PokerStrategist Mastermind Challenge - Q4

To be fair, I will post a new post with my edited answer.

One hint to all of you - read everything carefully.

I assume this means they play a Spin against each other?

If so, the answer is 5 tickets. If we give the player with the most tickets just 4 tickets or less, there will always be table setups where two or more players have the same number of tickets, where one is not necessarily higher than the other 2.

With 5 tickets, we can say that two players have one ticket and the others have 2, 3, 4 and 5 tickets respectively. That way no matter the table setup, someone will always have more tickets than the other two.

More detailed:

One more correction: the answer is 5.
There is only one player with more tickets than others at the table. The tickets are like this: 1,1,2,3 4, 5.
If there will be only maxium nubmer of tickets 4: then i can randomly sits players like this: 2,2,1, but this is incorrect. So the minimum is 5

Originally posted by AlpinaBG
One hint to all of you - read everything carefully.

Thanks for the hint!

Since someone came up with the right answer before me I'm making a new post.

If each player has at least 1 ticket and at least one player will have more than the other no matter how they are seated for their Spin then out of 6 players we can deduce the following.

In order to have 1 player with more tickets than the others no matter how they are seated in the 2 Spins then at least 4 players must have more than 1 ticket. So the remaining 4 must have at least 2 tickets to get a scenario of 1,1,2. In order to ensure that a player with 2 tickets does not share the most tickets earned 3 players must all have 3 tickets. In order to ensure that a player with 3 tickets never shares the most tickets earned 2 players must have at least 4 tickets. And to ensure that they never share top place one of them must have 5 tickets.

So the correct answer (Finally!) is 5 tickets.

Originally posted by HennieP
Since someone came up with the right answer before me I'm making a new post.

If each player has at least 1 ticket and at least one player will have more than the other no matter how they are seated for their Spin then out of 6 players we can deduce the following.

In order to have 1 player with more tickets than the others no matter how they are seated in the 2 Spins then at least 4 players must have more than 1 ticket. So the remaining 4 must have at least 2 tickets to get a scenario of 1,1,2. In order to ensure that a player with 2 tickets does not share the most tickets earned 3 players must all have 3 tickets. In order to ensure that a player with 3 tickets never shares the most tickets earned 2 players must have at least 4 tickets. And to ensure that they never share top place one of them must have 5 tickets.

So the correct answer (Finally!) is 5 tickets.

In case this was a trick question and you meant the smallest amount of tickets a player can have to be the player with the most tickets at his table then still following the reasoning above the answer would be 2 tickets.

Just hedging my bets here.

Interesting question. If you are considering the minimum tickets the players should have to valid the statement there will always be one player with more tickets than the other two, then it is 1,1,2,3,4,5. Considering this, the response to your question can be manifold in my opinion. 5 is the minimum tickets he will have for the above statement and still be the player with most tickets, 1 is the right answer for the fairness between players (disregarding the statement), or he can come with 2 tickets if you want the statement to be valid but he can chose to bet only the minimum tickets.

Let's assume the 6 players are Player A,B,C,D,E,F. The different combinations that can be exist while playing Spin n Go (AvsBvsC) (AvsBvsD) (AvsBvsE),(AvsBvsF)(BvsCvsD),(BvsDvsE),(BvsEvsF),(CvsDvsE),(CvsEvsF),(DvsEvsF).

The only way to fullfil the statement is Player A holds 1 ticket,B has 1, C has 2, D has 3, E has 4, F has 5 tickets and that way no matter who plays against each other. There will always be one player with more tickets than the other two player.

In my opinion, the question can be interpreted in different ways.

The question: Which is the smallest number of tickets that the player who wins the most tickets should have?

A(1)vsB(1)vC(2). A wins, A has 2 tickets, B has 0 ticket. C has 1 ticket. B is eliminated
A(2)vsC(1)vsD(3). A wins. A has 3 tickets. C has 0 ticket D has 2. C is eliminated.
A(3)vsD(2)vsE(4). A wins. A has 4 ticket. D has 1 ticket. E has 3 ticket.
A(4)vsF(5)vsE(3). A wins. A has 5 ticket. F has 4 ticket. E has 2 ticket.
A(5)vsD(1)vsE(2). A wins. A has 6 ticket. D has 0 ticket. E has 1 ticket. D is eliminated.
A(6)vsE(1)vsF(4). A wins. A has 7 tickets. E 0 ticket. F has 3 ticket. E is eliminated.
A(7)vsF(3)vs?. Spin n Go cannot be play properly due to limited amount of members.

The smallest number of tickets ( Player A) who wins the most ticket in the household is
7.

Originally posted by luckinsearch
9 tickets? If my answer is correct, I'll post the explanation.

We couldn't say if it's right or wrong before we reveal the results and count the points.
If you are the first one with the correct answer - you will get the 5 points, so no need to worry if somebody else c/p your explanation (because we look when the post is edit too).
But if you are first one with the correct answer and didn't explain it - sorry, but no points, because we will explain the results.

Unfortunately this time there are no right answer to the question.
The smallest number of tickets that the player who has won the most tickets should have are 15

We're looking for the smallest amount of tickets a player must have so that in any 3 players combination, 1 of them will always have more tickets than the other 2.

We are assuming that 2 players have 1 ticket and, from then on, we will analyze the worst case scenario (that in each spin participate both players with more tickets)

SCENARIO 1
Player 1 - 1 ticket
Player 2 - 1 ticket
so:
Player 3 - 3 tickets (1+1) +1

SCENARIO 2
P1 - 1 ticket
P3 - 3 tickets
so:
P4 - 5 tickets (3+1) +1

SCENARIO 3
P3 - 3 tickets
P4 - 5 tickets
so:
P5 - 9 tickets (5+3) +1

SCENARIO 4
P4 - 5 tickets
P5 - 9 tickets
so:
P6 - 15 tickets (9+5) +1

The smallest number a player must have so that there's always a Spin where one player has more tickets than the other two together is: 15

Semantics: "1 of them will always have more tickets than the other 2". Maybe you should've mentioned combined in that sentence, otherwise, 1.1.2 is as well correct...

exacly the statmenet is incorrect, it should be: one of them will always have more tickets than the other twos TOGHETER or more than a sum of tickets of two others players