Tossin a Monkey Wrench into Probability

[UtzChips]

From a theoretical standpoint, we base probability on the premise that over the long-term, the random distribution of the cards even out. That is, if say 1 billion Hold Em hands were dealt out to two players, including the Board, we surmise that everything we could compute probability wise, would be equal. That is, both players would get the same amount of individual pocket pairs, they'd flop an equal amount of Sets or better. The Flop would be monotone exactly 5% (or whatever the exact chance is)of the time; both players would make their Flushs and Straights; turn two pair into a Boat, etc. etc., an equal amount of the time, that is in line with the odds.

So, if that's true, then in a 10 player Hold Em game, over 1 billion hands (or whatever it takes for the randomness of the deal to even out), you would find, that on average, the 20 cards that were dealt out to the 10 players averaged 5 of each suit.

That being the case;

(9/47)*2 - (9/47*8/46) = .3497 = 1.8598:1 or 1.86:1

as we all know it, now becomes (since we "know" the distribution of the suits to our 9 opponents, ie. we have 2 clubs, so 3 more are distributed to my opponents and the true odds of catching our Flush by the River:

(6/29)*2 - (6/29*5/28) = .3768 or 1.6536:1

So, if you ever run into me in a game, and wonder why I'm willing to cap it on the Flop, with nothing but the Nut Flush Draw, you now know why. My way of thinking is......
well, I'll let you describe it. [img]/images/graemlins/cool.gif[/img]

[phishstiiix]

You are a little bit wrong.

Yes the expectation of clubs- knowing nothing else- is 5, but the expectation of clubs knowing that you hold two already is higher. In your case it's 2+ (11/52)*18= 5.8. That is you can mathematically expect a long-term average of 3.8 clubs to be dealt to your opponents, as well the two you hold yourself.

The number of clubs in total is not independent of the number of clubs you yourself have been dealt. If you held none, you would expect (13/52)(18)=4.5 clubs, If you had one, you would expect 1+(12/52)(18)= 5.15. So in the above example if you have 2 clubs, you expect there to be (13-5.8)= 7.2 clubs left in the deck, so the chance of making your flush is a little lower.

The reason the average of all hands is only 5 is because naturally you are not always dealt two clubs. If you take a billion examples of hands where you are dealt 2 clubs yourself, you would get a long-run average of 5.8 clubs in total.

[PantsOnFire]

[ QUOTE ]
...we have 2 clubs, so 3 more are distributed to my opponents...

[/ QUOTE ]
This is your flaw. You are stating that since 5 clubs will be distrubuted to 10 players and you have 2 of them, then only 3 must be among the other players. This is incorrect. Without even using any math, you can logically state the since you got two clubs this hand, there will be on average more than 5 clubs in the players hole cards for this hand. In other words, you got an above average number of clubs this hand so you are bringing up the group average which will now be higher than 5.

You can use the fact that you have two clubs to calculate how many the other players have though. However, you will find that it will be the exact same proportion as the other clubs in the deck.

[AaronBrown]

In addition to phishstiiix and PantsOnFire's excellent answers, let me give another way to look at it.

The other players could have anywhere from 0 to 9 clubs. If they have zero clubs, you have a 53% chance of making your flush, but this only happens 0.7% of the time. The table below shows the number of possible clubs held by the other nine players, the chance of making the flush given that number, and the chance that they will hold that many clubs.

0 0.53202 0.00735
1 0.48276 0.05670
2 0.43103 0.17525
3 0.37685 0.28446
4 0.32020 0.26668
5 0.26108 0.14934
6 0.19951 0.04978
7 0.13547 0.00948
8 0.06897 0.00093
9 0.00000 0.00004

As you point out, 3 is the most likely number of clubs held by the other players (28%) chance and if that is the correct number, you make your flush 38% of the time.

But notice that the chance of 4 is greater than the chance of 2, the chance of 5 is greater than the chance of 1, the chance of 6 is greater than the chance of 0; and 7 through 9 are possible, while they can't hold less than zero. The other players will hold more than three clubs twice as often as they hold fewer than three clubs (48% versus 24%). Therefore the average outcome is worse than the most common outcome (three clubs).

If you take the product of the two numbers for each line in the table and add them up, you get 378/1,081 = 0.34968, the correct chance of filling your flush.

[UtzChips]

All 3 of you did an excellent job of helping me see the error in my thinking. Thank you for taking the time.