Taking bet A doesn’t require any commitment. My argument just uses backward induction (+ignoring statewise incomparability), which you should generally use in sequential choice situations, or else you’ll be worse off in many situations, even with sharp probabilities.
It allows unsharpness. Having unsharp probabilities does not require sequential decisions to be made independently.
that very concession is what Elga reads as victory: the imprecision has stopped doing the one thing it was introduced to do.
The argument against unsharp probabilities is defeated. We just have to treat them in certain ways. The summary of the paper here missed one way we could treat them, and claimed too much against another (if we accept commitments or resolute choice in other cases).
Here is a video I found useful that explains how to use backward induction. Below is Claude’s reply to your comment after some iteration between us.
Thanks Michael — the backward-induction framing is the strongest version of the reply, and I want to grant what it gets right before saying where I think it’s still exposed.
It does defuse three things at once. It needs no commitment (you predict the future Bet B choice and fold it back, rather than binding yourself), it needs no complete ordering (it runs on statewise dominance, so the Bet B node can stay genuinely unsharp), and it isn’t ad hoc (backward induction is the standard discipline for sequential choice). So this isn’t PLAN in disguise. Fair enough.
But I think the argument turns on a step that quietly does more than “just backward induction.” Here is the full tree, with payoffs written as (if H / if not‑H). Bet A pays −10/+15 and Bet B pays +15/−10, so the four leaves are BOTH +5/+5, A-only −10/+15, B-only +15/−10, and NEITHER 0/0:
Notice both Bet B nodes are under-determined: at each, neither action statewise-dominates the other (BOTH vs A-only cross; B-only vs NEITHER cross). That is exactly the optionality unsharpness is meant to preserve, so dominance-pruning removes nothing at a Bet B node. To get a verdict on Bet A, backward induction has to fold each Bet B node back into a single continuation value — and the value of the reject-A branch depends entirely on which of its two (equally maximal) leaves you assume you’ll pick.
Crucially, the accept-A node is also under-determined — it can land on BOTH or on A-only. So to compare the two root actions I have to fix a policy over both identical Bet B nodes. There are only three consistent options:
The only statewise-dominance relation anywhere in the tree is BOTH ≻ NEITHER. In particular A-only vs NEITHER crosses — A-only is worse than NEITHER in the H-state (−10 < 0) — so accepting A does not statewise-dominate rejecting A. Under either consistent policy (always-accept or always-reject), both root actions stay admissible and there’s no dominance reason to prefer accepting A. And note that under “always accept B,” NEITHER is never reached on either branch, so there’s nothing for accepting-A to protect against in the first place.
The recommendation to accept A appears only under the third policy — the one that accepts B after accept-A but rejects B after reject-A. That is what produces the BOTH-vs-NEITHER pairing that makes accepting A look dominant. But that policy isn’t backward induction resolving each node on its merits; it’s a rule that makes your Bet B choice depend on whether Bet A preceded it, handing down different verdicts at two Bet B nodes that (for a money-only agent) are identical in every respect she cares about. That is precisely the SEQUENCE/PLAN pattern Elga’s Sally case is built to reject.
Put differently: the recommendation to accept A materialises only when you assume you’ll reject B specifically on the reject-A branch — i.e. you distrust your future self on one branch but not the other. That asymmetric self-distrust is either the sophisticated-chooser reading (treat your own future permitted choice as a hazard to steer around) or the differential treatment of identical nodes. Both are exactly the concessions at issue: if you’re rationally required to prevent your future self from exercising reject-B, then reject-B was never really optional — which is just SHARP’s verdict reached the long way.
So a sharper version of my earlier question: your derivation of “accept A” resolves the accept-A continuation to BOTH and the reject-A continuation to NEITHER. What consistent policy over the two identical Bet B nodes yields that pair? If “always accept B,” reject-A gives B-only and the dominance is gone. If “always reject B,” accept-A gives A-only and the dominance is gone. The only policy that yields it treats the two Bet B nodes differently — which is the thing an imprecise theorist owes an account of, and which Sally says you can’t have.
(One aside on “you’d use backward induction even with sharp probabilities, or be worse off”: agreed, but with sharp credences backward induction never has to override a node’s verdict — it agrees with local EV-maximisation, and the cases where skipping it hurts are cases of myopia, not override. This is the unique setting where the rule must reverse a choice the agent’s own decision rule calls permissible. That asymmetry is the tell.)
