What is the probability that 13 randomly drawn cards will come in a strictly increasing sequence of ranks in a common 52 card deck (239KA is strictly increasing, but A239K is not, and JJJKK is not)? Ignore suits. How do I go about doing this?

  • $\begingroup$ Hint: there's only one possible sequence of ranks. The answer will depend on whether the cards are drawn with or without replacement. Perhaps you could mention this detail in your question. If the answer does not become obvious, think about a six-card deck with just three ranks and two suits: you can enumerate all the possibilities. $\endgroup$ – whuber Jul 13 '16 at 18:39
  • $\begingroup$ Sampling without replacement (from the traditional 52-card deck) is standard in playing-card problems. $\endgroup$ – Kodiologist Jul 13 '16 at 18:58
  • $\begingroup$ All 13-card sequences that are strictly increasing / All 13-card sequences. $\endgroup$ – FisherDisinformation Jul 13 '16 at 19:31

I assume your cards are sampled without replacement, although if sampled with replacement the maths wouldn't be very different (but results may be very different).

I'll go in two steps: First I will try to find the probability of getting 13 cards with different numbers or figures, since you stated that a repeated number or figure breaks an strictly increasing sequence. Second, I'l try the probability of a sample of 13 different cards get well ordered.

For the first card, we can chose any of our 52 cards. For the second, we have to chose a different number, so we only have 48 out of 51 remaining cards to chose, for the third we have 44 out of 50, and so, until we have 4 out of 40 for the last one. That is:

$$P(\text{13 different numbers}) = \frac{52\times48\times44\times40\times36\times32\times28\times24\times20\times16\times12\times8\times4}{52\times51\times50\times49\times48\times47\times46\times45\times44\times43\times42\times41\times40} = \frac{4\times13!}{52\times51\times50\times49\times48\times47\times46\times45\times44\times43\times42\times41\times40}$$

For the second part, there is only a way to order 13 different numbers (or cards) to have an increasing sequence, and therefore the probability of getting such a sequence will be of $1$ on $13!$.


$$P(\text{13 cards in strictly increasing sequence}) = \frac{4\times13!}{52\times51\times50\times49\times48\times47\times46\times45\times44\times43\times42\times41\times40}\times\frac{1}{13!}=\frac{4}{52\times51\times50\times49\times48\times47\times46\times45\times44\times43\times42\times41\times40} $$

And that is about $1.01157\times10^{-21}$.

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