Forcedhand

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About Forcedhand

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  1. Forcedhand added a post in a topic   

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  3. Forcedhand added a post in a topic   


    The question should have stated that the player is skilled in the game and follows the best strategy. The answer shoud identify that strategy.
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  4. Forcedhand added a topic in New Logic/Math Puzzles   

    During my extensive search for new conumdrums I noticed a game often played in bars in South East Asia. There are 9 tiles numbered 1 to 9 on the back. Two dice are thrown and a tile is turned face up based on the following,

    CHoose a number from either die and turn over the corresponding tile.

    Turn over a tile that is the sum of the two numbers on the dice.

    E.G. if a 2 and a 4 is thrown then the player has a choice of turning over tile number 2, 4, or 6. However if a double 6 is thrown, the player can only turn over tile number 6. The player keeps throwing until all tiles are turned over and he wins the game and gets a free beer, or until he cannot turn over a tile because it has been turned over already, in which event he looses.

    WHat is the probibility of winning that free beer?
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  5. Forcedhand added a post in a topic   



    I think from a "duece" situation you are still left with 60/40. If we were to say that A had 100% chance of winning the next point, then you would say that they will have 100% chance of being the first to win two consecutive points. If (as with the coins) A has 50% chance of winning the next point then he will have 50% chance of being the first to win two consecutive points. Likewise if A has a 0% chance of winning the next point then he has 0% chance of being the first to win two consecutive points. Why does this logic not prevail for all other combinations of probabilities?
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  6. Forcedhand added a post in a topic   


    You're looking for the probability of each player winning two consecutive points. I go with 60/40 which is the same as each player winning a particular point.

    My logic is that we are only looking at the possibility of A winning two consecutive points or B winning two conecutive points. We can ignore all other combinations. If we were tossing a coin and the question was which side is most likley to turn up on two consecutive throws, you would have no problem saying they have equal an equal probability. Likewise here, they have a 60/40 probability.
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  7. Forcedhand added a post in a topic   

    Can we not look at it this way? THe question stated that "at any given point in the game A has a 60% chance of winning and B has a 40% chance". At duece, that is a given point in the game, hence A has a 60% chance an d heB has a 40 % chance of winning.
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  8. Forcedhand added a post in a topic   

    He could type it out in braille and give it to the blind guy to read.
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  9. Forcedhand added a post in a topic   


    It's not really semantics, substitute "encircle" for "go around". I did encircle the organ, but did I encircle the monkey?
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  10. Forcedhand added a topic in New Logic/Math Puzzles   

    A disc of 1 unit diameter is in contact and coplanar with a disc of 5 units diameter. If the smaller disc is rotated around the larger disc, how many rotations does it have to make before the same two points are in contact again?
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  11. Forcedhand added a topic in New Logic/Math Puzzles   

    I was walking in the city last week when I came across an old fashioned organ grinder. He had a monkey on top of the organ who would dance as the old man rotated the drum. The monkey stared at me as I stopped to look. I then slowly began to walk around the organ, all the time the monkey kept turning and staring at me. When I got back to the point I had started from it was clear that I had walked around the organ, but had I walked around the monkey?
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  12. Forcedhand added a post in a topic   



    This is the correct answer.
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  13. Forcedhand added a topic in New Logic/Math Puzzles   

    0,0,0,0,4,?,5,1,1
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  14. Forcedhand added a topic in New Logic/Math Puzzles   

    HIJKLMNO
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  15. Forcedhand added a post in a topic   

    My solution is

    Vs/Vc = Pi/4(Cos(x)+Sin(x)Tan(x))

    This may be the same as other solutions.
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