Yes.
Wow. Not my expected answer at all. Thank you.
I noodled on this for awhile and was pretty hung up on the idea of Alice’s decision hinging on two decision factors: Her raw p[sub]t[/sub][sup]A[/sup] vector, and an overall metric for her confidence in her skill. If her skill was known or believed to be 100%, then she’d use her vector unmodified. But if she lacked confidence, she’d bias (somehow) towards favoring those ts where p[sub]t[/sub][sup]B[/sup] > p[sub]t[/sub][sup]A[/sup]. With the favoring proportional somehow to the differences in p values. At the limit where she had no clue whatsoever (like my stockpicking history) she’d just use Bill’s p values as her own.
I suppose maybe I can turn this thinking around to fit your result. If Alice is a really crappy handicapper, then by betting her raw p[sub]t[/sub][sup]A[/sup] vector she’ll do as well as she can. But her outcome is likely to be poor.
OTOH, if she’s an awesome handicapper she should still bet her raw p[sub]t[/sub][sup]A[/sup] vector to do as well as she possibly can. But this time her outcome will (over the long haul) be better.
The assumption is that Alice believes in her final probability estimates. If she thinks she’s a poor handicapper and initially gives a 10% chance where Bill has only 5%, Alice will presumably modify her 10% to an estimate of, say, 8% and use that as her final estimate.
The more general question you pose is Given different probability estimates p[sup]A[/sup], p[sup]B[/sup], p[sup]C[/sup], … (along with statistics about the performance of A, B, C), then what is the best consensus probability estimate? I was interested in this problem two decades ago, but all I remember now is an interesting idea by Professor Rosalind Picard called “Exaggerated Consensus”: If two weathermen each predict a 70% chance of rain tomorrow and you know the two weathermen use very different methods, it might be appropriate to estimate a higher chance, say 75 or 80%!
They don’t have to be winners. A popular accumulator here is a “Yankee”. This is a 4 selection wager consisting of 11 bets: 6 doubles, 4 trebles, and a four-fold accumulator. The return for a small outlay can be considerable.
I haven’t read your link, but I can say for certain that a Yankee, per se, is not an accumulator bet.
It does contain a 4 horse accumulator, but in and of itself, a Yankee can not be described as an “accumulator”, any more than a “Heinz” or a "Canadian " can be described as an “accumulator” despite the fact that they contain multiple accumulators.
The correct term to describe a Yankee, a Heinz, or a Canadian is a “combination” bet.
Let me guess: A “Heinz” is a bet that a particular horse will start off behind, and then will ketchup?
Nice idea, but it’s because it combines 57 accumulators. All the possible doubles and higher accumulators on 6 selections.
If the horse does poorly does he end up as burgers?