1.

2.

The other known solutions are shown below:

22377222

224900277^{2} + 1036492627^{2} = 1124897100420037858

224900277^{3} + 1036492627^{3} = 1124897100420037858898449816

223772220^{2} + 1036200450^{2} =1123785379023930900

223772220^{3} + 1036200450^{3} =1123785379023930900736173000

42168172711^{2} + 108702731146^{2} = 13594438548384283562837

42168172711^{3} + 108702731146^{3} = 1359443854838428356283751827361567

70487560695^{2} + 111743231377^{2} = 17455045971305066199154

70487560695^{3} + 111743231377^{3} = 1745504597130506619915466723682008

118280864170^{2} + 1012043721225^{2} = 1038222856499747505289525

118280864170^{3} + 1012043721225^{3} = 1038222856499747505289525051686978625

920726348310^{2} + 1059829825078^{2} = 1970976266597131517562184

920726348310^{3} + 1059829825078^{3} = 1970976266597131517562184310094565552

3.

The answer is n(n–1)/2. This follows from the result that the expected number of rounds left at any stage is the sum of the pairwise products of the fortunes.

For example, if there are 3 players left with $A, $B, and $C, then there are 6 equally likely outcomes of a player losing $1 to another, and:

6(AB+AC+BC – 1) | = (A+1)(B–1)+(A+1)(C)+(B–1)(C) |

+ (A+1)(B)+(A+1)(C–1)+(B)(C–1) | |

+ (A)(B+1)+(A)(C–1)+(B+1)(C–1) | |

+ (A)(B–1)+(A)(C+1)+(B–1)(C+1) | |

+ (A–1)(B+1)+(A–1)(C)+(B+1)(C) | |

+ (A–1)(B)+(A–1)(C+1)+(B)(C+1) |

4.

5.

The tournament is as follows: A plays B a best-of-3, C plays D a best-of-3, and then the winners play. If at least one of the earlier best-of-3 matches takes only 2 games for a player to advance, then the winners play a best-of-3, other wise, they play a single game to decide the winner.

The probability that a best-of-3 match only lasts 2 games is (2/3)(2/3)+(1/3)(1/3) = 5/9. And the probability that the winner in this case is the better player is (4/9)/(5/9) = 4/5. This leaves 4/9 for the probability that a best-of-3 match lasts all 3 games, and in this case the probability the winner is the better player is 2/3. The chance that the better player wins a best-of-3 tournament in any way is (2/3)^{3} + 3(2/3)^{2}(1/3) = 20/27.

So the probability that the best players wins the tournament described is (5/9)(5/9)(4/5)(20/27) + (5/9)(4/9)(4/5)(20/27) + (4/9)(5/9)(2/3)(20/27) + (4/9)(4/9)(2/3)(2/3) = 3536/6561 = 53.9%.