# Conditional probability?

Let's consider this hypothetical game, in which there are 10 cards, and one of the cards is marked as a winning card. We put the cards in a stack and pass them around in a circle, each person choosing one card at random and passing the remaining stack on, until none are left.

What is the probability that I pick the winning card? Initially, for the first person choosing, the probability is 1/10. Does the probability increase with successive turns, because the chances that the winning card is picked is smaller at the beginning? Or is the probability the same 1/10 throughout the whole game? Is it more beneficial to go later than to go first? I tried to think about this through an expected value lens and got all confused.

• A bit confusing in the sense that it depends on how many friends are participating. For example if there are five friends, then each has 2 cards. Also, it depends on whether or not the cards are face up or face down. If the cards are face up, the odds must account for the additional information of not getting the winning card at each step. If the cards are hidden, for five friends, and are not looked at until all the cards are dealt, then the odds are one in five for each player. So, in effect, the conditional probability depends on the conditions of the game. – Carl Aug 23 at 7:30

The probability that the first player wins is $$1/10$$.

The probability that the second player wins is the probability to have the winning card still in the stack (i.e. the probability that the first player did not win - $$1-1/10=9/10$$) times the probability that the second player will draw the winning card from the stack of 9 cards (i.e. $$1/9$$). Multiply those and you see that you end up with $$9/10 \cdot 1/9 = 1/10$$, same as the first player.

Similarly, the third player has a probability of $$9/10 \cdot 8/9 = 8/10$$ to get a stack with the winning card in it, but if they get it, they have a probability of $$1/8$$ to draw the winning card; again, it multiplies to $$1/10$$.

This holds on all the way until the 10th player (of course, the 11th player has no probability to draw the winning card).

For your intuition, consider this: The later players have a lower probability to have the winning card still in the stack (because more players drew from it), but if it's there - they have a higher probability to draw it (because the stack is smaller).

For another bit of intuition, consider that this process is no different than assigning cards at random... would you expect any other probability than $$1/10$$?

The probability is always 1/10 unless the players look at the card and then you know that it's not a winning card (or is). This is not really a conditional probability problem because you don't condition on anything and no new information is made available until all 10 player, presumably reveal their cards after everyone has drawn.