Re: Probability Theory re: voids
> We agree. But I think Ryan's introducing an important conceptYes, 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
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 haveYes, 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
> let's say your P leads a low trump and the opponent plays theNot 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
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-------- In EuchreScience@y..., "Jed Taylor" <jed@o...> wrote:
From: Ryan Romanik [mailto:ryanromanik@...]
Sent: Sunday, December 01, 2002 14:02
Subject: [EuchreScience] Re: Probability Theory re: voids
> Actually, I think what I wanted to say is that for Joe, the
> he'll have the AH is reduced by the difference between the
> have a spade now and he'd have a spade by random, not by the full
> having a spade now. So it's less than 4/14, but only by the
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.
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