## Re: Probability Theory re: voids

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• ... Yes, but the playing of a card in and of itself does not change anything -- it s the fact that the play of that card may mean something because of the
Message 1 of 35 , Dec 1, 2002
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> We agree. But I think Ryan's introducing an important concept
> 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.

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

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

Not me. In fact, I say you should always play the A. There are
essentially 4 scenarios:

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

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. ... From: Ryan Romanik [mailto:ryanromanik@hotmail.com] Sent: Sunday, December 01,
Message 35 of 35 , Dec 2, 2002
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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