bonanova

Just as a quick response, Jester will survive, and beat you, if you don't eliminate him using Peasant. And when, other than Round 1 can you ensure the two will play each other? You used Jester to eliminate King in R1, but that leaves Jester and Duke probably still alive. I put Duke against King, and Peasant against Jester, to eliminate 2 of the 3 dangers [Jester, King, Duke] in R1, and achieved a better than 50% survival rate into the finals. Thanks... I'll go back and give Jester some insane strength number and redo some calculations. This is intriguing!

The function f(x,y) = x^y has a discontinuity at the origin: the limit along x=0 is 0, and the limit along y=0 is 1. There is some benefit if we define the value to be 1: for the binomial theorem valid everywhere, x^0 must = 1 for all x. There is no comparable benefit for making the function 0^x well behaved at x=0, so much of modern thought favors defining the value as 1. But others argue it should remain indeterminate, because x^y is in fact discontinuous at the origin.

And here's another.

Thanks for clarifying what you're having difficulty with. Your die analysis is correct, and in fact nicely illustrates the matter. You state that p[even] is 1/2. How do you arrive at that result? Simply by noting that p[2] = 1/6, p[4] = 1/6, and p[6] = 1/6; and these events are disjoint. So, paraphrasing my previous comment, with the disjoint property stated for clarity, The three events are rolling a 2, rolling a 4 and rolling a 6, and their individual probabilities are all 1/6. p[even] = p[2 or 4 or 6] = p[2] + p[4] + p[6] = 3p[2] = 1/2. When you combine p[even] with p[<=3] i.e. apples and oranges, it's a different matter, as you point out. Hope that helps.

UR, I see you pairing the King against Jester with Jester winning. Here's what you say about Jester: All you say explicitly is that [a] Jester beats me. Peasant beats Jester. Where do you get that Jester beats King? "only weakness" doesn't seem that specific. You might have said Jester beats everybody - except Peasant - if that's what he does. Can you clarify? I can do a new calculation with Jester against King then. Thanks for a nice puzzle...!

This puzzle is really a nice challenge. Kudos!! Jester's strength against anyone other than Peasant and Me is not given, making it mandatory to match them in the first round. If more info is given about Jester, that could change my best result. For now, it's 0.191 That's slightly better than the 7/54 = 0.1296 result. My analysis - in four steps ... . | L O S E R | +==========+=====+======+======+======+======+======+======+======+======+ | Winner | Str | bona | Jest | King | Duke | Queen| Peas | Earl | Prnce| +==========+=====+======+======+======+======+======+======+======+======+ | bonanova 100 | x | 0 | .250 | .333 | .500 | .500 | .667 |1.000 | | Jester - |1.000 | x | | | | 0 | | | | King 300 | .750 | | x | .600 | .750 | .750 | .857 |1.000 | | Duke 200 | .667 | | .400 | x | .667 | .667 | .800 |1.000 | | Queen 100 | .500 | | .250 | .333 | x | .500 | .667 |1.000 | | Peasant 100 | .500 |1.000 | .250 | .333 | .500 | x | .667 |1.000 | | Earl 50 | .333 | | .143 | .200 | .333 | .333 | x |1.000 | | Prince 0 | .000 | | .000 | .000 | .000 | .000 | .000 | x | +==========+=====+======+======+======+======+======+======+======+======+. +======+======+======+======+======+======+======+======+ | |---A---+ | +---A---+ |---B---+ | | +---A---+ |---C---+ | | | +---U---+ | |---D---+ | | | | +--A-- | | |---E---+ | | +---V---+ | |---F---+ | | | +---X---+ |---G---+ | | +---W---+ |---H---+| Round 1 Round 2 Round 3 Compute probabilities for all entrants to reach the next three levels: Each entrant has a certain probability of winning Round 1 and reaching Round 2. Get those numbers directly from the table. Each entrant has a certain probability of winning Round 2 and reaching Round 3. Get those numbers from the Round 2 probabilities for all potential opponents and then from table and adding them up. Each entrant has a certain probability of winning Round 3 and being champion. Get those numbers from the Round 3 probabilities for all potential opponents and then from table and adding them up. e.g., Probabilities for A getting to Round 2: P[A2] = Ab Round 3: P[A3] = Ab.Au = P[A2].[P[C2].Ac + P[D2].Ad] Round 4: P[A4] = Ab.Au.Ax = P[A3].[P[E3].Ae + P[F3].Af + P[G3].Ag + P[H3].Ah] 100 1.000 .500 .172 - bonanova 0 .000 .000 .000 - Prince 100 1.000 .500 .172 - Peasant - same as bonanova 0 .000 .000 .000 - Jester 300 .600 .471 .354 - King - better than bonanova 200 .400 .284 .190 - Duke - better than bonanova 100 .667 .189 .094 - Queen 50 .333 .055 .018 - EarlStrength Round 2 Round 3 Champion Peasant/Jester in lower bracket: better because Earl might knock off Queen, getting me into semifinals with .556 prob instead of .500 above. 100 1.000 .556 .191 - bonanova 0.191 seems to be the best winning chance. Only the King's is better. 0 .000 .000 .000 - Prince 100 .667 .333 .114 - Queen 50 .333 .111 .024 - Earl 300 .600 .450 .343 - King <- better than .191 200 .400 .267 .182 - Duke 100 1.000 .283 .147 - Peasant 0 .000 .000 .000 - JesterStrength Round 2 Round 3 Champion

That was the warm-up. Now for the Grand prize. [OP edited:] How many rectangles are there on a chessboard?

King Queen Prince Princess = 4 Duke, Earl, Jester, Peasant = 4 Then there's you. That's 9 total. But there are only 8 slots. Help.

Warmup question: How many squares are there on a chessboard? Edited: Grand prize: How many rectangles are there on a chessboard?

Prove or disprove: If p is a prime number > 3, then p2-1 is divisible by 24.

A B D O P Q ?

There's an old game children play that goes like this. 1. Two players take turns saying a number. 2. The first number called must be 10 or smaller. 3. Each number is greater than the previous number, but by no more than 10. 4. The player who first calls 50 wins. There is a coin toss to see who goes first. You win the toss. What's your strategy?

The solution is not ambiguous if I include the information that Al and Bob each made four statements. My bad ...

In the Land of Knights and Knaves [LKK], Knights always tell the truth, and Knaves always lie. Last night the LKK Community Center Committee threw a party for some of the children. Some adults were there, also, as chaperones. The six-person LKKCCC was talking about it afterwards: Alice: I counted all the Men and Children that came. There were 9 total. Barbara: I counted two more Women than Children. Clarice: I was playing Cupid and decided there were exactly 24 possible Man-Woman couples here tonight. Daniel: Of the groups of Men, Women and Children, I counted 4 in one group. Elliot: Of the groups of Men, Women and Children, I counted 6 in one group. Fred: Of the groups of Men, Women and Children, I counted 8 in one group. I guess you can tell that the Committee members weren't all Knights; ... but be kind, and give the benefit of the doubt to as many as possible. How many men / women / children attended the party? How many of the committee members are Knaves? Who are they? Please be considerate and conceal your answer in a spoiler.

A vine has grown from the ground to a height of 15 feet on a tree in my back yard. I counted four times that the vine encircled the tree as it grew. I measured the circumference of the tree to be 2 feet. How long is the vine?

There are 8 statements to work with. Did you use all of them?

Play the Prince twice - either of the two-game matches.