I'VE GOT THE BYTE ON MY SIDE

57005 or alive

3-player war

Jun 10, 2012 math

A friend encountered the following event recently when playing a game of 3-player war: Three distinct cards were played (their exact value isn’t important), then on the next hand the same 3 cards were played again, and by the same players (different suits, obviously). What are the chances of this happening?

The fact that there are 3 different players isn’t really relevant. The event is equivalent to randomly picking 6 cards from a deck and having the first 3 match the second 3, in order, and with 3 distinct values.

How many ways can 6 cards be drawn from a standard deck, with order considered? There are 52 ways to pick the first card, 51 ways to pick the second card, 50 ways to pick the third card, etc. The total number of possibilities is therefore 52 * 51 * 50 * 49 * 48 * 47 = 14,658,134,400.

So how many ways can 6 cards be drawn so that our event occurs? I see two ways to count the possibilities.

First way: The first card drawn can be anything, so there are 52 ways to pick it. The only restriction on the second card is that is doesn’t match the first. Since there are 51 cards left to select from, and we know 3 of those match the first card, there are 51 - 3 = 48 valid options for the second card. Similarly, the third card can’t match the first or second, so out of the 50 remaining cards, only 50 - 6 = 44 are valid possibilities. For each of cards four, five, and six, there are exactly 3 possibilities (they need to be one of the 3 remaining matches to cards one, two, and three, respectively). The total number of ways to draw all 6 cards is therefore 52 * 48 * 44 * 3 * 3 * 3 = 2,965,248.

Second way: First we count how many ways there are to pick 6 cards (ignoring order) that match our event. There are \(13 \choose 3\) ways to pick the 3 pip values that will be present. For each of the three pip values, there are \(4 \choose 2\) ways to select the two suits which will be present. So there are \({13 \choose 3}\cdot{4 \choose 2}^3\) ways to select the cards themselves. Next we count how many ways those 6 cards can be ordered so that the first three match the second three. The three pip values can be ordered in 3! ways. There are two cards of each pip, so there a 2 ways to order each of the 3 pairs. So there are \(3! \cdot 2^3\) ways to order our six cards. The total for this approach thus comes out to \({13 \choose 3}\cdot{4 \choose 2}^3\cdot 3! \cdot 2^3\) = 2,965,248.

That means the overall probability of this event is 2,965,248 / 14,658,134,400 = 0.0002023, just north of a 1 in 5,000 chance.


Comments