Probability Question for a Card Game?

Raymundo Kyser: Yes, both of you have the right answer.Just a small question:What if we consider sequence in continuance?i.e. Since 1 comes after 13, we take 1 as equidistant from 13 and 2 (in such case tie can happen even if one picks odd and one picks even)...Show more

Mozell Sponsler: For there to be a tie, the chosen card has to be exactly between the two chosen numbers.The numbers could be 2 apart (e.g. 2 and 4, with the chosen card being 3)The numbers could be 4 apart (e.g. 7 and 11, with the chosen card being a 9)The numbers could be 6 apart, 8 apart, 10 apart or 12 apart.So now we need to figure the probability of each of those outcomes:P(2 apart):Lower card (1 to 11)Upper card (3 to 13)That's 11 possibilities (e.g. (1,3), (2,4), ... (11,13))P(4 apart)Lower card (1 to 9)Upper card (5 to 13)That's 9 possibilities (e.g. (1,5), (2,6), ... (9, 13))...All the way down to 12 apart:1 way (1,13)I think you see the pattern.11 + 9 + 7 + 5 + 3 + 1 = 36 waysThe total n! umber of ways of picking two different numbers out of 13 is:C(13, 2) = (13 x 12) / (2 x 1)= 78 waysFinally you have to factor in the probability that the card chosen is in the middle. This only happens 1/13 of the time.36/78 * 1/13= 6/13 * 1/13= 6/169≈ 3.55%P.S. Here we assume that the two numbers chosen by the players are randomly assigned, rather than intelligently picked. If you don't make this stipulation, the game is skewed toward one player or the other. So they each pick a random card and that is their number. If they are the same, they draw again. Once the numbers are chosen, they return the cards to the deck and draw the target card.P(Player 1 wins) ≈ 48.225%P(Player 2 wins) ≈ 48.225%P(Players tie) ≈ 3.550%Edit: Regarding your follow-on question. If you allow the numbers to be considered in a "circle" then every pair of numbers will have a corresponding "tie" card. So regardless of what the two players pick (assuming they are different) there is a t! ie card with 1/13 probability.P(tie with circular "distance") ! = 1/13 ≈ 7.69%However, if you want to now allow for the two players to pick the same number, every card would be a tie card. Thus, there are two cases:1) Numbers are different (12/13 of the time) --> tie card = 1/13P(different, and tie) = 12/1692) Numbers are the same (1/13 of the time) --> tie card = 13/13 P(same, and tie) = 13/169P(tie) = 12/169 + 13/169 = 25/169 ≈ 14.79%...Show more