MTraveler. yes. that’s a good way to summarize the ongoing dispute.

thanks for the reference.

Tom initiated the confusion for me by stating in the solution:

TOM: So you’re saying the chances of winning are two to one in favor of red?

which isn’t true in general.

It’s not about red, it’s about the con man betting it’s the “same” color.

So Tom’s statement seems wrong to me.

It’s 2 to 1 in favor of “same”, in general. not red.

there’s 1/3 chance the card is red/red

there’s 1/3 chance the card is grn/grn

there’s 1/3 chance the card is red/grn

or, stated differently, there’s 2/3 chance the card you pick has the same

color on both sides, as Rushen and Dsorgnzd note.

Over the long run, the con man wins because he keeps switching the offer.

If red is shown first, the offer is that the other side is red.

If grn is shown first, the offer is that the other side is grn.

That will win 2/3 of the time.

The link from MTraveler also noted one of the confusions in this discussion:

Bertrand’s point in constructing this example was to show that merely

counting cases is not always proper. Instead, one should sum the

probabilities that the cases would produce the observed result;

i.e. some cases are weighted differently than others.

If you could convince the con man to keep betting that the other side is red,

but only when red is showing, and you win if green is showing,

then you’d win 2/3 of the time.

BTW.

I found littlemouse’s reasoning specious, at best, relying on the stereotype

of what motivates a con man, not the mathematics of why the scheme works

for the con man.

littlemouse even says:

You would not be offered the same wager (even money that the other side would

be whatever color) if the side showing was green.

I don’t believe that.

The offer will always be made in the real con (even though the puzzler doesn’t

talk about doing it again).

Given that color X is showing, the con man will bet that the other side is X.

Bayes’ Theorem works just fine here.

Triedaq, you said:

Littlemouse-- you are correct in saying that Barkydog is wrong for focusing

on the cards and not on the sides.

I was not so sure about that at first.

I thought since we’re picking cards, it was necessary to assign probabilities

to the cards, etc.

Apparently it doesn’t matter.

The reference from MTraveler seems to use either approach,

starting with cards, or sides, assigning probabilities,

and getting the same answer in either case.