Posted Fri Mar 19, 2004 6:26 pm GMT by lukasbradley
All,
On this site are a series of small articles about calculating odds. While informative, I have a question about the way things are worded. Perhaps I missed this part, but I'm sure one of you can clear it up.
On the following page, http://www.texasholdem-poker.com/odds1.php, the example is as follows:
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You start with a pair of Jacks in the pocket. Not too shabby. The flop however, doesn't contain another Jack.
Lesson 1: What's my chance of getting a Jack on the turn?
You need to just figure out the number of outs and divide it by the number of cards in the deck. There's 2 more Jacks. There's 47 more cards since you've seen five already. The answer is 2/47, or .0426, close to 4.3%.
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While good math, I'm not sure it's good statistics. What about the other 5 imaginary players (for this discussion)? Each also received 2 cards down, which brings our "out" cards to 37 instead of 47. Our chances are seemingly improved (2/37 = .054) when other people are playing.
However, this is not the case, especially in other examples, such as straight draws. What if someone has our Jack, with an Ace kicker?
Is the first example that doesn't take into account the others' cards the correct one for play, or is there a modified version that changes the odds to include the other out cards?
Thanks for any help.
Lukas
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Posted Sat Mar 20, 2004 1:40 am GMT by Sente
As long as do not know what you opponents have, the cards are considered to distributed randomly. There is an equal chance of Jack being any card in the deck or an opponents hand. There are 47 cards that you don't know about. So the odds of hitting a Jack on the turn are 2/47.
Let's pretend you have only one opponent in this hand. If you knew for certain that he didn't have a Jack, then your odds would be 2/45. If you knew he had one Jack, your odds would be 1/45. And if he had both Jacks, your odds would be 0/45.
Odds that your opponent has no Jack: 45/47 * 44/46
Odds that your opponent has one Jack: (2/47 * 45/46) + (45/47 * 2/46)
Odds that your opponent has both Jacks: 2/47 * 1/46
Odds that you get a Jack given that your opponent has no Jack: 2/45 * 45/47 * 44/46
Odds that you get a Jack given that your opponent has one Jacks: 1/45 * ((2/47 * 45/46) + (45/47 * 2/46))
Odds that you get a Jack given that your opponent has both Jacks: 0/45 * 2/47 * 1/46
Add those together to get your total odds that you get a Jack:
(2/45 * 45/47 * 44/46)+(1/45 * ((2/47 * 45/46) + (45/47 * 2/46)))+(0/45 * 2/47 * 1/46) =
(2*45*44)/(45*46*47) + (1*2*45)/(45*46*47) + (1*45*2)/(45*46*47) + 0 =
((2*45*44)+(2*45)+(2*45)) / (45*46*47) =
(2*44+2+2)/(46*47) =
(88+2+2)/(46*47) =
92/(46*47) =
2/47
The math for just one opponent is bad enough, I'm not going to do five opponents. The logic is the same. When you consider the odds that they have your card and the odds that they don't have your cards it always comes back to 2/47.
Whether the cards that you cannot see are in an opponent's hand, been burned, in the middle of the deck, at the bottom of the deck each is equally likely to be a Jack. 47 cards, 2 are Jacks. 2/47 that the turn card will be a Jack.
Posted Sat Mar 20, 2004 9:08 am GMT by JohnnyCache
Yah, the fact is, even tho in a big game, someone might have one or both of the jacks, that doesn't effect your odds of drawing one! Isn't zat veird?
Posted Wed Mar 24, 2004 9:34 pm GMT by Jonniedough
Lukas,
Odds are calcualted as if your the only person playing.
Just a shorter version of what the previous two stated.
Posted Thu Mar 25, 2004 9:33 am GMT by Ninja
Like mentioned above, you can't take into account other peoples hands, because, frankly, you have no idea what they are.
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