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

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

Please help.

Thanks

[diebitter]

Nah. You haven't seen 47, so 47 is the number you use.

Imagine this. You are presented with a pack of cards that have been shuffled randomly. You can draw one. what's the odds that you draw an ace?

4/52 or 1/13

Someone removes 50 of the cards randomly, so you have 2 left. what's the odds that you draw an ace?

It isn't 1/2, is it? Nope, it's 4/52 still, as you know nothing about the 50 cards removed.

Hope this clears this up for ya.

diebitter says it all, really; unless your opponents are kind enough to show you their cards, your out is just as likely to be in their hands as on the bottom of the deck, or the top. I know that some players count out odds on the basis of how their opponents are betting ("he's gotta have an Ace, so that's only three left"), but this is far too inexact a science to put any stock in.

Also, as for the subject, "the odds on [Sklansky] helping me with the odds": immense. A lock. Freeroll. I've read three books so far that upon arriving to the subject of counting outs and calculating odds, provide a brief summary, but then courteously direct the reader to SSHE.

[pudley4]

The previous posters explained the answer to you, but here's a simple example that might help you understand if you're still having problems:

Say you have a hat that has 3 marbles - one red, one blue, one white. You reach in and pull out a marble. What are the odds that it's red? 1/3 obviously. Now, suppose someone reaches in before you do and pulls out a marble. You don't know which marble he pulled out. Now what are your odds of pulling out a red marble? Still 1/3. Here's why:

When he pulls out a marble, there are 3 marbles he could get. What we'll do is figure out the probabilities of each of the scenarios, then add them all together to get our final answer.

1/3 of the time he pulls out the blue marble. If the blue marble is gone, you now have a 1/2 chance of getting the red one. Multiply these two numbers together and you get 1/6 - this is the probability that he pulls out the blue and then you pull out the red.

If he pulls out the white one (which again happens 1/3 times) you also have a 1/2 chance of subsequently pulling out the red. Again, this will happen 1/6 times.

If he pulls out the red marble, you have a 0/2 chance of pulling the red one. So this will happen 0/6 times.

To get the final probability, we add up all the intermediate steps: 1/6 + 1/6 + 0 = 2/6 = 1/3 - the same number we arrive at when no marble is taken out beforehand.

This shows that whether marbles (or cards) are in the bag (deck) or taken out (in players hands), it doesn't change the probabilities (as long as we don't know where the specific marbles/cards are)

[AaronBrown]

I understand and agree with the point the previous posters are making, but there is another aspect as well. You do know something about your opponent's hands. For example, most people will call preflop with an Ace. If you don't have an Ace in the big blind, and only 1 out of 9 opponents calls preflop, the odds are pretty good that (a) he has an Ace or two and (b) no one else does. That can be an enormous help in playing the hand.

Another example is holding 9[img]/images/graemlins/heart.gif[/img] 8[img]/images/graemlins/heart.gif[/img]. There are a bunch of calls before the flop, which comes down 2[img]/images/graemlins/heart.gif[/img] 4[img]/images/graemlins/heart.gif[/img] 8[img]/images/graemlins/diamond.gif[/img]. Someone bets, you call and everyone else folds. You have a pretty powerful hand here, but only if you use the information about other people folding. If your opponent is holding two hearts, he's almost certainly got you beat if you get a flush, but the odds of a flush in that case are lower than usual because there are only 7 outs, not 9. Those call/folds probably represent high card hands, reducing the likelihood that your opponent will pair one of these. Of course, I'm not saying you're a lock to win, just that your pair of 8's looks a lot better to hold up than it would if you go by the book statistics.

Certainly weaker players will call on any Ace, but better players (of which there are enough out there) know well enough to almost always throw away A-3o. So a call could mean the Ace, or a K-Qs, or a 2-7o for the looser amongst us.

If we deal a ten-handed round, each individual's odds of getting at least one Ace is around 15%. On the whole, there's about an 87% chance that at least one Ace is out there, but does that mean that if everyone folds to the button, that the button is any more likely than 15% to have an Ace? I'd say no: his cards were independent of the others dealt, and while no one having an Ace is quite unlikely, no one before the button having an Ace is unlikely as well, so it's not like the odds need to "compensate". In some sense this resembles the reasoning of those who think they're "due" after a poor run of cards, though your argument is certainly better reasoned.

I like your second example though, and agree that post-flop, there is much more information at hand. Still, now we're talking more about reads, and less about probability.

This is a great example, thanks for sharing it. I've tried to explain this concept many times, and some people have a real hard time with it - I'm sure your example will help!

[AaronBrown]

[ QUOTE ]
If we deal a ten-handed round, each individual's odds of getting at least one Ace is around 15%. On the whole, there's about an 87% chance that at least one Ace is out there, but does that mean that if everyone folds to the button, that the button is any more likely than 15% to have an Ace? I'd say no: his cards were independent of the others dealt, and while no one having an Ace is quite unlikely, no one before the button having an Ace is unlikely as well, so it's not like the odds need to "compensate". In some sense this resembles the reasoning of those who think they're "due" after a poor run of cards, though your argument is certainly better reasoned.

[/ QUOTE ]
Thanks for the kind words.

I agree the argument sounds like players thinking they're due, but the difference is one player's hand is not independent of another's on the same deal; the way one deal in independent of the last deal. If none of the first nine players has an Ace, the probability that the last player does goes up from 1 - (48/52)*(47/51) = 15% to 1 - (30/34)*(29/33) = 22%. With fewer non-Aces available, the chance of getting an Ace goes up.

To extend that to my example, assume that players call 80% of the time with an Ace and 20% of the time without an Ace. I know that's math, not Poker, but bear with me. In that case, a player would call 29% of the time, 12% of the time because they had an Ace (15%*80%) and 17% of the time without one (85%*20%). So if the first player calls, you figure there's 12%/29% = 41% chance he has an Ace.

But suppose it comes all the way around to you in the big blind, and only one person has called. You don't have an Ace. This should happen 15% of the time. One possibility is no one has an Ace (13% chance) and one person out of 9 called without one (30% chance) for an overall 4% chance. But it's more likely that one person had an Ace (37% chance) and called with all the non-Ace hands folding (13% chance) for an overall 5% chance. There's 1% chance a person has a pair of Aces and called, 3% chance one Ace was out but only a non-Ace hand called and 1% chance that more than one Ace is out and one of the Ace hands called.

The odds that you're facing an Ace have gone up from 41% to 55%, and the odds that there are at least three Aces left in the deck have gone up from 50% to 78%.

Obviously in real Poker, there are a lot more factors. Some players play loose, others tight. Some hands will raise. Some will call late position but not early. And so on. Still, I think there's value in considering the hands that have folded.

To me, reading is guessing what your opponent has based on his play. Probability is using inferences from the folded hands and your opponents' to guess what's left in the deck.

[mudbuddha]

wow are you a math professor?

[AaronBrown]

I used to be a finance professor, which is actually more helpful for this kind of thing.