Ok, assuming I have 2 cards, a card that's black on both sides, and a card that is black on one side and white on the other. I get the double black, and you get the black white. We place both cards in a bag, and draw one, with the black facing up. If a black is not facing up, it doesn't count cos the suspense is lost. So the unknown black card is placed on the table, and then opened. If it is your card, you get a point. If it is mine, I get a point. First to 10 points win. So, there are 2 possibilities each round. The bottom is either black or white. Therefore chance is 50-50.
Seems like a fair game? It's not. Think about it. There are 4 possible ways to draw the card, 3 blacks and 1 white for both cards. I discard the white option, so essentially, only 3 ways are accepted. 2 of which is my card and 1 of which is yours. So, the actual chance of winning is 66.67-33.33. That's not the main thing. The main question is now this: what is the chance that you win the game? The number of draws is not fixed, since if I'm really lucky, 10 draws are enough to win the game. If you are lucky enough to break even, 19 draws are required to end the game. The number of draws depends on both my outcome and your outcome. Thus, it is no geometric distribution right? The chance of winning each round is 1/3 for you, but each game there's a different number of rounds. So how do you calculate? The answer is below, but don't spoiler yourself till you're really stuck. By the way, if you are wondering, the chance of you winning is about 6.48%. see if you get that.
It is disturbingly simple. Regardless of how many times you try, in 19 rounds, only one will win. That is the person who gets 10 points or more. So, the entire model is essentially a sum of 10 binomial distributions: binomcdf(19,1/3, 10,19) or the chance that you get 10 points in 19 tries, 11 points in 19 tries.. then sum the lot. Now I really feel like an idiot.
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