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To: Harvey Allen who wrote (5904)	10/27/2001 12:07:57 AM
From: mishedlo	Read Replies (2) \| Respond to of 6873

No - you can not add them that way. A simple example. The odds of a coin flip being heads are 50/50 So if you add them you get say 50% + 50% on two flips. Of course you know that the odds are not 100% of a head turning up in two flips. The correct figure is 50% + 50% -(50% * 50%). You must subtract the fact that they both may be heads (or tails). In my simple example (I sure hope this is correct) you get .5 + .5 - .25 or 75%. In this case each flip is independant of the others. In your case if you get the card you need on the first try it is irrelevant (for your purposes) but not for mathematics. You have to discount the fact they both could be the card you need. 4/47 + 4/46 - (4/47 * 3/46) It has been a long time and I am not sure this is correct but you have to discount the fact that both could be the card you need. In this case that is about 1/2 % difference. Your way if you kept adding 4/47 + 4/46 + 4/45 + 4/44 you would end up with a % over 100% before its time was due. I am glad there is not a third card cause it gets more complicated. M

To: Harvey Allen who wrote (5904)	10/27/2001 3:52:21 PM
From: jjeannie	Read Replies (1) \| Respond to of 6873

The probability of drawing 2 fours is (4/47)*(3/46).

To: Harvey Allen who wrote (5904)	10/28/2001 11:14:39 AM
From: Michail Shadkin	Respond to of 6873

Harvey here are the calculations on your question: Easiest way is to figure out the percentage of not hitting a 4 on either card and than subtract the percentage from 100 to get the percentage of hitting a 4 on either card. 100-(43/4742/46)100=16.46624% the odds are 16.46624% of hitting a 4 on either card. Which is roughly 6 to 1 Whether a player stays in depends on many factors, but pot odds and not the number of players left in the hand are crutial. Lets say there was 1 raise before the flop and 7 players called in a 10/20 game. Thats a 20*7=$140 if the next round of betting is only going to cost you $10 then you should call, because you are getting pot odds regardless if there are 7 or 3 players left. your $10 to win 140+10for each player staying in. so lets say that 4 players stay in to see the 4th card (including yourself) that means you are getting 170 to your 10 which by far is a calling hand. On the last card the same calculations are made to see if you want to call again. Keep in mind you might also hit an Ace and win the pot or 2 running 3s. Also, there is such a thing as implied value (meaning if you do hit and there are very wild players in that will not fold and might even raise you with some mediocre hands) Under such circumstances you might even want to take the worse of it early to get paid huge later. That why you need to know your customers and know how they will react in the future. Math is only part of poker. The larger the stakes, the smaller the use of math and larger your gut feelings play in your decisions. Best Wishes Michail P.S. keep in mind that is was a very basin explanation and in reality the decisions are much more complex.

To: Harvey Allen who wrote (5904)	10/29/2001 4:52:15 PM
From: patron_anejo_por_favor	Read Replies (1) \| Respond to of 6873

Harvey, one other small point to add to Michail's excellent discussion on your poker question is that if a 4 hits on the turn or the river, you may fill your gutshot draw, but not have the "nut" straight, ie, you would lose to someone holding 3-6. Therefore the implied odds the pot needs to give you to stay in the hand should be higher than what the calculated pot odds are. In the example you've cited, admittedly, it's a minor point (in a game with skilled players, few would hold 3-6 anyway unless they were on the big blind and no raises occurred). But stranger things have happenned.