> We agree. But I think Ryan's introducing an important concept

Yes, but the playing of a card in and of itself does not change

> here that's vital to understand, since it quantifies the very

> thing you're talking about as it relates to the game of Euchre:

> as the game unfolds, the probabilities change based on the rules

> combined with the knowledge we've gained by what's already been

> played.

anything -- it's the fact that the play of that card may mean

something because of the rules of the game (e.g. trumping in implies

no cards in the suit lead).

> For instance, we agree that immediately after dealing, you have

Yes, but only because a fixed card (i.e. the top one) is turned

> a 5/24 probability of holding the AH. But when the top card of

> the dummy is turned over, if it's the AH, that probability drops

> to 0, and if it's not, that probability increases to 5/23.

over. If the dealer had the ability to examine the kitty and turn

over whatever he wanted, this would not be true.

Ryan touched on this in an earlier post. He said an often unspoken

assumption is that Monty will not reveal the door with the prize.

This is equally important when playing cards. The cards, in general,

are not being exposed according to a uniform distribution -- they are

being exposed according to a "Euchre distribution." As such, it is

not valid to assume holdings based on random exposure (e.g. the fact

that I've played 4 of my cards does NOT make me any more or less

likely to hold the Ah than at the beginning of the hand).

The fundamental flaw we've been making all along is saying Joe has a

4/14 chance, while Mary has a 3/14 chance, because these are the

number of cards left in their respective hands vs. the total number

of unexposed cards. Yes, we can conclude Mary is more likely than

others to hold the Ah because she's void in spades, but we cannot

conclude she's less likely to hold it because she has fewer remaining

cards.

> let's say your P leads a low trump and the opponent plays the

Not me. In fact, I say you should always play the A. There are

> K. You hold the R and A, and you need to make both of them good

> to win the hand.... I think some people would say without

> question that you should always play the R.

essentially 4 scenarios:

1) Neither opponent has the L, in which case it doesn't matter what

you play (you've made it).

2) The opponent to your right has the L, in which case the A is the

proper play.

3) The opponent to your left has the L, AND

3a) has to follow suit to the lead, in which case the A is the

proper play; or

3b) does not have to follow suit to the lead, in which case it

doesn't matter what you play (you're euchred).

So we've got two votes for the A, and two don't cares. In other

words, playing the R in this situation will NEVER fare better than

playing the A.- Please see my new thread for the purpose on reaching an agreement on this.-----Original Message-----
**From:**Ryan Romanik [mailto:ryanromanik@...]**Sent:**Sunday, December 01, 2002 14:02**To:**EuchreScience@yahoogroups.com**Subject:**[EuchreScience] Re: Probability Theory re: voids`--- In EuchreScience@y..., "Jed Taylor" <jed@o...> wrote:`

> Actually, I think what I wanted to say is that for Joe, the

probability that

> he'll have the AH is reduced by the difference between the

probability he'll

> have a spade now and he'd have a spade by random, not by the full

amount of

> having a spade now. So it's less than 4/14, but only by the

difference.

>

I guess, but it would actually be more like "reduced by the

difference between the probability he'll have his expected value of

spades in spades..." or something. I.e. he can have more than one

spade, or maybe even less than one.

> In any case, are we using the correct method to get to the actual

> probabilities? Or are the probabilities of the constrained deal

the same as

> what you've calculated?

>

I think I might have made an error in my post about the expected

number of hearts in each persons hand, but I made the error in giving

Mary too few. Bimbert made me realize it with something he posted,

although I don't think he actually responded to that post. Anyway, I

was basically accounting for Mary having what is like 3 cards in her

hand, but she still should have had essentially has 4, because she

selectively chose the card.

I'm not sure if Rob's accurate critiques of the method used for

computing probability apply to my "expected value" method of

computing average number of trump or not. I have to think long and

hard about it.

-Ryan!`To unsubscribe from this group, send an email to:`

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