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Message: [QUOTE="Four Dogs, post: 928100, member: 2668"] Correct, assuming you see both the turn and the river. But this is your drawing odds, not your pot odds. If your pot odds are greater than your drawing odds then this is said to have a positive EV or EXPECTED VALUE. There's also implied odds and expected odds to consider, but let's save this for later. Well, no. Looks like you're including your $20 call into the pot. The pot is $100 not $120. Assuming you're on a draw, it will cost you $20 to win $100 so your POT ODDS are 5:1 read 5 TO 1. This is a RATIO. to get the fraction you would add the 2 numbers together and divide the smaller by the larger. In the ratio 5:1 there are a total of 6 parts. so 5:1 = 5/6 = .83 or 83%. If either he or you is all-in, then that is all you can win. If you were on a str8 draw or a flush draw, that's about what you'd need to call in a cash game with only 1 card to come. With both the turn and the river to come you'd be getting more than enough odds to call. No, you DO NOT take other peoples cards into account. You must calculate the odds on the number of UNSEEN cards, not the number remaining in the deck. Think of all the unseen cards, dealt or not as a bowl of M&M's. Each the entire bowl is the entire deck, minus your hole cards and the board. Let's say the green ones represent the cards you need to improve your hand. Now lets say the proportion of green M&M's to all other colors is 4:1 and the bowl is perfectly mixed (I'm not exactly sure how many different colors there are, but humor me). If you have 5 guests and give them each a handful, clearly the number of M&M's in the bowl will be less, but the proportion will ALWAYS be 4:1. This is why it doesn't matter that some of your out cards may be unavailable. Your odds of drawing the cards you need remain the same. In fact, if you could somehow see everyones hole cards and sa saw that one of your OUTS had been dealt to another player, your odds of drawing one of the remaining OUTS could actually improve. If your looking for any diamond for example and there are 9 left out of 46 unseen cards your drawing odds (not pot odds) would be 9:37 or about 24%. If you could see that one of the 8 other players had mucked a diamond, you would have to remove that card as a possibility, but you would also remove the 15 other non diamonds from the solution. SO your odds of catching an OUT would now be 8:22 or about 36%. This actually happens all the time in STUD. I hope this was readble. My kids are all over me for story time so I gota run. Think I'll read them chapter 2 of [B]Multivariate Statistics 5th edition.:) [/B] [/QUOTE]

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