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 in our investigation.
Yes you can initially reduce the twelve trials to the six unique combinations, but ONLY because each of the six appears the same number of times as each of the other six (they are equally probable). In a set which favors some combinations, reducing the probability set to the possibility set will lose any meaningfulness.
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 in our investigation.
Yes you can initially reduce the twelve trials to the six unique combinations, but ONLY because each of the six appears the same number of times as each of the other six (they are equally probable). In a set which favors some combinations, reducing the probability set to the possibility set will lose any meaningfulness.