Conference rusure::math

    Number of cases where 1st player has >= 2nd player = 21 *win*
    Number of cases where 1st player has <  2nd player = 15 *lose*
                                                         --
    Total cases                                          36

    Therefore the probability of 1st player winning is 21/36 = 7/12 or
    0.5833...

	Mike

Mike,

Thanks for your help. There was a piece that I was not told about the problem. 
Player one can keep going as long as he beats player two's roll. In other
words, if player one rolled a "one," and then player two beats him with a
"two," player one can keep going if he rolls a "three," then player two beats
him with a "four," player one keeps going by rolling a "five," then player two
rolls a "six," at which point player one cannot win beacause six is the
highest.  Is is possible to figure out this probability?

Tim  

> Two players each roll one die.  If the second player's roll is greater
> than that of the the first player, the second player wins.  If it is equal
> or less, the first player wins.

> Player one can keep going as long as he beats player two's roll. 
> [...] then player two
> rolls a "six," at which point player one cannot win beacause six is the
> highest.

But doesn't player one 'beat' the six by rolling another six?

re .3
    
>                      -< Need to get the rules straight >-
    
    I think what the base noter means is as follows.
    
    The two players take turns rolling the dice. The loser is the player
    who first fails to raise the other player (by rolling a higher number).
    
    Expressed as an algorithm...
    
    last_roll := 0		{ avoid special casing first roll }
    last_player_number := 2
    
    do
    	t := roll of the dice
        if t <= last_roll then
    	    break
        endif
    	last_roll := t
    	last_player_number := 3 - last_player_number
    enddo
    win for Player last_player_number

    If the above is what is meant then
    
    prob(win for Player 1) = 31031/46656 = 0.665 (3dp)
    
    		      1
    N.B. this prob. = - sigma(n=1,6) (5/6)^n
    		      5

It seems that since the chance of either die being less than the other die
is equal, we need only compute the probability that the two dice are EQUAL.

There are only 6 such possibilities:

	1-1  2-2   3-3   4-4  5-5   6-6


That's 6 out of 36, or 1/6.


So, 5/6 of the time, one of the 2 dice is larger.  We've been told that
if die A is larger, it's a win, so 1/2 of this 5/6 is a win.  Add to that the
wins for the dice being equal.  So we have

	1/6    +    1/2 * 5/6


Is that the answer that was given ?

/Eric

Unfortunately, no answer was given.  I assume the answer will appear in the
next issue.

Tim

re .5
    
>	1/6    +    1/2 * 5/6

    This is equal to the answer in .1 but I don't think either takes into
    account the fact that player 1 can roll again if he is beaten and try
    to keep the game alive.

Possible games:

2 moves, won by the 1st player:
1,1	6,any
2,<=2	5,<=5
3,<=3	4,<=4
probability 21/36

3 moves, won by second player:
1,2,<=2
1,3,<=3   2,3,<=3
1,4,<=4   2,4,<=4   3,4,<=4
1,5,<=5   2,5,<=5   3,5,<=5   4,5,<=5
1,6,any   2,6,any   3,6,any   4,6,any   5,6,any

probability (5*6 + 4*5 + 3*4 + 2*3 + 2)/6^3
	= 35/108
 :
 :

6 moves, won by the 1st player:
1,2,3,4,5,<=5	1,2,4,5,6,any
1,2,3,4,6,any	1,3,4,5,6,any
1,2,3,5,6,any	2,3,4,5,6,any
probability 5*(1/6^5) + (5/6)*(1/6)^5
	= 35/7776

7 moves, won by the 2nd player: 1,2,3,4,5,6,any
probability 1/6^6 = 1/46656

It seems as if there should be a nice formula for the probability
that a game will last exactly n moves, but I can't find it.

    re .8
    
    Suggestion...consider the probability of a win by player 1
    partitioned according to his first throw.
    
    e.g. probability(player 1 wins and throws a 6 first up) = 1/6
         probability(player 1 wins and throws a 5 first up) = ?
         :
         :