I was playing this card game and noticed what seems like a very unlikely event. I was wondering how one would compute the probability of it happening.

The scenario was that there are ten categories of cards, each category having five identical cards.

The deck is given to a person and they place a card down on the ground. They then pass the deck on to the next person, who also places another card down on top. If it is a matching card of the same category, that person picks up and keeps the pile.

The unlikely event that occurred was that everybody kept placing cards down and nobody picked up anything, because there were no consecutive cards of the same category being placed down, until the very last card. So essentially, the very last person (who had the 50th card) ended up picking up the entire pile because their card matched the second last card!

How would one compute the probability of such an event occurring? I've no idea where to begin, but my gut tells me that inclusion/exclusion may be needed here.

  • $\begingroup$ So you win the pile only when the two consecutive are from same category? $\endgroup$
    – Kamster
    Commented Sep 22, 2014 at 16:09
  • $\begingroup$ That is correct. $\endgroup$
    – Trogdor
    Commented Sep 22, 2014 at 16:10

1 Answer 1


suppose the second player places matching card, the probability is: $$\frac{\binom{50}1\binom41}{\binom{50}1\binom{49}1}\approx0.0816=8.16\%$$

Select one card of 50, then 1 out of 4 remaining matching same type.

At the first turn only.The probabilities for other later turns depend on location of cards and is a little-little bit lengthy.Hope you got the idea.

  • $\begingroup$ That's where I have issues, for later turns. More specifically, I have issues with computing the probability for the very last 2 cards matching in categories. $\endgroup$
    – Trogdor
    Commented Sep 23, 2014 at 3:11

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