Suppose we have six different coloured decks of cards numbered 1 to 10, ordered ascendingly with 1 on top. Now we're going to draw cards, no replacement. The choice of deck is independent (let's say a die decides). This leads me to believe that, in any given round, the probability of drawing the top card is $1/|decks|$, so $1/6$ until the first deck is emptied, $1/5$ from then until the second deck is emptied and so on. Is this correct?

Now, the probability of drawing, say, four cards from the same deck out of four draws, is the probability of drawing the first card followed by the second card and so on. But what if, instead, precisely one of those cards was from any other stack?


  1. blue
  2. blue
  3. not blue
  4. blue

First let's say we don't care when in the sequence the different coloured card is drawn. What is the probability of this outcome? The probability of drawing three same-coloured cards in sequence plus the probability of not drawing that colour? That doesn't sound right to me but with mutual exclusivity I'm not sure how to limit the outcome as specified. Does this throw conditional probability out the window?

Now we'll assume that the alien card was neither the first nor the last card; or more generally, we are given a set of draws that may have resulted in a different colour and another set that definitely did not. The constraints are otherwise the same. How does this change the game?


1 Answer 1


I’m assuming that color equates to deck in your description. There are $6$ ways to choose the deck that gets picked three times and then $5$ ways to choose the deck that gets picked once, for a total of $6\cdot5=30$ different combinations of decks. The odd card can appear in any of $4$ places in the sequence, so there are $120$ different draws that result in $3$ cards from one deck and one from another. Since a deck cannot be exhausted in $3$ draws, there are $6^4$ possible sequences of $4$ draws. The desired probability is therefore $$\frac{120}{6^4}=\frac{120}{1296}=\frac{5}{54}=0.0\overline{925}\;.$$

I’m not sure that I understand your last paragraph. The probability of drawing three cards from one deck and one from another if the odd card is either the second or third drawn can be calculated similarly; the only difference is that thre are only $2$ places in the sequence in which it can appear instead of $4$, so there are only $60$ different draws that meet the conditions, and the probability is only half as much.

  • $\begingroup$ That's what I was looking for. $\endgroup$
    – Quail
    May 31, 2012 at 11:36

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