RammerJammer
Visionary
Silver Level
A long and rancorous debate has broken out on another thread concerning poker probabilities. (And probabilities in general.) After chastising for not doing their homework those in disagreement with what I knew to be a standard mathematical equation for determining the odds of succession, I decided to practice what I preach. I looked it up. Given that it's good stuff for everyone, I decided to post it here in a new thread instead of 4 pages deep on the old one.
Surprise! We're ALL right! As I repeatedly asserted in the original thread, it completely depends on how you approach the question. It's as much a matter of perspective as pure math. Read on.
Brian Alspach's Mathematics & Poker Page
Professor Emeritus of Mathematics & Statistics
Simon Fraser University
www.math.sfu.ca/~alspach/index.html
"Probabilities for Successive Memorable Hands"
Back-to-Back Identical Hold 'Em Hands
"Back-to-back hands consisting of the same two cards is something a player notices. Another successive occurrence players notice is being dealt back-to-back special hands. For example, back-to-back pocket aces usually draws comments when it happens. When this happens, sometimes you will hear a player ask, 'What are the chances of being dealt successive hands of pocket aces?'
"The problem with the question as asked is that it makes no sense, because it may be interpreted in several ways. The probability of being dealt A-A on a particular deal is 1/221. Thus, if you interpret the preceding question to be asking for the probability of being dealt A-A on two particular successive deals, the answer is
1/221 x 1/221 = 1/48,841
"If, on the other hand, you interpret the preceding question to be asking for the probability of being dealt A-A on the next hand, given that you have just been dealt A-A, then the answer is 1/221. This interpretation is not terribly interesting."
So, the reason why we have stalemated over this probability problem is that you can take either view and be correct. Your odds of drawing a pair of pocket aces is the same on every deal...221 to 1 on that deal. But if you're asking what is the probability that someone actually will be dealt pocket aces on two consecutive deals, it is 1 in 48,841.
Surprise! We're ALL right! As I repeatedly asserted in the original thread, it completely depends on how you approach the question. It's as much a matter of perspective as pure math. Read on.
Brian Alspach's Mathematics & Poker Page
Professor Emeritus of Mathematics & Statistics
Simon Fraser University
www.math.sfu.ca/~alspach/index.html
"Probabilities for Successive Memorable Hands"
Back-to-Back Identical Hold 'Em Hands
"Back-to-back hands consisting of the same two cards is something a player notices. Another successive occurrence players notice is being dealt back-to-back special hands. For example, back-to-back pocket aces usually draws comments when it happens. When this happens, sometimes you will hear a player ask, 'What are the chances of being dealt successive hands of pocket aces?'
"The problem with the question as asked is that it makes no sense, because it may be interpreted in several ways. The probability of being dealt A-A on a particular deal is 1/221. Thus, if you interpret the preceding question to be asking for the probability of being dealt A-A on two particular successive deals, the answer is
1/221 x 1/221 = 1/48,841
"If, on the other hand, you interpret the preceding question to be asking for the probability of being dealt A-A on the next hand, given that you have just been dealt A-A, then the answer is 1/221. This interpretation is not terribly interesting."
So, the reason why we have stalemated over this probability problem is that you can take either view and be correct. Your odds of drawing a pair of pocket aces is the same on every deal...221 to 1 on that deal. But if you're asking what is the probability that someone actually will be dealt pocket aces on two consecutive deals, it is 1 in 48,841.
