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Outline:

  • 2 players
  • Consecutive draws without putting anything back
  • 5 total marbles
  • 2 white, 3 black

Given solution:

-In the first game the white marble wins and for the person starting to draw the probability of winning is 3/5 -In another game the black marble wins and the probability of winning of the person starting to draw is 7/10

My thoughts on first game:

The person starting to draw has a 2/5 chance of winning the game in the first round. If the first player draws a black marble the game continues. In the second round the second player has a 2/4 chance of winning. The second player also draw a black marble (the same chance for the first player that the game will continue). In the third round the first player has a 2/3 chance of winning. If the player draws a black marble the game continues. In the 4th round the second player is left with only white marbles, so the chance of winning is 2/2.

For the first player I calculated:

2/5 * 2/4 * 2/3 = 2/15 (compare to given solution 3/5)

For the second game I calculated:

3/5 * 3/4 * 3/3 = 9/20 (compare to given solution 7/10)

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1 Answer 1

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For the first game you should sum up the probabilities when you read "or" and mulitply them (as you did) when you read "and". So, the person starting the game can win, if

  1. He draws a white marble in the first draw, with probability $$2\over 5$$ or
  2. If he does not draw a white marble and the second person does not draw a white marble and he draws a white marble in his second effort, with (total) probability $$\frac35\cdot\frac24\cdot\frac23$$

and there are no other possible ways, since otherwise the second person wins for sure in his subsequent draw. Summing up the probabilities of cases 1. and 2. we find that the probability that the person starting the game wins, is equal to $$\frac25+\frac35\cdot\frac24\cdot\frac23=\frac25+\frac15=\frac35$$

Analogous argumentation applies in the second game (where black marble wins)

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  • $\begingroup$ Now I get it. Thanks allot! $\endgroup$ Commented Mar 20, 2014 at 14:02
  • $\begingroup$ @Spießbürger you are welcome! $\endgroup$ Commented Mar 20, 2014 at 14:06

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