Probability of winning a card game against yourself?

A

AndySavage

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I have no idea how difficult this problem is, but I couldn't figure it out. I have played this game with some friends and it seems like the probability of winning is pretty low.

The reason I'm here is because I still have not managed to win this game, and my friend beat the game on his first attempt. He is no longer my friend. I bet a lot of money against him that the probability of winning is less than 1%, but I have not been able to calculate the odds. Can anyone out there be of assistance?

This is how the game works:

-One person.
-One deck of cards face down.

The player turns one card over at a time. For each card that is turned over the player says one, two, three, one, two, three, one, two, three...etc
If the card matches what the player is saying out loud he will lose the game.

To elaborate:

The first card can not be an ace
the second card can not be a deuce
The third card can not be a three
the fourth card can not be an ace
the fifth card can not be a deuce
the sixth card can not be a three
the seventh card can not be an ace
the eighth card can not be a deuce
the ninth card can not be a three

And so on...

What is the probability that the player turns over all the 52 cards and wins the game?
 
Maybe someone else here can come up with a nice formula for calculating the odds, but the only way I know of to solve it is just to consider every possible outcome. Since the only draws that can make you lose are 1, 2, and 3, you can make it a lot easier on yourself by saying that there are only four types of cards in the deck (i.e. there are 4 aces, 4 twos, 4 threes, and 40 other cards)
 
Thanks, but I run into problems really fast when I have to consider the probability of A,2 and 3s appearing in the cards I allready have drawn without losing the game.
 
Unfortunately, there's really not much that can be done about that problem. You can alleviate someof the strain of remembering by making an indented list, like so:


  • Draw an ace (4/52) - Lose
  • Draw a two (4/52)
    • Draw an ace (4/51)
      • And so on...
    • Draw a two (3/51) - Lose
    • Draw a three (4/51)
    • Draw other (40/51)
  • Draw a three (4/52)
  • Draw other (40/52)

Such a list has the benefit of being able to trace the path back to see how many of each card you've already drawn. Plus, you can just look at the list and immediately see that if you start with an ace, you lose. That means that 1/13 or ~7.69% of the games are losers. And then if you draw 2 twos in a row, that's also a loser. That means (4/52)*(3/51) or ~0.45% of the games are also losers.
 
AndySavage said:
I have no idea how difficult this problem is, but I couldn't figure it out. I have played this game with some friends and it seems like the probability of winning is pretty low.

The reason I'm here is because I still have not managed to win this game, and my friend beat the game on his first attempt. He is no longer my friend. I bet a lot of money against him that the probability of winning is less than 1%, but I have not been able to calculate the odds. Can anyone out there be of assistance?

This is how the game works:

-One person.
-One deck of cards face down.

The player turns one card over at a time. For each card that is turned over the player says one, two, three, one, two, three, one, two, three...etc
If the card matches what the player is saying out loud he will lose the game.

To elaborate:

The first card can not be an ace
the second card can not be a deuce
The third card can not be a three
the fourth card can not be an ace
the fifth card can not be a deuce
the sixth card can not be a three
the seventh card can not be an ace
the eighth card can not be a deuce
the ninth card can not be a three

And so on...

What is the probability that the player turns over all the 52 cards and wins the game?
Click to expand...

One way to get an approximation is to write a computer program. Doing this and making 10000 deals, you lose about 99% of the time [9928, 9921, 9927 out of 10000]. EDIT:Correct dumb misteak in program. Off to the corner for 9928 deciseconds.
 
Last edited:
AndySavage said:
This is how the game works:
-One person.
-One deck of cards face down.
The player turns one card over at a time. For each card that is turned over the player says one, two, three, one, two, three, one, two, three...etc If the card matches what the player is saying out loud he will lose the game.
Click to expand...
Conceptually there is a straightforward solution, but the computation is a nightmare. I agree that a computer is a good aid in the solution.

The solution evolves the inclusion/exclusion rule. If we layout the fifty-two cards in a randomly ordered string in fifty-two numbered positions, that is equivalent to any such game.
Let \(\displaystyle \mathcal{A}\) be the event where the game is lost with a ace, \(\displaystyle \mathcal{B}\) be the event where the game is lost with a two, &
\(\displaystyle \mathcal{C}\) be the event where the game is lost with a three.
Event \(\displaystyle \mathcal{A}\) can happen on \(\displaystyle 1,4,7\cdots49,52\) draw, eighteen in all.
Event \(\displaystyle \mathcal{B}\) can happen on \(\displaystyle 2,5,8\cdots47,50\) draw, seventeen in all.
Event \(\displaystyle \mathcal{C}\) can happen on \(\displaystyle 3,6,9\cdots48,51\) draw, seventeen in all.

The horror comes in putting the various combinations of those three events in order to use inclusion/exclusion.

You need to calculate Event \(\displaystyle P[\mathcal{(A\cup B\cup C)^c}]\) or \(\displaystyle 1-P[\mathcal{(A^c\cap B^c\cap C^c})]\)
 
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Denis said:
Agree, probability is less than 1%; it is close to .0092.

Ran 2 simulations of 1 million each: you won(!) 9171 and 9152 times.

[...]

As far as a "fully calculated probability" goes, I'll bravely state
that you have a probability of 4/52 of losing on 1st card; I happen
to be too busy(!) to calculate in full, and will leave that for
our resident expert Sir Pka to work that out (at no charge) for you.
Click to expand...

Wow! thank you for your help!
 
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