That would seem to give an unevenly high probability that the
last few cards of the last suit are in the same hand -
depending on your deck order that's either the high clubs or the low clubs.
Alternatively, you can put the deck in some other order,
which might be more even from a bridge-playing perspective but
still gives away information.
At 10:43 AM 6/25/99 -0500, Matt Crawford 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.
>
>I can do it in 101 bits trivially. Start with deck in a fixed order,
>use two random bits per card to decide which hand gets that card. If
>the selected hand is full, give the card to the next unfilled hand
>clockwise. That's 100 bits for the first 50 cards. The 51st card
>needs at most one bit and the 52nd needs none.
Thanks!
Bill
Bill Stewart, [EMAIL PROTECTED]
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