dellbarnes
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dellbarnes has not written any posts yet.
dellbarnes has not written any posts yet.
I have an interest in gaming management and practical probabilities. I have a great interest in economics as well. I stumbled onto this site and tyhe "Drawing 2 aces" post. I struggled with it for about a week, and then wrote a few things. The thread is old, but I look forward to any helpful responses.
Even further: the probability that I can guess your hand is 1/5. However, the probability that you have 2 aces is 1/3. No?
Further: write out your 12 possible trial pulls. Raw odds of ace-ace =2/12. Once an ace is pulled, do two things: cross out the two null-null pulls. The odds of ace-ace appear to be 2/10, but... We didn't draw out Schrodinger's ace; it is either ah or as. pick one (it doesnt matter which, they are symmetrical in distribution) and cross out the combinations that do not have this ace. As the waveform collapses the true odds of ace-ace appear- 2/6. Do you see that? We didnt draw an equally hearty or spadey ace, it had to be one of them or the other, which made combinations without it no longer a factor... (read more)
Sorry to zombie this thread, but I could use some help.
Hmm.. I'm going with 1/3 for 2 reasons. i might be wrong, but maybe I can explain how I am wrong well enough for someone to help me see it, because I'm stumped here.
reason one, from the bottom end: when the answers are yes-yes, there are only 3 possible types hands, ace-ace, and (2) ace-null, each as likely. Weighting probabilities based on how one may answer "yes-no" and still have ace-ace seems erroneous when the answers are yes-yes.
reason two, from the top end: It seems that a false set is being used in the explanations favoring argument 2. here's how I think this... (read more)
If we have an ace in the hand, 2 of those "equally likely cases" are no longer possible. (2 of those cases involve the other ace and a non-ace card.)