Repeated roll of a die - probability of success

#1
I repeatedly roll a 6 sided die. I win when the number 1 comes up for the ninth time. Game over. My opponent wins when the number 2 comes up for the tenth time. Game over. What is the probability that I will win? How is the probability calculated?
 

katxt

Active Member
#2
Is this a real life question?
It's probably too hard to do analytically (for me anyway). A Monte Carlo simulation gives just under 60%.
 
#3
Thanks katxt. You might say the question is 'real life.' I am trying to put together a six horse race game with the horses having different numbers of rectangles they must travel through on their way to the finish line. The horse with 9 (slightly longer) rectangles (horse #1) should win more often than the horse with 10 rectangles (horse #2) which would give horse #1 better initial odds than horse #2. Your simulation confirmed this! Of course, the final payout will depend on the bets placed.

I could not come up with an analytical solution and I was hoping someone else could point me in the right direction. I will probably end up using a Monte Carlo simulation as you suggested.
 

katxt

Active Member
#4
I played a similar when I was a student. A track with 5 horses. Two tracks with 3 steps, 1 each with 4 5 and 6. I forget what odds were offered, but I think they were $2.50, $2.50, $5, $10 and $20. Throw a dice and move the appropriate horse up. A fast and exciting game with a good edge for the bookies.
 
#5
Im thinking that if you conceptualize a play as rolling dice until 1 or 2 (ie equal chance of winning a play is 50%) then you winning by the yth play should be negbin(.5,9), as i recall? Opponent wins only if y > 19. Does that sound like productive line of argument? I haven't actually run the numbers on that so might be way off the mark here.

Rcode looking promising agreement with monte carlo above:
1 - pnbinom(9, 9, .5, lower.tail=T)

??
 
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Dason

Ambassador to the humans
#6
Im thinking that if you conceptualize a play as rolling dice until 1 or 2 (ie equal chance of winning a play is 50%) then you winning by the yth play should be negbin(.5,9), as i recall? Opponent wins only if y > 19. Does that sound like productive line of argument? I haven't actually run the numbers on that so might be way off the mark here.

Rcode looking promising agreement with monte carlo above:
1 - pnbinom(9, 9, .5, lower.tail=T)

??
That's exactly where my mind went too. We can ignore the non relevant stuff and since each turn gives one person a point each player has a round they must win by or else the other person has already won.