# Shuffling a deck of cards "n" number of times to get an organized order

• May 28th 2011, 04:02 PM
Masterthief1324
Shuffling a deck of cards "n" number of times to get an organized order
The scenario is that a fair deck of 52 playing cards is shuffled a certain number of times (52! times) . (a) What is the probability of shuffling a correctly ordered (Ace to 2) deck of cards? (b) What happens to the probability of shuffling a correctly ordered deck of cards as the number of shuffles increase?

Here's what I did so far:

1. Probability of shuffling once and getting the correctly ordered deck = 1/52!. (52! arrangements of 52 cards, 1 permutation that works)

2. Probability of shuffling twice and getting the correctly ordered deck:
P(2) = 1/52!^(2) so in general: P(n) = 1/52!^(n); We see that as the number of trials increase, P(n) approaches zero. So for (b) the probability of shuffling a correctly ordered deck decreases as "n" approaches infinity. (Counterintuitive here so I'm a bit skeptical that this is right)

3. P(52!) = 1/52!^(52!) = very close to 0.

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My friend disagrees maintaining that:

The probability of it NOT being the right deck is (1- 1/52!) If we shuffle again and again, and we do NOT get the deck we are looking for, then the probability of NOT getting the right deck after n shuffles is (1-1/52!)^n. As n approaches infinity, this probability approaches 0, meaning that we WILL eventually break the chain of not getting the right deck.

He thinks that the probabilities do not change for any specific shuffle. However, they do if you consider n shuffles, and are looking for the probability of success or failure at least once in these shuffles.

He interprets it in reverse Q(n) as opposed to P(n) and it makes sense intuitively. But i'm confused as to why my friend's interpretation contrast with mines even though we seem to be modeling the same scenario albeit in terms of complements.
• May 28th 2011, 04:37 PM
bryangoodrich
Your assessment of (1) is correct. There are 52! permutations without replacement that can occur with a deck of 52 cards. It should be clear that the first card can come in 52 possible ways, the second from (52-1), the third from (52-2), ..., etc. Thus, we have 52*51*...*1 = 51! permutations of the entire deck. The case of a completely ordered deck is just one of those 52! and therefore the answer is, as you indicated, 1 / 52!

As for part 2, you need to be clear about what the probability is about. You do not want to commit the Gambler's Fallacy. The probability found in (1) will not change just because we shuffle the deck, no matter how many times we do it. The odds of a given hand in poker, for instance, does not change if the deck is randomized each time (and that is the key here). However, if you're taking the experiment to be "what are the odds of getting a completely ordered deck in n shuffles" then we're looking at a binomial experiment. In each of the n trials you either get it or you don't. The probability of a success comes from (1). From there it is a direct application of the binomial distribution.

Therefore, your friend is correct in that "the probabilities do not change for any specific shuffle." That is what it means to have determined (1). But that is not the same thing as talking about observing a certain number of successes (in this case, one) in n such shuffles. Each shuffle is considered a Bernoulli trial, and as I indicated above, the binomial distribution is the extension of those trials. The probability of a joint event as represented in a binomial distribution is clearly going to be different than any one Bernoulli trial.
• May 28th 2011, 06:49 PM
bryangoodrich
Quote:

Originally Posted by Masterthief1324
(b) What happens to the probability of shuffling a correctly ordered deck of cards as the number of shuffles increase?

I'll add that the way this question is worded sounds to me like asking what happens to the probability found in (1) as you increase the number of times you repeat that experiment. The answer would be it doesn't change the probability at all. You could ask a number of other questions, too, as I alluded to above. You can ask how many times you can expect to see the ordered deck in n shuffles, or you can ask how many shuffles are required before you can expect to see an ordered deck.
• May 29th 2011, 10:46 AM
Masterthief1324
Thanks, I can see your point about ambiguity. To clarify, I wanted to ask about the probability of 1 success in 52! trials. I figured that this was a case of the Binomial experiment so I used the formula P(r) = (C_n,r)(p^r)(q^(n-r)) where n = number of trials, r = number of success, p = probability of success, to obtain the probability of 1. In a binomial experiment, the probability of success in each trial is the same as you have suggested.

I mentioned to my friend that his method was incomplete since this was a binomial trial and this scenario simplifies to P(1) = n*p*q^(n-1) and not q^(n). r=1 is insignificant in the case that "n" is very large but not so insignificant that it should be disregarded because small "r" is significant in the case that n is small. Was I correct?
• May 29th 2011, 11:31 AM
bryangoodrich
You can use latex to state the equation thusly:

\$\displaystyle _nC_k p^k (1-p)^{n-k}\$
[tex]_nC_k p^k (1-p)^(n-k)[/tex]

or

\$\displaystyle \binom{n}{k} p^k (1-p)^{n-k}\$
[tex]\binom{n}{k} p^k (1-p)^{n-k}[/tex]

In the case of one trial, we have

\$\displaystyle P(k = 1) = \binom{n}{1} p^1 (1-p)^{n-1} = npq^{n-1}, q = 1-p\$

Of course if n increases so does the probability. Even unlikely events become more probable to happen if we consider them as a joint process. Consider a biased coin that turns up heads 1/51! of the time. In any one toss we would not expect a heads, but if we consider the process (experiment, event, etc.) of heads showing up in \$\displaystyle 10^{10^{10}}\$ tosses, it becomes a lot more likely. We can model that process with the binomial distribution which looks at such processes that consist of those single events with probability p.

As for significance, you need to define what is 'significant'.