Card Game

[jasonHoldEm]

Given: <ul type="square">[*]One standard deck of 52 cards[*]One new car*[*]51 other people[*]You[/list]
Problem #1

Everyone lines up (the first person in line is player 1, second is player 2, etc). Player 1 selects a card from the deck of 52, if it's the ace of spades he wins the car. If player 1 is unsuccessful his card is discarded and player 2 will now select a card (from a deck of 51), this process will continue until someone pulls the ace of spades and wins the car.

Where do you want to stand in line to maximize your chance of winning the car?

Problem #2

Everyone lines up again. Player one goes to the deck and selects one card, if it is the ace of spades he wins the car, if not his card is replaced and the deck is reshuffled. Player two will now go to the deck and select TWO cards if either of them are the ace of spades he wins the car, if not his cards are replaced and the deck is reshuffled. This process continues where each player selects cards equal to their place in line (third person selects 3, fourth selects 4, etc).

Where do you want to stand in line to maximize your chance of winning the car?

(I'm not sure of the answer to the second problem. [img]/images/graemlins/crazy.gif[/img] )

Have fun, post your answers before you peek at anyone else's.

Peace,
J

* You won't actually win a car, this is just for fun.

[junkmail3]

1. Last
2. Last

Both chances you have a 1/1 chance of winning. It's the best probability.

Now I don't know if I would really want to, but you know. From a statistical standpoint, you have the best chance here.

[junkmail3]

Oh, and what kind of car. It might change my answer.

[GrannyMae]

just a guess:

1. all have an equal chance if the deck is shuffled properly, so i would guess the last position is best.

2. same as #1 for same reason

[BradleyT]

1. First position is the only one that is guaranteed to always have a chance to win the car.

2. Again first position is the only one guaranteed to have a chance however the ability to redraw in a later position probably increases your odds. The 52nd person would never win because the odds of the 51st, 50th, 49th, 48th, etc not picking the ace are astronomical.

No clue how to solve the 2nd but I'd assume you'd say something like:
Player 1 has 1/52 chance
Player 2 has 2/52 chance
If I'm player 3 I have 3/52 chance on my picks but to get my chance players one and 2 have to fail so my odds are X%.

As a guess I'd say position 4.

[TheRake]

#1 = 27th
#2 = Not sure, but seems like it should be 27 also.

Truthfully I am not sure of either of the guesses [img]/images/graemlins/smile.gif[/img]

TheRake

[jasonHoldEm]

Answers (?)
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In the first problem everyone has an equal chance of pulling the ace of spades (odds will be 51:1 for everyone because the cards are not replaced). I'm pretty sure it doesn't matter where you stand, although the argument to go first seems logical.

The second problem requires more math than I can handle, but here is my general logic. Odds for the first player are 51:1, the second player will have the combined odds of 51:1 and 50:1 but he will only get to pick 51 times in 52 (because the first player will have won the other time).

I'm sure there is a way to figure out your max EV by comparing your combined odds of pulling the ace of spades with the chances that someone in front of you will win first. I'm almost positive this problem will have a specific answer for a position in line, but we might need bruceZ to give us the answer. [img]/images/graemlins/blush.gif[/img]

J

[jek187]

1) All spots are equal:
1st person: 1/52
2nd person: 1/51*51/52
3rd person: 1/50*51/52*50/51
etc
These all come out to 1.92%

2nd:
7th
You have an 8.8% chance of winning in this spot, which is the best you can ask for. I just used a brute force method of computing all the chances w/a spreadsheet and am too lazy to type it out here.

[B00T]

I am no math major nor do I like typing out fractions but whatever...

For the first + second one, I want to be standing right before the chance = 50% of all the previous pullers chances combined.

First person has 1/52 chance, 2nd person has 1/51 chance, I wanna be standing before the person who first has the 50% chance in that plan. What spot it is I have no idea, but I want to go before the 50% threshold is assumed in all the previous people picking. I'm guessing I would want to be like the 23rd person on line.

For part 2, the chances of each person getting it increases after the first one. So once the total pulls = 26 taken,, so I'd say I want to be the 8th person in line or so.

Someone can do the math, but I think thats the right answer.

[daryn]

i disagree that all spots are equal in the first one. assuming there is only 1 car to give away, not all spots have an equal chance to win.

the first guy is 51:1 against winning, but the second guy is worse. of course if he gets to pick he is 51:1 also, but he won't even get to pick 1 out of 51 times.

you want to be first.