carfax

UnLuckyLloyd wrote:

I hate to be smart, but your chances here are incredibly miniscule. I would just stick on my coat and head for the door tbh.

But to those who wish to do the calculations, enjoy...

Agree with that, the value of knowing the odds in this situation is pretty minimal but as a maths exercise fair enough.

dingle

UnLuckyLloyd wrote:

I hate to be smart, but your chances here are incredibly miniscule. I would just stick on my coat and head for the door tbh.

Thank you for the insightful comment, but as I said in the original post I'm not interested in finding out the result of the example but rather the correct method to generate results.

Mellor wrote:

No of possible deals = (43*42)/2 = 903

no of deals were you win = (2*31) = 62........ Jacks,
______________________= (3*2)/2 = 3........running sevens

Mellor, how did you come up with the numbers in your calculation i.e the number of deals where I get Jacks and does this number include deals where I get Jacks and still lose?

Mellor

lenny_leonard wrote:

Jx: x cant be a spade, 5 or Ace or 7 (we've already calculated the J7 combs). That's 14 cards, so we win with 43-14 = 29 cards. There are two jacks left, so we have 58 Jx combs.

I think you made it over complicated, and you'rer double counting here I think. the X cant be any of the 14 you listed, it also can't be a the other J as you've already counted that, 15 cards it cant be, and you subtract this from 42 as the 43rd card is the first J.

So thats 27 cards, and two jacks make 54, plus the first ten is 64/903
Its easier to count Jx alltogether and incluced JJ and J7 as the 7 and 2nd J make no difference here.

Mellor

The first card is one of two remaining jacks, the second is one of the 31 cards thats not a 5 spade or ace, 42 (cards in deck) - 11 (A,5, spade)
Then you minus 1 as JaJb is the same as JbJa, an is counted twice

Hitman Actual

Mellor wrote:

I think you made it over complicated, and you'rer double counting here I think. the X cant be any of the 14 you listed, it also can't be a the other J as you've already counted that, 15 cards it cant be, and you subtract this from 42 as the 43rd card is the first J.

Yeah, I missed something, but there should be 68 combinations and not 64. I just can't see where.

handsfree

wins JJ = (3/46*2/45)=0.0029
JXx2 =(3/46*14/45)*2= 0.0899
J7x2 =(3/46*3/45)*2=0.0087
77 = (3/46*2/45)=0.0029

minus JAx2 (3/46*2/45)*2=0.0058
J5x2 (3/46*2/45)*2=0.0058
JXs x 2 =(3/46*7/45)*2= 0.0203

probability = 7.83%

Hitman Actual

Missed the (not so obvious) J5 combination which gives us the higher full-house. So we win with:

Quads: JJ (1)
Full House: 77 (3), J7 (6), J5 (4)
Trips: J2, J3, J4, J6, J8, J9, JT, JQ, JK (54)

Gives the 68 combs for the correct figure 68/903 = 7.53%

NickyOD

Did you win?

carfax

NickyOD wrote:

Did you win?

Hee hee.

Mellor

lenny_leonard wrote:

Yeah, I missed something, but there should be 68 combinations and not 64. I just can't see where.

I thought so as poker stove gives full results and not just a monte carlo simulation,
but J5 was had to spot, well done