Yet here you explicitly agree with that statement.

]]>The probabilities of each hand *given fairness* are equal, so these two probabilities alone do not constitute evidence of cheating. That is the correct conclusion of your argument. But no Bayesian would reason this way – this is frequentist-style. The probabilities of each hand given cheating are not equal, and this can be used as evidence of cheating.

As always, put it in Bayes theorem!

]]>There seems to be a mistake here.

The following is a sound argument:

Premise: The probability of being dealt a straight flush by a fair dealer is identical to the probability of being dealt the hand 2h, 3s, 7s, jd, qd by a fair dealer.

Premise: The probability of being dealt the hand 2h3s7sjdqd does not constitute evidence of a non-fair dealer should such a hand actually be dealt.

Conclusion: The probability of being dealt a straight flush does not constitute evidence of a non-fair dealer should such a hand actually be dealt.

(In other words, briefly: Since the probabilities of the two hands are identical, it cannot be probability alone which makes it right to suspect cheating.)

But you have insisted that the probability of being dealt a straight flush DOES constitute evidence of cheating should such a hand actually be dealt.

You can’t be right–as the mantra goes, the conclusion of a sound argument is always true, and what you’ve said contradicts that true conclusion.

But have you misspoken?

]]>Nice; you took the words right out of my mouth.

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