Another question about pot odds....Never seen this one.

[dollyfan]

I know that after the flop with a flush draw (in hearts) you have a 1:4.22 chance of catching your flush on the next card. This is because there are 9 hearts left in the deck that can make your flush and 38 cards that won't. so 9/38 = 1/4.22. The problem is that there is 99.44% chance that at least one of the cards needed to catch your flush is in the hand of one of your opponents.

The chance of your opponents having no hearts is ~ .75 to the 18th power or .56%. 18 being the number of cards held by your opponents at a table w/ 10 people. 20 cards - your 2 cards = 18 cards. .75 being the chance that one card is not a heart.

So with .56% chance that there are no hearts, there would be a 99.44% chance that there is at least one heart in my opponents hand that is out of play.

To find the true odds of a heart coming out on the next card, this information would have to be considered, wouldn't it? And if so how would I use this information in decision making.

All responses welcome, except for the people who are just going to nit pick my math by saying that there is a slightly greater chance than 75% that one card will not be a heart because I have two in my hand. I know that. Other than that nit pick away.

[waffle]

Let's assume a "standard" flush draw: 2 hearts in your hand and 2 on the flop. The easiest way of taking your opponents cards into consideration is to lump them in with the deck, as unknowns. You haven't seen 47 cards, and of those, 9 will complete your flush. So you have a 19.15% chance of hitting it on the turn, or 1:4.22 odds.

[stigmata]

I think you are getting confused by over-thinking the problem. If we start discounting the outs that other players may have, then we also have to discount the non-outs that they are holding. But because we have no idea what the other players have, it is far easier to treat all the cards that we can’t see as a single deck. Therefore, you correctly calculate above that there are 9 cards out that can make the flush, and 38 that can’t. E.g. 9:38.

If we use your method and estimate the chance other players have one of your outs, we come to the same result:

There are 20 cards from 47 remaining cards that other players hold. 20/47 = 0.42, e.g 42% of the pack is held by other players.

Therefore 0.42 x 9 = 3.83. Therefore on average, 3.83 of your outs will be in the hands of other players.

So now we have 9 – 3.83 = 5.17 outs, but this is the average number of outs in the 27 remaining cards:

5.17:27 = 9:38

So we get the same result. Essentially the easy method just calculates from the remaining 47 cards, and the hard method splits the pack into cards held by other players (20, in fact I just realised that this should be 18, but you get the point) and cards in the deck (27). Both get the same result.


Err sorry. I have just realised that my post is stupidly confusing and even more nonsensical than the question I attempt to adress.

Apologies.

[dollyfan]

I understand what your saying. I didn't take into consideration that the remaining cards in the deck have the same percentage of my outs on any given example. thanks for clearing that up for me.

[hustalasta]

My take:

# of cards dealt to all players divided by 4 = average number of cards dealt of any particular suit.

In a ten handed game if you are dealt 2 hearts and 2 more come on the flop I would count your outs (to a flush) as 6.

13 of any suit - (20 cards dealt pre-flop/4) - 2 on the board = 6 outs remaining on average.

Anyone see any problem with this logic?

[stigmata]

Yes, I have a problem. There are definitely 9 outs if you are 4-to-a-flush. You are making the same mistake as dollyfan.

There are 47 cards that you cannot see. Therefore any individual card has the same chance of being exposed on the turn. 9 hearts in 47 remaining cards = 9:38 odds.

[skibum]

to just use the odds tables calculated by our better's.

[hustalasta]

Even though you cant see X number of cards because they were dealt face down to your opponents, you can acurately predict how many of each suit are dealt face down (on average). Not using that information is a mistake IMO.

By your logic your opponents never get dealt a card of the same suit as you. Logic should tell you that 1/4th the cards dealt pre-flop will be of each suit.

[hustalasta]

Test numbers

Ran 50 hands where I placed Ah & 2h in my hand, 3h 4h Ac as the flop on the board. Dealt 50 hands with remaining cards.

With 4 other opponents (5 handed) my opponents held an average of 1.66 of my flush outs. 7 out of 50 times my opponents held no hearts and I had a full 9 outs to my flush. 22 out of 50 times they held 2 or more of my flush outs.

With 9 other opponents (10 handed) my opponents held an average of 3.6 of my flush outs. 1 out of 50 times my opponents held no hearts, 24 out of 50 times they held 4 or more of my heart outs.

While 50 hands is not a huge sample set, I think this shows you shouldnt count 9 outs to your flush when playing 5 handed and it would be a huge mistake when playing 10 handed.

[stigmata]

I am not saying that my opponents never get the same suit as me. I’m saying that for the turn, there is a 9 in 47 chance it will be the suit I need (if I have 4 to a flush). There will also be a 9 in 47 chance that any other card in the deck or somebodys hand will be the same suit.