#1
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mathematical probabilities
not looking for pot odds or any of the sort:
Does anyone know where I can find the probabilities or how to find the probabilities of getting any given hand in Texas Hold em such as a pair or two pair or just high card? I'm examining the change in probability as you go up in the poker hand order for science fair this year (yes, Im in high school) and it would be much appreciated if anyone could provide me with some solid info/sources. BTW, there are two ways I can go about doing this, but each has its set of problems. I can do it like so: Probability of a pair (or better) in the pocket: 52/52 x 3/51 But finding out the probability of finding out pair or better with two different hole cards is tougher: 52/52 x 48/51 x 6/50 <--- I'm not sure if this is right I can also calculate all the possible different poker hands by computing 52 given 7. (a math computation, if you don't understand it you might as well stop here) But I know this is not right because it's the best 5 card hand but I know 7 is involved somehow because you have a choice of 7 different cards so I dont know what to do in this situation. If I could somehow find out how to do it this way, it would be much easier because I could just do the total possible hands for a pair or for a straight or for trips over the total number of possible hands in hold em. (a very large number that I cant figure out although I know I am close with the above 52 given 7) Thanks ahead of time. |
#2
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This is from Ken Warren Teaches Texas Hold em:
(He's got a great section on probability and stats) 1,326 ways to be dealt a two card hand. 1 in 52 odds for the first card, 1 in 51 for the second. 52 x 51 = 2,652. 2652 / 2 because cards can come in either order (i.e. AJ or JA). If you disregard suits, you can make 169 different two card hands. Of 1,326 possible hands, 78 are pairs. 1,326/78 = 17 so you will be dealt a pocket pair 1 out of 17 times. The rank does not increase, nor decrease the odds of you holding the hand. You have a 220 to 1 shot of being dealt any specific hand on the next deal. I'm not gonna give you every stat in the book, but let me tell you, there is more than enough to do a high school science project. I may have given you enough here. Here's one more for good measure: You have two different hole cards, you'll flop a pair 40.408% of the time (don't know the math, just the facts). That's 1.47 to 1. 1/3 of those will be on the board, and not one of your hole cards. You flop a pair with one hole card 26.939% of the time, or 2.71 to 1. Happy Birthday Kid. |
#3
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i think the odds of one hand of poker in holdem is something like this:
Pr{2 card hand you are looking at}/ ( 52 choose 2) * 50 choose 5... maybe... |
#4
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??????
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#5
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lol thats so made up, nice try tho
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#6
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to defend my previous quote against all you mean people.
I can tell you the probability of any 5 card hand using the (n choose k) notation. This notation is used to represent the following formula: n!/(k!(n - k)!) For example, to find the probability of, say, a pair we have: ((13 choose 1)*(4 choose 2)*(12 choose 3)*(4 choose 1)^3)/(52 choose 5) A B C D E A: first we have to choose one of the 13 ranks of cards for our pair. B: next, out of that rank we just chose, we choose 2 of them. AFter all, out of those four cards there are 6 possibilites and we have to consider that. C: We must now exclude the rank that we already chose, and choose three different ranks out of the remaining 12 in order to avoid 2 pair. D: then we have to choose one suit out of those 3 ranks. E: Finally, all that has to be over the possibility of every 5 card hand. And we get: (13*6*220*64)/(52 choose 5 [a really big number like 2599388ish]) =0.422569 or 42.2% So as i was saying in my previous quote...wait no i was wrong. BUT, to find the probability of a finished hand( i.e. after the river) you multiply the probability of the 5 card board times the probabilty of your hole cards (52 choose 2) and that should be your answer. THank you Georgia Tech Probabilty with Applications. Any questions? |
#7
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If you just want answers
__________________
If aces didn't get cracked they would be writing books about me! |
#8
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Mike Caro is the man, but
How can I have a better chance of being dealt pocket Kings, rather than pocket Aces?
At least that's what I'm reading from that link. To me it's simple math. There are 4 of each of the 13 ranks. 52 cards. To hit any pair I've still gotta hit two of the 52. How does the rank come into play? |
#9
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Im gonna have to agree that that table looks wrong. Because the probability of getting Aces or dueces is exactly the same. Can you explain?
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#10
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The odds expressed mean any of the pairs between Kings and Jacks
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#11
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Exactly, that part means getting a pair of Kings, Queens or Jacks, any of the 3 as opposed to one particular pair. 220-1 is the number I have heard to be dealt a specific pair AA or 22 or any in between.
__________________
If aces didn't get cracked they would be writing books about me! |
#12
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And by the way, never question Caro when it comes to numbers.
__________________
If aces didn't get cracked they would be writing books about me! |
#13
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Yes, what he said. Obviously the odds of being dealt KK, QQ or JJ are going to be 3 times the odds of being dealt AA. 220:1 / 3 = 73ish:1.
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#14
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Agreed!
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#15
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Yeah, i got a second look at the chart and finally put two and two together. Then, I didn't give a fuck because the chart doesn't help me anyway.
Wasn't questioning Caro, but either I was stoned or the information wasn't presented in the best manner. I guess I was questioning myself as to what the hell I was missing? |
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