[mycroft]

A poker site gave the following claim, given the power of large numbers (which he demonstrated):

Quote

If a random-looking combination of 52 cards had ever occurred more than once, throughout the entire history, then the combination in question was definitely not a random one.

But one thing that was conspicuously absent from his demonstrations of the size of 52! was the Birthday Paradox. Which led me to wonder:

• How many random deck permutations lead to a 50% chance of two identical? quick approximation (from wikipaedia) is sqrt(2*52!*.5)= sqrt(52!) ~ 9E33, no better.
  
• How many lead to a .5% chance? sqrt(52!*1E-2) - okay, 9E32...
  
• One in a million chance of it being coincidence...9E27 or so - still heat death of the universe level.

How about bridge hands? There are a lot fewer! less than a million hands leads to one repeat at p ~ .5. There are certainly players who play a million hands in their lifetime, even if it's in the realm of years. So: the chance that you've played an exact hand before in your life, if you play effectively fulltime, is better than even. In my case, I'm sure it looks like KT86 842 Q9 J754. Remembering it? NOYL.

Of course, the chance that it was a function of bad "shuffling" is much higher still :-) And the chance of a bridge *deal* being identical, never mind a whole set - pretty much zero from chance, as we'd hope.

But going back to the poker site - it's true in general, but if you're a specialized heads-up Hold'em player, the chance that you see the exact same "hand" in your life (the 9 relevant cards identical) is pretty much 1. If there are 25 relevant cards, at a 10-player table, much less so. Interesting...

[TylerE]

Birthday paradox doesn't apply to this situation at all, I'm not even sure I follow your logic at all really.

In addition, your math is way wrong.

Number of bridge hands = C(52,13) = 635,013,559,600

If you were capable of playing bridge from the moment of birth, played 20 hands an hr, 24 hours a day until you were 100, you'd play about 17.5M hands in your lifetime. And again, that's at a totally unsustainable pace of 420 hands a day. That would mean you would play less than 1/36000 of the possible hands.

[winkle]

Nowadays you can play more than one hand a minute with robots on BBO. I saw Leo Lasota once played 500 hands in ~7 hours trying to catch whoever was ahead of him in masterpoints that day.

[Zelandakh]

TylerE, on 2014-April-10, 14:18, said:

Number of bridge hands = C(52,13) = 635,013,559,600

No forgetting to allow for vulnerability and seating in your calculations. If we also include rubber bridge then you can regard different part-score situations as separate deals too.

[ArtK78]

I believe that mycroft's point is that you only have to play a fraction of the possible hands in order for the odds to be better than 50% that you will have played the same hand twice.

The so-called birthday paradox is the answer to this question - how many people do you have to have in a room before the chances that any two or more of them will have the same birthdate (day and month, not year)? Despite the fact that the number of possible birthdates is 365 (ignoring February 29), the answer is 23. Once you have 23 or more people together in the same place, the chances that any two of them have the same birthdate exceeds 50%. This is a very basic probability problem.

Similarly, despite the fact that there are in excess of 635 billion possible bridge hands, you only need to play a small fraction of that number before the likelihood of your having played two identical hands will be over 50%. I leave it to others to determine what that number of hands is.

[mycroft]

Tyler: correct. And to a first approximation (again, re: wikipaedia), the number of hands required for a birthday paradox to have a 50% chance of happening is n = sqrt(2*hands*.5) or sqrt(hands). Which, on something approximating 1E12, is about 1E6, i.e. a million.

There is a great deal of difference between the time it takes you to repeat any *particular* hand, and the time it takes you to repeat any hand you have played.

Zel: hand != deal; there are monstrously more deals, and even square root of that is in the billions. Add vulnerability and dealer (and system and opponents' system and ... having said that, vul/dealer is only x2^4. That means only 4 times as long with the sqrt(), not really a big deal when we're talking 2^30 range) and correct, we're way over the top. I did mention that in my original.

[TylerE]

Zelandakh, on 2014-April-10, 14:53, said:

No forgetting to allow for vulnerability and seating in your calculations. If we also include rubber bridge then you can regard different part-score situations as separate deals too.

No, that's just the number of combinations of 13 different cards...hands not deals. Deals would be hugely higher.

[PhilKing]

Birthday paradox obviously applies.

[billw55]

Yes, this is the birthday problem with higher numbers.

I have never heard it called a "paradox" before this thread. To me it is not a paradox, it is just math.

[barmar]

From Oxford American Dictionary:

Quote

paradox: a seemingly absurd or self-contradictory statement or proposition that when investigated or explained may prove to be well founded or true

The claims in the birthday problem is very different from most people's intuitive expectations, so it's "seemingly absurd", but the math shows that it's true.

[barmar]

Extrapolating from his recent activity, Leo plays about 80,000 hands on BBO per year. So it would take him at least a decade before he will have probably held the same hand twice.