Probability of 10 Cards

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lalakelly

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I'm having a lot of trouble with the following problem. I'm not even sure where to start. My teacher said it should use the multiplication rule P(A and B) = P(A) * P(A|B)

I have ten cards: a golden card and nine regular cards. I want to place them in a stack and have ten players draw from the stack. The one that draws the golden card wins. I want to rig the game so there is no longer a 10% chance each person wins. In other words, manipulate the shuffle in a way such that at least one card has a better/worse odds of being selected (no zero probabilities & the card’s position must be unknown to you).

–Find a method of ordering the cards that has at least two different odds (up to ten different odds) dependent on my position, and explain your method and calculate probability for each spot drawn.

He also mentioned that this would probably be a two step process. I'm very confused and just don't know where to go with this. Any help would be awesome.
 
It doesn't really matter that it's gold or regular, just that one is different from the other nine.

I thought the same thing but when I talked to my teacher today he didn't really address that. I think it just means that initially every player has a 10% chance.
 
lalakelly said:
I'm having a lot of trouble with the following problem. I'm not even sure where to start. My teacher said it should use the multiplication rule P(A and B) = P(A) * P(A|B)

I have ten cards: a golden card and nine regular cards. I want to place them in a stack and have ten players draw from the stack. The one that draws the golden card wins. I want to rig the game so there is no longer a 10% chance each person wins. In other words, manipulate the shuffle in a way such that at least one card has a better/worse odds of being selected (no zero probabilities & the card’s position must be unknown to you).

–Find a method of ordering the cards that has at least two different odds (up to ten different odds) dependent on my position, and explain your method and calculate probability for each spot drawn.

He also mentioned that this would probably be a two step process. I'm very confused and just don't know where to go with this. Any help would be awesome.
Click to expand...

Talking through the solution, look at the winner being in position number j. A fair shuffle would have a 10% chance of the golden card being in position number j but you want it to be x chance [as a decimal, not percentage]. Since the probability the card is not in the jth position is 1-x, we have, given the distribution of the other positions winning is uniform, the probability of someone other than the jth position winning is (1-x)/9.

So, if you have a computer program shuffle the cards, then you could
  • Define the function {t, f(t)}: {(1, 1), (2,2), (3, 3) ... (j-1, j-1), (j, j+1), (j+1, j+2), ..., (9, y); y=10 unless j =10 in which case y=9}. That is, take every position but j, 'count them', and assign the counting number to that position.
  • Shuffle the 9 regular cards in the normal fashion.
  • Choose a random number r between, say, 0 and 1.
    [*=1]If r<x then the golden card goes in the jth position.
    [*=1]Otherwise, randomly choose a number between 1 and 9. Use that as the counting number t for the f(t) position which gets the golden card.

If you can't know beforehand what the position is going to be, you could do the same process yourself. That is do a false shuffle of just the nine cards, etc.
 
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