Distribution of suits question

[SA125]

What is the likely distribution of suits for a full ring 10 player holdem game before the flop?

My basic math skills say 4 suits into 20 cards gives an average of 5 per suit per deal.

If that's true and you hold 2 of the same suit, when you hit another 2 of your suit on the flop that means the average number of your suit to be out would be 7.

I'd really appreciate any light you can shed on that. Thanks.

[elitegimp]

okay, you have 2 spades (for example), and there are 2 spades in the flop. At this point, you can not account for 9 of the spades, and there are 47 cards you haven't seen. From this, you should think that there is a 9/47 (19%) chance of turning a spade.

What you are asking about is conditional probability: since there are 9 spades un-accounted for, and 18 cards dealt to your opponents, there is a 0.003% chance that all the spades are in you opponents hands and that would give you a 0% of getting your flush on the turn.

There is also a 0.09% chance that 8 of the spades are dealt out, giving you a 1/29 chance (3 or 4%) of hitting your flush.

Anyway, you find the probabilities of X spades being dealt to your opponents, multiply that by the chance of hitting your flush with 9-X spades left, sum it all up, and you should get 9/47, just like if you don't know any of your opponents cards. Statistics is cool like that.

edit:
Here's a handy table I made... note that probabilities are not given as percents, and are rounded to 5 decimal places.
<font class="small">Code:</font><hr /><pre>
N P1 P2 P
- ------- ------- -------
0 0.00735 0.31034 0.00228
1 0.05670 0.27586 0.01564
2 0.17525 0.24138 0.04230
3 0.28446 0.20690 0.05885
4 0.26668 0.17241 0.04598
5 0.14934 0.13793 0.02060
6 0.04978 0.10345 0.00515
7 0.00948 0.06897 0.00065
8 0.00093 0.03448 0.00003
9 0.00004 0.00000 0.00000</pre><hr />

N is just a counter,
P1 is probability that your 9 opponents account for N of the spades,
P2 is probability that you hit one of the 9-N spades left in the deck on the turn
P is P1*P2

If you sum up all the P values, you do get 9/47

[SA125]

Thanks. I'm sure it adds up like you say. I'm still stuck on the probability of the avg # per suit being dealt in the first 20 cards and how that has no bearing when calculating unseen cards.

It doesn't really matter because I looked at it this way, even if it's wrong. The avg per suit per 20 cards is 5. 2 more on the flop is 7. There's 6 left in the 29 remaining cards. That's 23-6 or 4-1. The same for the wrong reason. Thanks again.

[elitegimp]

[ QUOTE ]
I'm still stuck on the probability of the avg # per suit being dealt in the first 20 cards and how that has no bearing when calculating unseen cards.

[/ QUOTE ]

Just because you _expect_ 6 spades to be dealt among your opponents on average does not mean that 6 spades will be dealt out every time. Also, the chance of you turning a spade changes depending on how many spades are out -- so you have to look at the interaction between how often a certain number of spades are dealt to your opponents and how likely you are to hit your flush given that however many spades are left in the deck.

I'm trying to clear things up, but I have a feeling I'm just adding to the confusion. Sorry!

not sure about this. but wouldn't they generally be just as likely yet to be dealt as in the opponent's hands (relative to the the respective total number of cards).

i.e. if you say, but there's expected to be a few spades in my opponents hands (folded or live), then you have to subtract the 16 cards from your card count and it should get you back to the same sort of percentages.

obviously if you have two spades and flop comes two spades and lots of people stay around or are aggressive, then yes, maybe there are quite a few spades in opponent's hands.

i asked a similar question. namely if 7-8 see the flop and you have AK, can you really assume the ace has three cards to pair with i.e. i'd assume maybe 1 or 2 aces are out. although i guess you could assume there's another ace out when a couple of early callers come in.

all the more reason not to play A7o, not only kicker problems, but maybe with 6-7 callers, there is only one ace left in the deck.

vis-a-vis your question, before the flop your opponents are just as likely to be playing suited connectors of other suits.

[AaronBrown]

Let's simplify things to test your logic. I take the four Aces and deal one to you and one to me, face down and for the moment neither one of us look. On average, there is one red card and one black card between us. Now I deal a card faceup from the two remaining Aces. It's red. So on average, there are two red Aces accounted for.

What is the probability that the remaining undealt card is red? Using your logic (which I admit I pushed to an extreme) it's zero since two red Aces are already accounted for.

There are two ways to get the correct answer, and many years ago Reverend Thomas Bayes proved they always give the same answer.

First is to ignore the two down cards in front of each of us and treat them as still in the deck. Then it's obvious that one of the three unknown cards is red, so the chance of the next face up card being red is 1/3.

Second is to consider the three possibilities. We can't both have red cards, because a red card came face up. I could have red and you could have black, in which case the next up card will be black. I could have black and you could have red, in which case the next up card will be black. Or I could have black and you could have black, in which case the next up card will be red. So the chance of another red upcard is 1/3.

With 20 cards in 10 hands you can do either of the same two things: treat all the unknown cards as still in the deck or consider every possible distribution of suits (not just the expected distribution).