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Probability & Statistics/Probability of repeating same results


I took P&S in college but only have a rudimentary knowledge of the subject. Suppose you took a deck of playing cards, threw them on the floor and then picked them up (one at a time) without looking at them. I know if you did it enough times, they would end up in exactly the same order as they were the first time. I also realize that any similar experiment would always, at some point, have the original result because it could be repeated an infinite number of times. Does the theory for this sort of thinking have a name? I used to think it was the "law of large numbers" but I don't think that's exactly right. Can you help me? Thank you for your thoughts!


ANSWER: There are a large number of ways to rearrange a deck of cards. Let's pretend, temporarily, that there are only 4 cards in the deck (say 1-4 of spades).

The number of ways to rearrange the cards (throwing them down and picking them up in a "random" order is just a rearrangement). This is called a permutation. The mathematical way to count permutations is using a factorial, which is denoted by the exclamation point.

With 4 cards, the number of permutations is 4! = 4󫢪1 = 24.

Now, if you do this once, the probability of picking up the cards in the correct order is p = 1/24, because we only want one specific permutation out of the 24 possibilities.

If we do it twice, what is the probability that neither was correct?

Well, if p is the probability we get it, 1-p is the probability we don't. (This is a complementary probability.) What is the probability we fail twice? It's (1-p)^2, because you multiply the probability of failure for each failure you want.

So to fail N times, it's (1-p)^N.

As time goes on, (1-p)^N gets close to zero. Our probability of at least one success would be 1-(1-p)^N. For example, if p=1/24 like in our ongoing example, and N=30, then you get:

1 - ( 1 - 1/24 )^30 = 1 - (23/24)^30 ≈ 1 - 0.958333^30 = 1 - 0.278932 = 0.721068

That means there is 72.1068% (nearly 3/4) chance that one of these 30 drops would result in a correct ordering of the four cards.

If you did 100 drops, you'd get:

1 - ( 1 - 1/24 )^100 ≈ 0.985821

That's over 98% probability of getting it right after 100 drops. Technically, you can never be 100% certain -- someone who is very unlucky might still not get 1-2-3-4 after 100 drops. And an even unluckier person after 200 drops. But it gets closer and closer to 100%, the more drops you do.

Now, we were using 4 cards. If there are d cards (d for "deck" sounds good to me), and you do N drops, then the formula is:

1 - ( 1 - 1/d! )^N

If you use 52 cards and do 100 drops, you get:

1 - ( 1 - 1/52! )^100 ≈ 0.00000000000000000000000000000000000000000000000000000000000000000124...

That's a TINY number. With 52 cards, it is very unlikely to get the order right. Even after 100 drops, it is nearly impossible. You would have to get very lucky after 100 drops.

However, if you an enormous number of drops, it gets more likely:

1 - ( 1 - 1/52! )^N = 1/100

For a large value of N, you could guarantee a one in 100 chance. For an even larger value, you might want something like:

1 - ( 1 - 1/52! )^N = 99/100

But how big should N be to guarantee that dropping 52 cards N times will give you a 99% chance of picking them up at least once in order?

Well, if you solve this equation for N, you get:

N = log(1-99/100)/log(1-1/52!)

Unfortunately, this is an enormous number. It is roughly:

N ≈ one billion trillion trillion trillion trillion trillion

To give you some idea, this number is approaching the number of atoms in the universe.

If you had a computer processor to even simulate this many trials, you would need a trillion trillion trillion computers with 8 cores running at 3 GHz, assuming each CPU cycle could compute one simulation of dropping and reordering the cards. Even then, with all those computers, it would take you 10 million times the age of the universe to do it.

So, in an abstract mathematical sense, yes -- if you do enough trials of this 52-card-pick-up game, you will "eventually" be almost certain (99%, 99.9%, whatever certainty you want) to have picked up the cards in order. However, in practice, with 52 cards, there are just too many possibilities to expect this to happen in a reasonable amount of time.

---------- FOLLOW-UP ----------

Your answer is fascinating, as well as mind boggling and I appreciate you taking time to explain the process in such an interesting way. I understand the factorial aspect of the experiment and also that achieving the same result twice would be extremely unlikely. My only other question is this: If, in spite of the extraordinary odds, the same result would occur with enough tries, is their a name given for such an experiment? Thank you, Clyde!

I'm not aware of a term that specifically denotes this type of event, but there are a few ideas related to this.

If you consider the limit of the probability 1-(1-1/d!)^N, where N increases without bound, it must be true that either the likelihood of success is either "with almost certainty" or "virtually impossible." Regardless of the practical implications, we have seen that it is almost certain (just not in a realistic timeframe, since large values of N are not practical). This is one instance of a law called Kolmogorov's zero-one law. The probability of many such sequential processes is (as the process continues indefinitely) either very close to zero (virtually impossible) or one (eventually must happen).

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Clyde Oliver


I can answer all questions up to, and including, graduate level mathematics. I do not have expertise in statistics (I can answer questions about the mathematical foundations of statistics). I am very much proficient in probability. I am not inclined to answer questions that appear to be homework, nor questions that are not meaningful or advanced in any way.


I am a PhD educated mathematician working in research at a major university.


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BA mathematics & physics, PhD mathematics from a top 20 US school.

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