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View Full Version : Sklansky: What's the odds on you helping me with the odds?

08-22-2005, 09:59 AM
The Theory of Poker, page 36, 2nd parargraph.

Figuring the odds for making a hand is done on the basis of the number of unseen cards and the number among them that will make the hand........

Here is my question, unless you are playing poker by yourself, then at least 5 of those "unseen" cards (the ones dealt to your opponent) have been removed from the deck.
Leaving only 42 cards instead of 47. Keep in mind this is
assuming you have only one opponent. If there are only 42 cards left in the deck how can you base you calculation on 47 cards? I just do not understnad why he is using unseen cards instead of the actual number of cards left in the deck.

In addition how do correct you calculation for the possiblility that some of your hand making cards may have been dealt to your ooponents? For example if you hold 4 to a flush, how can you assume that there are 9 cards (9 of the same suit) that are left to make your hand? If five
of your flush cards have been dealt to your opponents than you have a serious miscalculation of your odds of making the flush. I do not understand how the calculation of the odds can be correct when it is based on an assumtion which can be wrong.

The correct calculation of odds, I would think, would go as follows.

Actual number of cards left in the deck / True number of cards left in the deck that will make you hand

Thanks

PairTheBoard
08-22-2005, 11:20 AM
The unseen cards in your opponents hands are no different than the unseen cards at the bottom of the deck. Meditate on this. You will have meditated long enough when you no longer need to ask this question.

PairTheBoard

Tom1975
08-22-2005, 11:35 AM
This question comes up a lot. I know I've gotten it a lot from some of my freinds when I've been explaining odds to them. You're probably going to get a lot of answers which just say 'an unseen card is an unseen card, it doesn't matter if it's in the deck or your opponent's hand', so I'll try to explain a little further.

Let's take the example of you playing 5 card draw by yourself. You deal yourself four clubs, and draw one to attempt to complete your flush. There are nine clubs left in the deck of 47, so the odds are 9/47=19.1%. Agree so far?

Ok, now let's say you play with a friend, you both get dealt five cards, and again you get four clubs and draw one. Now there are only 42 cards left in the deck. The problem now is that we don't know how many clubs your friend has. If he has no clubs, your odds are now 9/42 (21.4 %), which is better than the first example. However, if has five clubs, your odds are only 4/42 (9.5%). He could also have 1,2,3, or 4 clubs. To summarize:

0 9/42 21.4%
1 8/42 19.0%
2 7/42 16.7%
3 6/42 14.3%
4 5/42 11.9%
5 4/42 9.5%

So if your friend has no clubs your odds are actually better than if you were playing by yourself, but if he has one or more, then they are worse. But, you don't know which is the case, so how can you calculate your odds at this point? You would have to take the chance of your friend being dealt 0,1,2,3,4,or 5 clubs times each of the percentages that you make a flush in that scenario and add them up. So your 'true' odds at this point are
(chance your friends was dealt 0 clubs) * 21.0% +
(chance your friends was dealt 1 clubs) * 19.0% +
(chance your friends was dealt 2 clubs) * 16.7% +
(chance your friends was dealt 3 clubs) * 14.3% +
(chance your friends was dealt 4 clubs) * 11.9% +
(chance your friends was dealt 5 clubs) * 9.5%

If you were sit down and do all the math it works out to exactly 9/47=19.1%, which is the same if you were playing by yourself. Think of it this way - there are five fewer cards in the deck, so if your friend has no clubs, this obviously helps you, compared to you playing by yourself. However if your friend has 1 or more clubs then this hurts you. The effect of these two possibilities effectively cancel each other out, so you can just pretend there are still 47 cards in the deck, nine of which are clubs. I hope this helps, if you still can't grasp the concept, you're just going to have to take it on faith that David Sklanksy and a lot of other smart people have done the math and this is in fact correct.

VivaLaViking
08-22-2005, 11:58 AM
Much better that I would have done. My explanation would be that unless you can see your opponents cards they are unseen. Lets work on that problem though, seeing your opponents cards. lol

BTW: I doing some math on the fractional outs problem.

SheetWise
08-22-2005, 03:06 PM
[ QUOTE ]
I just do not understnad why he is using unseen cards instead of the actual number of cards left in the deck.

[/ QUOTE ]

I see your point. And why is he using all of the cards when at most we will only see five of them? Why count the cards at the bottom of the deck? Answer me that! I suggest we use only the five cards we're going to see when calculating the odds. Send me your preliminary calculations, we can talk about a book deal. We'll make millions.

GTSamIAm
08-23-2005, 02:14 AM
What about the times you have reason to suspect your opponent has specific cards? Say you have AsKc. You and villain see the flop heads-up. The flop is 29J of spades. You bet, and the villain raises. You know for a fact he only bets or raises a three flush flop when he has a pair and a good flush draw. Assume he always slowplays a made flush.

So in that case, you know that one of his cards is a spade and one is not. So you know the suit of eight (13-5=8) cards. I now can say my chances of hitting the nut flush are (13 - 5) / (52 - 7) = 8/45 instead of the usual 9/47.

An expansion of this would be that the identity of the opponent's cards is in the real world a summation of probabilities. So in fact an adjusted equation is developed (in the context of similar flush draws):

(13 - 4 - 1*p_1 - 2*p_2) / (52 - 7)

13 spades in all, 4 I can see for sure, 1*the probability he has one spade, and 2*the probability he has two, divided by 52 - 7 known cards. Reduces to:

(9 - p_1 - 2*p_2) / 45

Performing odds calculation assuming his cards as completely unknown is not using the most information possible. Correct?

mosdef
08-23-2005, 06:11 AM
[ QUOTE ]
Performing odds calculation assuming his cards as completely unknown is not using the most information possible. Correct?

[/ QUOTE ]

correct, but you'll almost never have that kind of info on an opponent. even if it's someone who has told you how they play, now you have to play the "i know that he knows that i know..." game, and you can really only guess some probabilities as to what his hole cards might be. this is usually extremely murky.

GTSamIAm
08-24-2005, 02:38 AM
But what about the case where you know how they play, but they don't know that you know?