Taking bet A doesn’t require any commitment. My argument just uses backward induction (+ignoring statewise incomparability), which you should generally use in sequential choice situations, or else you’ll be worse off in many situations, even with sharp probabilities.
It allows unsharpness. Having unsharp probabilities does not require sequential decisions to be made independently.
The argument against unsharp probabilities is defeated. We just have to treat them in certain ways. The summary of the paper here missed one way we could treat them, and claimed too much against another (if we accept commitments or resolute choice in other cases).
Here is a video I found useful that explains how to use backward induction. Below is Claude’s reply to your comment after some iteration between us.
Thanks Michael — the backward-induction framing is the strongest version of the reply, and I want to grant what it gets right before saying where I think it’s still exposed.
It does defuse three things at once. It needs no commitment (you predict the future Bet B choice and fold it back, rather than binding yourself), it needs no complete ordering (it runs on statewise dominance, so the Bet B node can stay genuinely unsharp), and it isn’t ad hoc (backward induction is the standard discipline for sequential choice). So this isn’t PLAN in disguise. Fair enough.
But I think the argument turns on a step that quietly does more than “just backward induction.” Here is the full tree, with payoffs written as (if H / if not‑H). Bet A pays −10/+15 and Bet B pays +15/−10, so the four leaves are BOTH +5/+5, A-only −10/+15, B-only +15/−10, and NEITHER 0/0:
Notice both Bet B nodes are under-determined: at each, neither action statewise-dominates the other (BOTH vs A-only cross; B-only vs NEITHER cross). That is exactly the optionality unsharpness is meant to preserve, so dominance-pruning removes nothing at a Bet B node. To get a verdict on Bet A, backward induction has to fold each Bet B node back into a single continuation value — and the value of the reject-A branch depends entirely on which of its two (equally maximal) leaves you assume you’ll pick.
Crucially, the accept-A node is also under-determined — it can land on BOTH or on A-only. So to compare the two root actions I have to fix a policy over both identical Bet B nodes. There are only three consistent options:
The only statewise-dominance relation anywhere in the tree is BOTH ≻ NEITHER. In particular A-only vs NEITHER crosses — A-only is worse than NEITHER in the H-state (−10 < 0) — so accepting A does not statewise-dominate rejecting A. Under either consistent policy (always-accept or always-reject), both root actions stay admissible and there’s no dominance reason to prefer accepting A. And note that under “always accept B,” NEITHER is never reached on either branch, so there’s nothing for accepting-A to protect against in the first place.
The recommendation to accept A appears only under the third policy — the one that accepts B after accept-A but rejects B after reject-A. That is what produces the BOTH-vs-NEITHER pairing that makes accepting A look dominant. But that policy isn’t backward induction resolving each node on its merits; it’s a rule that makes your Bet B choice depend on whether Bet A preceded it, handing down different verdicts at two Bet B nodes that (for a money-only agent) are identical in every respect she cares about. That is precisely the SEQUENCE/PLAN pattern Elga’s Sally case is built to reject.
Put differently: the recommendation to accept A materialises only when you assume you’ll reject B specifically on the reject-A branch — i.e. you distrust your future self on one branch but not the other. That asymmetric self-distrust is either the sophisticated-chooser reading (treat your own future permitted choice as a hazard to steer around) or the differential treatment of identical nodes. Both are exactly the concessions at issue: if you’re rationally required to prevent your future self from exercising reject-B, then reject-B was never really optional — which is just SHARP’s verdict reached the long way.
So a sharper version of my earlier question: your derivation of “accept A” resolves the accept-A continuation to BOTH and the reject-A continuation to NEITHER. What consistent policy over the two identical Bet B nodes yields that pair? If “always accept B,” reject-A gives B-only and the dominance is gone. If “always reject B,” accept-A gives A-only and the dominance is gone. The only policy that yields it treats the two Bet B nodes differently — which is the thing an imprecise theorist owes an account of, and which Sally says you can’t have.
(One aside on “you’d use backward induction even with sharp probabilities, or be worse off”: agreed, but with sharp credences backward induction never has to override a node’s verdict — it agrees with local EV-maximisation, and the cases where skipping it hurts are cases of myopia, not override. This is the unique setting where the rule must reverse a choice the agent’s own decision rule calls permissible. That asymmetry is the tell.)