bridge question
Jauder Ho
hand is void in at least one suit. Since I don't play bridge I do not know
what is a void combination in bridge, in addition any hints to solving this
problem would greatly appreciated. Thanks.
--Jauder
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Jauder Ho http://www.med.umich.edu/~jauderho/
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char...@castlebbs.com
IJ> I have a question which I have to find the probability that a bridge
IJ>hand is void in at least one suit. Since I don't play bridge I do not know
IJ>what is a void combination in bridge, in addition any hints to solving this
IJ>problem would greatly appreciated. Thanks.
IJ>--Jauder
According to The Official Encyclopedia of Bridge, Mathmatical Tables,
the odds against having at least one void is "approximately 19 to 1."
The book contains a lot of other information on how to calculate odds
against various holdings and combinations.
P.S. A "void" is the holding of zero cards in a particular suit.
== Charles ==
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Tom Banks
> I have a question which I have to find the probability that a bridge
>hand is void in at least one suit. Since I don't play bridge I do not know
>what is a void combination in bridge, in addition any hints to solving this
>problem would greatly appreciated. Thanks.
>
>--Jauder
According to the Bridge Encyclopedia, the odds against a hand having at least
on void are approximately 19 to 1. No explanation of how this was calculated.
---
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Tom Banks email: ba...@hal.com
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The opinions expressed are my own, not those of HaL Computer Systems.
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Henk Uijterwaal (Oxford)
Tom Banks writes:
>Jauder Ho writes:
>> I have a question which I have to find the probability that a bridge
>>hand is void in at least one suit. Since I don't play bridge I do not know
>>what is a void combination in bridge, in addition any hints to solving this
>>problem would greatly appreciated. Thanks.
>According to the Bridge Encyclopedia, the odds against a hand having at least
>on void are approximately 19 to 1. No explanation of how this was calculated.
Simple: there are
52 52! 1*2*3*...*52
( ) = --------- = --------------------------
13 13! 39! 1*2*3...*13 * 1*2*3...39
ways to take 13 cards out of a deck of 52. This is the total number of
hands. For a void, one has to get these 13 cards from a deck of 39 cards,
there are
39
( )
13
ways to do this. One has to multiply this by 4, as you can be void in 4
suit. The fraction of hands with a void then is:
39 52
4*( )/( ) = 0.051
13 13
which is equal to 1 out of 19.5.
Henk.
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Marc 'Merle' Wallace
<char...@castlebbs.com> wrote:
>
>
>IJ> I have a question which I have to find the probability that a bridge
>IJ>hand is void in at least one suit. Since I don't play bridge I do not know
>IJ>what is a void combination in bridge, in addition any hints to solving this
>IJ>problem would greatly appreciated. Thanks.
>
>IJ>--Jauder
>
> According to The Official Encyclopedia of Bridge, Mathmatical Tables,
> the odds against having at least one void is "approximately 19 to 1."
> The book contains a lot of other information on how to calculate odds
> against various holdings and combinations.
>
> P.S. A "void" is the holding of zero cards in a particular sui
[Note this is really Bill_C posting from someone else's account--don't
reply by email.]
It's kinda sad that the two responses to this question simply
quotes a figure without explaining how to arrive at this figure.
Any bridge player who has understood advanced high school math can learn
how to do these type of calculations--personally I have found this skill
to be useful at the bridge table, so I will demonstrate it. Also, there's
a bit too much appeal to authority without explaining why things are true
nowadays, but that's another topic...
Background knowledge needed: The number of ways one can choose k objects
from a set of n objects is C(n,k) = n!/[k! (n-k)!], where n!=1x2x3..n.
Example: there are C(52, 13) possible bridge hands.
Anyone who does not understand the above proabably does not have the
background to understand the rest of this.
Ok first let's compute the probability of having specifically a spade
void. Then your hand must be constructed from the other 39 cards.
There are C(39,13) ways of doing this. So the chance of having a
spade void is C(39,13)/C(52,13). Of course this is also the
probability of having a void in any other particular suit.
As a first approximation the number of hands where you have
a void is around 4*C(39,13)/C(52,13). OWe have just added the probabilities
of finding a club, diamond, heart, and spade voids. You willl note that this
number is already pretty close to the right answer.
At this point the lecturer turns towards the class and asks
"Does anyone know why this is not the exact answer?"
