Statistics geeks..... show me the brains?

Question

I once played a card game called war. In case you are not familiar with it, you can search the web. I found the rules at http://boardgames.about.com/od/cardgames/a/war.htm (not including original variant).

I won in the absolute least number of rounds, 26. I realize no skill is involved, but the odd / probability of winning all 26 hands in a row must be tiny. I always wondered just how tiny. If your good in this area, and want to take a stab at it, I would enjoy hearing your answer.

Answer 1

From the rules you posted, it seems like it is possible to win the game in fewer than 26 rounds. If there is a tie, additional cards are placed on the table and more than one card can be won at a time.

However, assuming there are no ties, a player must win 26 rounds in a row. A generalization made to simplify calculations is that each round is an independent event, the cards are dealt randomly, and that both players have an equal chance of winning any round.

If this is the case, a player has (1/2) probability of winning a round and must do it consecutively 26 times, thus the probability is (1/2)^26, which comes out to a microscopic 0.0000000149 (that seven zero's after the point). In other words, harder than 1 in 67 million.

If you wanted to calculate the probablility that either player (you or your opponent) wins in 26 rounds, simply double that number: making the odds harder than 1 in 33.5 million.

Answer 2

multiply ever possibly outcome together nd such and ull get it. its 1 in however many different outcomes u can have. 52 is to large a number for me to calculate at 11 at night but good luck u can do it!

Answer 3

i could do the math and it would come out to be thouands or just say 50/50 since its just a game of chance and you could either win it or not!!

Answer 4

I know this has something to do with combinations/permutations and factorials but its been awhile since stats for me.