Re: probability problem


mensanator@xxxxxxxxxxx wrote:
> Greenleaf@xxxxxxxxxxxxxx wrote:
> > How do you do a question of the type, 2 players are throwing dice,
>
> Does "dice" mean 2 six-sided dice?
>
> > taking turns, and the first to get a 6 wins the game.
> > What are the chances that the first person to throw the dice
> > will win,
>
> Slightly higher since the first player goes first (because this is
> random selection WITH replacement, i.e., the numbers on the
> dice don't change from player to player).
>
> >and of the second person?
>
> Lower. The chances could be equal if you do random selection
> WITHOUT replacement, say by placing 6 slips of paper in a hat
> and draw and discard until someone gets the 6.
>
> That's important if you find yourself in a Russian Roulette
> tournament. If they spin the chamber before EACH trigger pull,
> you will want to go last. If the chamber is spun only once, then
> it doesn't matter.
>
> >
> > I get an answer of 7/12 for the first person,
>
> That's why I asked if there was one or 2 dice. Either way, I don't
> know how you came up with 7/12. The odds of 2 dice showing 6
> is 5/36.
>
> > but an answer of 5/12 + (1/12) * (5/6)^n for the second person.
>
> But the (5/6)^n makes me think perhaps you're using just one die.
>
> >It makes
> > sense if n tends to infinity but if not (and in a real game it wouldn't)
> > then how would you interpret the result, since then the chances
> > that either of the two players will win is greater than 1.
>
> A clue!
>
> > I think there was a mistake or something.
>
> Yep, for 2 dice the odds of a 6 showing on the first throw are 5/36.
>
> Same odds for the second throw. BUT...
>
> ...there is only a second throw IF the first throw is NOT a 6. And
> the odds of that are 31/36. So to win on the second throw you must
> roll a 6 (5/36) AND the previous throw must NOT be a 6 (31/36).
>
> Same for throws 3, 4, 5, 6, etc. To win on throw N, the odds are
>
> (5/36) * (31/36)^N

Starting with N=0.

>
> you must roll a 6 and ALL previous throws must NOT be 6.
>
> Now to figure up what A's total chances are, sum to infinity all the
> odd numbered throws. And for B, sum the even throws.
>
> These sums will not exceed 1. In fact, the two added together will
> equal 1.
>
>
> >
> > Thanks in advance.

.

Relevant Pages

  • No result is likely - Part IV (conclusion)
    ... "When do the dice know that it's time to give the player the win so ... To sum up: ... to be able to calculate the probabilities of different outcomes, ...
    (rec.gambling.craps)
  • [Review] Stack Market
    ... That being said, I'm a bit disappointed by the actual game play, ... on the dice to represent windows in the towers; ... and a clumsy player will be at a loss, no matter what else they do. ... their investment to any open space in any of the companies. ...
    (rec.games.board)
  • [Review] Wildside!
    ... Maybe the game felt ... with each player given a "Dice Pad" in front of them, ... six-sided dice (showing a star, moon, raindrop, lightning bolt, and two ... "wild" sides). ...
    (rec.games.board)
  • Re: probability problem
    ... Does "dice" mean 2 six-sided dice? ... Slightly higher since the first player goes first (because this is ... The chances could be equal if you do random selection ... And for B, sum the even throws. ...
    (sci.math)
  • Re: probability problem
    ... >> How do you do a question of the type, 2 players are throwing dice, ... > dice don't change from player to player). ... The chances could be equal if you do random selection ... And for B, sum the even throws. ...
    (sci.math)