I was playing a card game called Mille Bornes with my daughter. She dealt a hand while I was getting a cup of coffee in the other room. I came back and we started playing. She immediately played the four safety cards. These cards not only prevent the other player/team from playing bad cards on you, you also get an extra turn after playing them. I told her that it wasn't fun for me to play when the other person is cheating and that we could either play fair or not play at all. My math's a bit rusty, so while I'm confident enough that the odds for her drawing that hand are slim-to-none, to put it mildly. I was wondering what the exact odds were.

To sum up:

  • 106 cards in the deck
  • Each player is dealt 6 cards
  • What are the odds of getting the 4 best cards in your initial hand?
  • $\begingroup$ In making a calculation, it is natural to assume that all choices of $6$ cards are equally likely. This assumption is false: shuffling is often an ineffective randomization device. When in tournaments players started to be dealt cards by a good pseudo-random number generator, they complained they were getting more "weird" hands than in the past. $\endgroup$ – André Nicolas Apr 1 '13 at 16:53

The probability is: $$\frac{\binom{102}{2}}{\binom{106}{6}}=\frac{6\cdot 5\cdot 4\cdot 3}{106\cdot 105\cdot 104\cdot 103}\approx \frac{3}{1,000,000}$$

So this happens about three times in a million hands.

That means every deal, with two players, the odds is about 6 in a million that somebody gets these four cards. But the odds that any specific player (say the dealer) gets a hand like this is still $3$ out of a million.

In the above expression, the numerator is the number of hands with these four cards - you can pick any other two cards from the remaining 102 cards. The denominator is the total number of possible hands - the number of ways of picking six cards out of $106$.


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