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There are 49 red cards and 1 white card in a deck. If the white card is picked the game is won. If a red card is picked then the game continues with one less red card, until the white card is picked. What is the probability that the white card will be picked on exactly the k-th turn?

Can anyone guide as to how I should approach this problem?

Stefan M.
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  • Yes, in the $k-1$ turns you took a red card. What is the probability of taking a red card in the first round? What about the second round? Multiply them. – Phicar May 12 '19 at 17:53
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    So is the probability for picking the white card at any k-th turn equal to 1/50 because that is what I find when I do such a multiplication? – Stefan M. May 12 '19 at 18:03
  • Not really, at the beginning of the process it is indeed 1/50, if we assume that we are in the second turn, then it means that in the first turn you picked up a red card and you take it out, so now you have $1/49$ – Phicar May 12 '19 at 18:42
  • So, then when we multiply 49/50 with 1/49 to get 1/50 right? That's what I meant with 1/50 being the probability of drawing the white card at k-th turn. – Stefan M. May 12 '19 at 18:51
  • Oh ok, i see. Then yes. – Phicar May 12 '19 at 18:54
  • similar problem: https://math.stackexchange.com/questions/2336591/three-white-and-one-red-ball-probability/2336690#2336690 – farruhota May 14 '19 at 12:18

1 Answers1

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If the cards have been randomly arranged, the white card is equally likely to be in any position of the deck. In particular, the probability that it is in the $k$th position is $1/50$.

N. F. Taussig
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