Query - Poker Odds

[ron8]

Straight forward poker odds do not take into account the cards which the other players are holding, ie: A gut shot on the river is calculated as 4 divided by the 46 cards left in the pack = 11.5 to 1 shot. What I don't understand is this takes no account of the other players cards. For instance when playing a ten player table there would be 18 cards which are counted as being as part of the pack. These 18 cards represent almost a third of the pack and I would think there is a strong probality that one or more of the outs you are looking for are among these cards which would alter the odds drastically.
I would be very grateful for any help with this matter
All the best
Ron

[mute]

Unless you have some information that suggests that some of your outs are in your opponents hands, then their cards are just as unknown to you as the ones still in the deck.

[stone_7]

If I take a deck of cards and move 10 or 18 cards to the bottom of the deck how does that change the odds of the next card being an ace? I may have moved all 4 aces to the bottom but I have no way of knowing so I must assume that the odds are still the same.

Having said that you are correct in that if you can make a good estimate as to what your opponents are holding you can further clarify their/your outs.

Example from a tourney last night:
Sorry this is from memory so details will be approximate but the gist of what I have to say is here.
First Round
Everyone hass appx 1500 chips
Hero is in mp3 with T T
FOlded to hero who raises 3x bb
2 callers and the SB and BB fold
Flop is 6 7 8 Rainbow
Here bets 1/2 the pot
called by next player
reraised all in by third player

Hero thinks for a minute and decides that villain must have at best a pair and the 9 perhaps 2 nines. Let's assume 8 9 which is the hand I feel is most likely. Here are the outs the villain is counting

2 8's
3 9's (probably discounting this one)
4 Tens (but I have 2)
4 5's

This makes 13 which in his mind makes him the favorite with 2 cards to come. Plus I may have some other hand he is beating now such as AK or AQ which I may fold. This is usually a great situation to double up or bust out early with for the villain. However, I had 2 tens reducing his outs to 11 greatly swinging the odds in my favor.

Given this read I called and so did the next guy. They showed 89 and 69 respectively.

[DiceyPlay]

If there are 46 unseen cards, each one has the same probability. If you need the 4 of clubs (assuming it's not in your hand and there's 46 unseen cards). Each of the 46 unseen cards has a 1/46 chance of being the 4 of clubs. This includes the next card to come off the deck and all the cards in your opponents hands.

So the chances of you getting the 4 of clubs is 1/46 until you get more information which changes the odds to 0 (you see the card and it's not in your hand) or 1 (you see the card and it is in your hand) or somewhere in between (you haven't seen the card, but have more information - i.e. you've seen more cards).

-DP

[ron8]

Although I love a game of poker, I have made my living backing horses in England for many years. Now I love to play poker but these odds still worry me. I can see what you are saying but not taking any account of the other nine players cards is a bit like me betting in a race when I am only allowed to see two thirds of the field. Far from not making any difference because I have not seen the form of the other third of the field it would make a huge difference which would have to be built in to the odds I could accept.

[dealer_toe]

I was trying to explain odds to my friend and he had the same problem with the concept. I told him poker is a game of incomplete information. You need to take all the information you have and do your best with that. There is no way you can accurately account for folded cards. You can try, but I don't think it'll be very succesful.

I understand exactly what you mean as I've been asking myself the same question:

If you have an ace, then your odds on the flop of getting the other one on the turn is normally: 3/(52-5)=3/47.

Why not go one step further and ask:
what are the odds that one of my 3 aces is already in someone else's hands? Because over a 100 hands, your 3 aces won't always be in the deck ready to be drawn, there might only be 2.7 aces or whatever it exactly is.

Maybe it's negligeable, and we don't need to be that exact.
Maybe it isn't, maybe it's a tiny leak to overlook this.

Ron, look at it this way. Let's agree there are six cards seen that are not your Jack for the gutshot. That leaves four Jacks out of 46 if the rest of the cards were available. They are not available, in fact let's take out your 18 cards that you know got dealt which leaves 28. Now, out of those 18 cards gone, 18*4/46 or 1.565 of them were Jacks. So there must be 2.435 Jacks left out of 28. Same odds as 4 out of 46 that we started with, but lots more trouble.

Yes there actually will be from four to none available out of 28. But you just don't know which, so use what you know.

[DiceyPlay]

You guys are over-thinking probability 101. Read what I wrote and think about it instead of moving on to some other thoughts. Then read what permafrost wrote, he simply extrapolates and tells how many of the cards you need are expected to be in your opponents hands and how are exprected to still be in the deck.

Any card you haven't seen has the same probability of being the card you need. All the ones in your opponents hands and the next one to come off the deck.

This is a very basic concept ... and it's crucially important you think about it right and not confuse yourselves which will result in silly unnecessary(sp) mistakes at the table (errr keyboard).