outs calculation

It is 9/47 on the turn and 9/46 on the river, because that is the expected chance of hitting a flush. In reality it might be 0/31 if every space is dealt to the players already, or it might be 9/31 if zero spaces have been dealt to other players than you - you will never know for sure.
When calculating your outs, I believe it is best to only focus on what you know. You dont know your opponents cards, therefore you have to see it as there is 47 unknown cards left and 9 of those will make you hit the flush.

Hi Goget,

We don't have to take those cards villains might hold into account. It is true that your opponents will hold some amount of spades on average, but it doesn't matter. There are 47 unknown cards and 9 of those are spades. So the probability that the next card is a spade is 9/47.

You can basically take that into account if you want to but it makes this too complex without a reason. It is true that your villains will be dealt 4 spades on average if all cards are unknown. Since there are 4 cards known already (your hand + 2 on the flop) there are only 9 flush cards left. From those 9 flush cards your villains will be dealt only 3 cards on average, not 4. And since we take those cards into account we have to remove villains 16 cards from the deck also. This leaves us total of 6 outs on average in a deck on 31 hands. 6/31 is roughly the same than 9/47.

I'm not very good at math. I hope I'm correct when I say this and I hope it is not too difficult to understan.

Regards,

la55i

Originally posted by Malendes
It is 9/47 on the turn and 9/46 on the river, because that is the expected chance of hitting a flush. In reality it might be 0/31 if every space is dealt to the players already, or it might be 9/31 if zero spaces have been dealt to other players than you - you will never know for sure.
When calculating your outs, I believe it is best to only focus on what you know. You dont know your opponents cards, therefore you have to see it as there is 47 unknown cards left and 9 of those will make you hit the flush.

It is 9/47 OTF and 9/46 OTT. On the river we don't have any outs anymore since there are no cards to come. There are 52 cards in the deck we have 2 and 3 are on the flop so total of 47 unknown.

Originally posted by la55i

Originally posted by Malendes
It is 9/47 on the turn and 9/46 on the river, because that is the expected chance of hitting a flush. In reality it might be 0/31 if every space is dealt to the players already, or it might be 9/31 if zero spaces have been dealt to other players than you - you will never know for sure.
When calculating your outs, I believe it is best to only focus on what you know. You dont know your opponents cards, therefore you have to see it as there is 47 unknown cards left and 9 of those will make you hit the flush.

It is 9/47 OTF and 9/46 OTT. On the river we don't have any outs anymore since there are no cards to come. There are 52 cards in the deck we have 2 and 3 are on the flop so total of 47 unknown.

Yeah I know. I said it wrong I guess - what I meant was "it is 9/47 chance on hitting OTT and 9/46 on hitting OTR"

Hi Goget,

Malendes has it right. The cards that are dealt to your opponents are distributed at random, so it doesn't matter if the unknown cards are in the hands of your opponents or at the bottom of the deck. They are unknown, and thus treated as such.

If you have a good reason to believe that a player has one of the spades in his hole cards (the most obvious example would be a player accidentally flipping over a card in a live game), then you may of course include this in your calculation. Otherwise, you'll have to treat them the same way as any other unknown card in the deck.

I wondered the same thing when I was younger. And I calculated how often opponent has 0, 1 or 2 spades, what are the changes then and in the end the average % was the same as with 9/47.

If you don't believe, just do the math yourself. But treating unknown cards as unknows is the correct thing. Don't make it too complicated.

TO ALL:
Thanks for responding guys, but all you say is nothing new to me.
of course all cards are unknown, and villians can hold from 0 to 9 spades. that's why i say at 9-max they will hold 4 spades on average on the assumptions that every 4th card is going to be a spade on average.
most important: the probability of coming spades from the left deck/pack is some probability. the probability of coming spades from the dealt cards is zero, but in yr approach you claim there is some.
i want to offend noone, but what u all have said so far is not convincing to me.
the difference in 2 types of calculation (example of flush draw) is about 6% for turn + river!!! 39% vs 33% which is considerable.
i am really eager to look at opinions of people who really excel in theory of probability; i think mr. Collin Moshman is one of them.

It doesn't matter how many cards villains do hold on average.. It is irrelevant when calculating probabilities. Only thing that matter is the unknown cards.

You can do it the more difficult way as I explained earlier.

You said that villains have 4 spades on average in their total of 16 hands. Where did you get this? I guess you thought that 1/4 of the cards are spades so from villains 16 cards 25% -> 4, are spades.

Actually when you already know 4 of them because you have a flush draw, there is 9 spades left in the 47 unknown cards. So from those, villains do not have 1/4 spades, they will have 19.1% spades, which is 3,056 spades on average..

So your total outs are 9 - those villains have on average (3.056) = 5,944
Cards left on the deck = 31

5.944/31=0,1917
9/47=0,1914

So as you can see, this is the same calculation. You are only making it too complex without any reason. It can be achieved much easier.

So the difference between your two calculations is there because you thought there are 4 spades on average out there. But that would be true only if all of the spades would be unknown = you don't have a draw.

thanks.

I really made a mistake with the number of spades villains have on average in their total of 16 cards.
With both calculations the numbers are not even close, they are the same - 19,1489361702128%