Study 8J57 Main
by Richard Pavlicek
Have you ever held the same hand twice? Probably not, barring a technical foul-up such as replaying a previous duplicate board, because there are 635,013,559,600 possible bridge hands. Even the remarkable feat of playing 100 hands a day for 80 years would reach only 3 million, so 635+ billion is a long way off.
But wait! You dont have to hold every bridge hand to get the same hand twice, as it could happen anytime by chance. So what are the actual odds? Or to be more specific, consider this poser:
|How many hands must be held for the probability of the same hand twice to be greater than one-half?|
This is analogous to the Birthday Problem of how many people are required for the likelihood of any two to have the same birthday (month-date). Surprisingly, the answer is only 23 people. Since there are 366 possible birthdays (February 29 included) the probability of 23 unique birthdays is:
365/366 × 364/366 × 345/366 × 344/366 = 0.493676988
Since the above is less than one-half, most of the time there will be at least one duplicated birthday. To verify, remove the last factor to evaluate for 22 people, which is 0.525249354 , meaning that all unique birthdays would then be expected.*
*Birthdays are not equally likely (especially February 29) so probabilities are not exact, but close enough for practical analysis.
Now back to bridge hands. The necessary evaluation is easy to formulate: (H = 635013559600, N = number of hands)
(H-1)/H × (H-2)/H × (H-N)/H < 1/2
That is, to find the smallest N for which the probability of all hands being unique is less than one-half.
Argh! Thats easier said than done, as the daunting calculation might have millions of factors, blowing away any calculator. Even my super calculator with its 131,000+ digit precision cant handle it in one piece but I was able to break it down into blocks. Each block consists of 11,000 factors, a convenient number because the denominator (H11000) and the factorial-style numerator can be calculated exactly (at most 129,831 digits each) then a division produces the probability factor, which is shown here to 36 decimal places. For example, the first block of 11,000 factors yields:
(H-1)/H × (H-2)/H × (H-11000)/H = 0.999904722321315609612091283835306942
This shows that after 11,000 hands (not counting the original) there is a very high probability of no duplicates. Completing 85 blocks (11,000 hands each) produced the following table:
Each of the above probability factors is the product of 11,000 individual probabilities, so the product of them all will give the probability of 935,000 (85 × 11,000) successive hands without a duplicate. My super calculator will spare us the agony, as it multiplies to:
Were getting close! At some point within the next group this running probability will drop below the 0.500 mark, which will indicate the number of hands required to expect a duplicate. A little trial and error reveals the cusp numbers to be:
938,250 hands = 0.500000504948087435169100660973335833
938,251 hands = 0.499999766182801755987543645849931122
So there you have it. After 938,251 hands (or 938,252 counting the original) you are more likely than not to have held the exact same hand twice. Relative to 635+ billion, this is amazingly small, so dont be too surprised if it happens to you. Of course, nobody remembers every hand they held (particularly the spot cards) so most duplicates would go unnoticed.
This study presumes each hand is from a separate deal. Obviously, any two hands from the same deal could never match.
|Study 8J57 Main||Top The Same Hand Twice|
After completing the above calculation, it occurred to me that it might be possible to do the same for complete deals; that is, to find the number of deals required for the likelihood of getting the same deal twice. Whoa! Not in my lifetime. A two-day test run by my computing engine reached over 10 billion, and the probability was still at 0.9999999999 Is there any way to solve this?
It really bugs me when a solution cant be found when you know one exists, so I turned to my longtime cyber friend and math guru, Charles Blair. If he didnt know, I might have to wait until my afterlife on some ethereal abacus in the sky. Yes! There is a general formula, which he explained as shown below. (N = number of possibilities, Ln 2 is the natural logarithm of 2 = 0.69314718 )
Square root of (2N × Ln 2)
Testing the formula on my hands calculation produces:
Sqrt (2 × 635013559600 × Ln 2) = 938251.4145
Wow! Right on the money! Obviously, you cant hold a fractional hand, so it would require 938,252 hands for the likelihood (probability greater than one-half) of the same hand twice exactly as I discovered the hard way.
Applying the formula to complete deals produces:
Sqrt (2 × 53644737765488792839237440000 × Ln 2) = 272703864050461.0721
Thats 272,703,864,050,461 (272+ trillion) plus a fraction, so rounding up means that exactly 272,703,864,050,462 deals are required for the likelihood of playing the same deal twice. Dont try this at home!
Thanks, Charles, for the education and to think of all the time I could have saved by consulting you sooner.
Addendum: Charles modestly reminded me that he doesnt deserve any credit; he was simply aware that a formula might exist, and he looked it up on Wikipedia. Okay, then consider yourself discredited you bum!
Acknowledgments also to Sandro Bellini of Milan, Italy, who noted that the formula is only an approximation but a very good one. Using different tools he confirmed that my solutions for both hands and deals are exactly correct.
As with most of my probability studies, I take great care to ensure that any information you obtain here is completely useless. But hey, thanks for keeping me company!
And in conclusion, we see that protocol takes precedence over procedure.
This parliamentary point of order develops a centrifugal force as a catalyst,
hastening a change in reaction that remains indigenous to its inception.
-Professor Irwin Corey
|Study 8J57 Main||Top The Same Hand Twice|
© 2019 Richard Pavlicek