Well note that we have counted twice hands where we have TWO void suits,
and thrice the hands where there are THREE void suits. ok no
problem, how many hands have two void suits? Let's try two
specific void suits--spades and hearts. Then there are 26 cards left to
form your hand, so the answer is C(26,13). There are also 6 different
ways to choose 2 suits out of 4 {(C(4,2)=6} hence there seem to
6*C(26,13) ways to have voids in two suits. Of course we have again overcounted
thrice the number of ways we can have voids in three suits.
Hence 4*C(39,13) - 6*C(26,23) is the number of hands that have
at least one void, not counting the hands with three voids.
Fortunately there there are only 4 hands with all 13 cards in one suit.
Hence the exact probability that a hand has a void (or more) is:
[4*C(39,13) - 6*C(26,13) +4]/C(52,13).
Bill Chen.
Jerry Grossman
This is not QUITE right. You have overcounted the hands that contain two
or three voids. Since these are so rare, the numerical answer won't
change much, but the EXACT answer stated above needs to be modified
(especially if the original question was a homework problem!).
David Jones
[ Re: computation of P[holding void in any suit]]
>Any bridge player who has understood advanced high school math can learn
>how to do these type of calculations--personally I have found this skill
>to be useful at the bridge table, so I will demonstrate it. Also, there's
>As a first approximation the number of hands where you have
>a void is around 4*C(39,13)/C(52,13). OWe have just added the probabilities
>of finding a club, diamond, heart, and spade voids. You willl note that this
>number is already pretty close to the right answer.
Could you also demonstrate how to reduce this expression to a number, at the
table, subject to the following constraints:
- Calculators are not permitted at the table, need to do it in your head
- You must be able to do it in less than the 7 minutes you get to play a hand
Do you have the combinatorial tables memorized? What precision is required for
the results to be useful?
Marc 'Merle' Wallace
David Jones <d...@eecg.toronto.edu> wrote:
>In article <36q39q$l...@agate.berkeley.edu> wal...@beirut.berkeley.edu (Marc 'Merle' Wallace) writes:
>
>[ Re: computation of P[holding void in any suit]]
>
>>Any bridge player who has understood advanced high school math can learn
>>how to do these type of calculations--personally I have found this skill
>>to be useful at the bridge table, so I will demonstrate it. Also, there's
>
>>As a first approximation the number of hands where you have
>>a void is around 4*C(39,13)/C(52,13). OWe have just added the probabilities
>>of finding a club, diamond, heart, and spade voids. You willl note that this
>>number is already pretty close to the right answer.
>
>Could you also demonstrate how to reduce this expression to a number, at the
>table, subject to the following constraints:
>
>- Calculators are not permitted at the table, need to do it in your head
>- You must be able to do it in less than the 7 minutes you get to play a hand
>
this turns out to be 4* (39!*39!)/(52!*26! = 39/52*38/51*...*27/40
Let's group the terms this way :
4*39/52 * [38/51*27/40] * [37/50*28/41] * ... * [33/46*32/45]. Note that
each term in the bracket is approx. 1/2. So an approximation is:
4*39/52 * (1/2)^6 = 4* 3/4 *1/64 = 3/64 = approx. 1/21. The answer is around
1/19? This took longer to write down than to do.
>Do you have the combinatorial tables memorized?
Well,o--but I am pretty confident i can come up with some sort of
approximation to most probabilities in bridge. The above computation
is probably MUCH more complicated than most of the problems that come up,
eg. when you have to decide whether south having the hK or a 4-2 d split
is more likely given the cards that have shown up.
>What precision is required for
>the results to be useful?
Well having an approximation for me at least is useful when trying to
determine the better of two lines of play.
Bill
Robert_Bryan_Lipton
Personally I rarely worry about chances under five per cent unless I am
grossly underbid -- in which case I take low probability safety plays. I
am sure there are great players -- Dorothy Truscott and Bobby Goldman come
to mind -- who decide to take the 57.43% chance rather than the 56.92%
chance, but this falls well into the range of table presence for me. I
don't have the combinatorial tables memorized and still work the chance of
a double finesse failing twice as 25% rather than the 24% I am told it is.
If the opponents fight you up to the 6 level and you have 35 points
between you, then you can worry about two or more voids in a hand. I've
never met anyone with three voids in one hand who wasn't willing to play
it at the 7 level.
Bob