[itachi-san]

Bill and Ted each have a standard pack of cards (no jokers). They deal their cards one at a time, from the top of the deck, checking for an exact match. Bill wins if no matches are found. Ted wins if at least one match occurs. What is the probability that Bill wins?

[Guest]

(1-(1/52))^52 = 36.43% chance that Bill wins

[unreality]

Zarball Spectators

The answer, perhaps counterintuitively, can be approximated by

1/e or 36.8%. As the number of objects grow, it gets closer and closer to 1/e, alternating above and below the exact number [it never actually reaches 1/e]. As Llam4 says, with 52 objects it is ~36.4% that no matches are made

Edited January 2, 2009 by unreality

[Guest]

Bill and Ted each have a standard pack of cards (no jokers). They deal their cards one at a time, from the top of the deck, checking for an exact match. Bill wins if no matches are found. Ted wins if at least one match occurs. What is the probability that Bill wins?

50/50 first card 4/52 4/52

1/13 chance of selecting an King

1/169 of choosing another King

52 chances of hitting 1/169

And I am lost from there lol Valiant effort tho !

[Guest]

(1-(1/52))^52 = 36.43% chance that Bill wins

You did 1/52, Im assuming a pair is meant KK, not King of spades to King of spades

A pair, like every card game, is face equal, not face+spades, etc..

then again, this is a question cuz I have no idea how you formulate that equation

looks awesome tho lookin at it

[bonanova]

Spoiler for understanding the inclusion-exclusion principle:

Suppose there are two people in a room - a man and woman. We ask the probability of a woman.

There are [2] people. Exclude men[1]. 2 - 1 = 1.

p[woman] = 1/2.

Add a dimension:

Suppose there are four people: a blond man and woman, and a brunet man and woman. We ask the probability of a blond woman.

There are [4] people. Exclude men[2]. Exclude brunets[2]. 4 - 2 - 2 = 0. We excluded brunet men twice. Include that set.

|blond women| = |4 persons| - |men| - |brunets| + |brunet men| = 1.

p[blond woman] = 1/4.

Add a dimension:

Suppose two groups of four: one short the other tall. We ask the probability of a tall blond woman.

Eight[8] people. Exclude men[4]. Exclude brunets[4]. Exclude short[4] 8 - 4 - 4 - 4 = -4. We excluded those who match two categories twice. Include those sets.

Include brunet men[2]. Include short men[2]. Include short brunets[2]. 8 - 4 - 4 - 4 + 2 + 2 + 2 = 2. We included those who matched three categories twice. Exclude that set.

[tall blond women| = |8 persons| - |men| - |brunets| - |short| + |brunet men| + |short men| + |short brunets| - |short brunet men| = 1

p[tall blond woman] = 1/8.

Iteratively we exclude, include, exclude, include, ... until we get to the highest possible degree of overlap.

Exclude the deals with one match.

Include the deals with two matches.

Exclude the deals with three matches.

....

Include the deals with 52 matches.

Deal with no matches = 52! - ₅₂C₁ x 51! + ₅₂C₂ x 50! - ... + ₅₂C₅₂ 0!

Divide by all the deals

p[no match] = 1 - 1/1! + 1/2! - 1/3! + 1/4! - ... + 1/52! = 1/e.

Start with all the deals