Main   Study 8J57 by Richard Pavlicek  # The Same Hand Twice

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 don’t 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! That’s 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 can’t 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:  BlockProbability Factor
10.999904722321315609612091283835306942
20.999714211511904247859185748219455195
30.999523736997020587582013310287723042
40.999333298769750727132764117396221760
50.999142896823182081802941778652433006
60.998952531150403383572539508225011955
70.998762201744504680859264036144562184
80.998571908598577338267807277499255104
90.998381651705714036339165750930898688
100.998191431059008771300007737338791896
110.998001246651556854812088169700432533
120.997811098476454913721711244919877209
130.997620986526800889809240748616282782
140.997430910795694039538658083766888957
150.997240871276234933807167994120431705
160.997050867961525457694851973298706856
170.996860900844668810214369350505732541
180.996670969918769504060706043765688168
190.996481075176933365360970971612536285
200.996291216612267533424240114155962069
210.996101394217880460491448214449993142
220.995911607986881911485328111092390168
230.995721857912382963760397692984625992
240.995532143987496006852994467183998162
250.995342466205334742231357730781146347
260.995152824559014183045758337737972550
270.994963219041650653878676051622688070
280.994773649646361790495024475180436877
290.994584116366266539592423547679670448
300.994394619194485158551519600976174197
310.994205158124139215186352965238370332
320.994015733148351587494773115279246404
330.993826344260246463408901348441982877
340.993636991452949340545640984988076798
350.993447674719587025957235081938482059
360.993258394053287635881871651320009863
370.993069149447180595494336373770955714
380.992879940894396638656712798461641756
390.992690768388067807669130020287285339
400.992501631921327453020557825292326489
410.992312531487310233139649295287068425
420.992123467079152114145630862619206362
430.991934438689990369599239806064540665
440.991745446312963580253709178802890856
450.991556489941211633805800159446947158
460.991367569567875724646881817093516030
470.991178685186098353614058281368335666
480.990989836789023327741343308437356577
490.990801024369795760010882233959101206
500.990612247921562069104221303954435039
510.990423507437469979153624374571799839
520.990234802910668519493436971727677489
530.990046134334308024411497701603770462
540.989857501701540132900597002984102112
550.989668905005517788409983232416956857
560.989480344239395238596916073188296882
570.989291819396328035078267259095008185
580.989103330469473033182168604008044686
590.988914877451988391699707328217254688
600.988726460337033572636668672551389197
610.988538079117769340965325791268506544
620.988349733787357764376276914713702299
630.988161424338962213030329772742807758
640.987973150765747359310433269912414193
650.987784913060879177573656403438293658
660.987596711217524943903214414926000437
670.987408545228853235860542166879150149
680.987220415088033932237414734992586169
690.987032320788238212808115207239355312
700.986844262322638558081649680762126702
710.986656239684408749054009447581399391
720.986468252866723866960480360134555630
730.986280301862760293027999367661527662
740.986092386665695708227558214454556602
750.985904507268709093026654290991232294
760.985716663664980727141788628971713069
770.985528855847692189291011031282734003
780.985341083810026356946512327912721654
790.985153347545167406087263748844042302
800.984965647046300810951703404950119404
810.984777982306613343790469867926864408
820.984590353319293074619182840289573106
830.984402760077529370971270906468147450
840.984215202574512897650846356035210182
850.984027680803435616485627070103386704

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:

0.502403088210042507619291517429218950

We’re 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 don’t 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.

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## The Same Deal 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 can’t be found when you know one exists, so I turned to my longtime cyber friend and math guru, Charles Blair. If he didn’t 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 can’t 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…

That’s 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. Don’t 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 doesn’t 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. ### And in conclusion…

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

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© 2019 Richard Pavlicek