The "card war" game is played with 2 players and 8 decks (52 cards each, 416 total), the purpose of any player is to score a higher value card, where the values are ordered: ace < two < ... < ten < jack < queen < king. Before the game starts, the dealer draws the top card and removes the following [card value] cards (for example if he drew king then he'll remove the first 10 cards). Now the game begins, player 1 gets a card and then player 2. Each round, whoever has the highest value wins. If both players have the same value, no one wins

What's the probability of player 1 winning at the n'th round, given that the cards that were played are known? My idea is simple, take into account all the possibilities of the cards burned at the beginning and then multiply it by all the possibilities of p1 getting a better value than p2 (of course removing the drawn cards each round from the deck). I was wondering if there was and easier way.

  • 1
    $\begingroup$ By symmetry, the probability of the first player winning any particular round is half the probability that the two cards are not the same value. The probability they are the same value is complicated if you have seen some of the cards, but if not then it is $\frac{8 \times 4 -1}{8 \times 52 - 1}$ making the probability of the first player winning any particular round $\frac{192}{415}\approx 0.46$. If the initial cards are burned without revealing them, then this probability does not change $\endgroup$ – Henry Sep 12 '18 at 13:31
  • $\begingroup$ But what about the fact that some cards are removed at the start, and the fact that the cards drawn are known? For example, if many high value cards were drawn, then that's got to affect the probability that a high value card will be drawn next. $\endgroup$ – CodeHoarder Sep 12 '18 at 13:35
  • 1
    $\begingroup$ What Henry described here is the marginal/conditional probability, with no extra information given. In your comment, the situation is that you are given the cards removed, so the conditional probability of course is different from the marginal in general and will depends on the removed cards. So you can calculate them case by case, with the same methodology. $\endgroup$ – BGM Sep 12 '18 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.