At 11:44 AM 6/25/99 -0700, bram wrote:
>> > > > > There are 52! bridge hands, so a random hand has
>> > > > > log2(56!) = 226 bits of entropy or 68 decimal digits worth.
>> > No, just 52! / (13!)^4 hands, which is around 2^96.
>> The interesting part is to come up with an algorithm that only uses 96
bits.
>
>Take the 96 digits as a really big number base two,
>find it's value modulo 52! ...
(Actually 52!/(4*13!))
Doesn't work, though - for values higher than 52!/13!*4 you need to
reject the random number and draw again. Otherwise you've got an
excessively high probability of repeating the first
2**96 mod 52!/13!*4 hands.
>The real point, though, is that you never, *ever* need more than about 80
>bits of entropy for *any* amount of random numbers if you use a
>crypographically strong pseudo random number generator.
It depends on the application - for encryption keys, it's probably ok,
at least for the next N years, unless the structure of selecting your keyspace
interacts with the crypto algorithm in a way that decreases the strength
of the resulting encryption. It's unlikely in the general case,
but it can happen.
But for bridge games, if you don't use at least 52!/13!*4 bits,
or more if you're using them wastefully, there are hands that _won't_ happen,
and those hands can be predictable in ways that are useful to the players,
and therefore bias the results of the bridge game as well as complicatng play.
If you know the system will never generate a hand where more than one player
has more than 10 cards of one suit, and you're holding 10 clubs,
this can be fun, but it's less emotionally satisfying than bidding that slam
when you're worried that your opponents have 11 spades because the
deck wasn't shuffled right :-)
Thanks!
Bill
Bill Stewart, [EMAIL PROTECTED]
PGP Fingerprint D454 E202 CBC8 40BF 3C85 B884 0ABE 4639
