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Roll the Die Higher (Posted on 2005-04-07) Difficulty: 3 of 5
Two players alternate throwing a six-sided die. The first player who fails to roll a higher number than the preceding roll loses. What is the probability that the first player wins?

What if the die is n-sided?

See The Solution Submitted by Charlie    
Rating: 3.8571 (7 votes)

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Solution w/o expl. and a question | Comment 2 of 13 |

for n=6 the solution is

1/6 + 5/6^2 + 5^2/6^3 + 5^3/6^4 + 5^4/6^5 + 5^5/6^6 = 31031/46656 = .6651020233

One can easily see the pattern that extends this to any n-sided die.

My question:  what value does this tend to as n increases?

The highest I can go is n=56 solution = .6354298564 which is not very close to anything I recognize.


  Posted by Jer on 2005-04-07 20:18:35
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