[Proofrock]

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Hi Proof:

One possible flaw here is that if you do play the queens and now it's a little later and you have that $22,000 and the opportunity presents itself to double up again, you probably can't do it even if you wanted to since your opponent is most likely to have less than you.

In the initial example with the queens, since it's the first hand your opponent has the chips so that you can get all in. But I don't think that will usually be the case shortly after you have doubled up.

Best wishes,
Mason

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I'm not sure I was clear in my original post. Let's say you chose to fold the first hand. Matros assumes that you then sit with t10000 until you double up at time T. However, if you had doubled up, the assumption is that your stack will have grown to t22000 at time T. Now, if you have twice the stack of anybody else at the table, the extra t10000 are of no use as far as cEV is concerned, so I can only assume that whatever you had done to accumulate the extra 2000 chips by time T you would have been able to do with a t10000 stack. So I then proposed that it would make more sense to assume that you would have t12,000 at time T if you folded QQ on the first hand. So then, using Matros's own argument, "you can answer that by solving this equation: " x(24,000) = (.538)(22,000). By Matros' argument, this implies that you can pass up QQ the first hand, then get all-in as a dog (49.3% to win) at time T to equal the expecation from calling the first hand.

Thanks to your post, however, I've revisited this and have since come to believe that Matros's attempt to use this argument to support his claim is logically flawed: it assumes that there would be a situation where you would double up with a t10000 stack whereas you would fold preflop with the t22000 stack -- otherwise the comparison of what happens at time T via the equation x(20000)=(0.538)(22000) makes no sense.

Thankfully, though, the original article doesn't rest entirely on the shoulders of this argument.

-J.A.