Hypothetical question

In the course of creating my EwN bot, I've come up with interesting question. Let's assume, that two equally skilled players face each other, BUT one of them always knows what will be the outcome of the next dice roll. What is your estimation, regarding impact of such knowledge on the overall winrate between those players?

Will depends on the number of points you need to win.

In a 50 points game, I guess if you play thousand games, the bot who knows gonna always win.

it's a BIG handicap. I think it will give a tremendous advantage to the players who knows.

I'm not sure it gives a huge advantage, as there are bad rolls and good ones, I prefer to have good ones instead of knowing.

Say one player can chose his next dice roll, and the other just knows what it will be in advance, like if you say "I will get a one" in plain voice, and have it, the other player knows but rolls his dice. Who will win?

If you have done simulations with your bot I'm interested in the results.

Sometimes you made a very risky move and you wait for getting lucky. IF you know that u are not gonna be lucky, u prevent to play aggressive.

What I meant was win in a single game.

In a single game I don't expect a "huge" advantage.

 impact of such knowledge on the overall winrate between those players?

you need many games to estimate strength difference ?

Many games, but many "one point" games, it is what he means cause I asked for it.

Florian, probability to win n-game match game can be implied from probability of winning a 1-game match, therefore in my opinion question regarding match length is irrelevant.

Precisely, this is exactly the reason of the question :D

I bet it's at least 60%, but not as high as 90%.

Guesses                                                     both players same alternate start                   50%

Advantage only if pieces are about to cross maybe 2 in 20 moves                        x 0.1

other player has maybe 2 good choices maybe 90% of time (10% limited)         x 0.1

so maybe 50.5% - 49.5%

I've just completed a little java program to get some results in this area. It's not fully verified, but I think I'm going to take a break to work on other projects, which means I may never get back to it, so I figure I may as well post what I have.

Working from the position which results if the first 3 moves are captures (i.e. only the front-line pieces are left), I found the following:

P1, normal conditions: 59.2531235%
P1, can see future: 67.5281014%
P1, opponent can see future: 52.9756135%

In other words, the probability of winning is:

(67.5281014 + (100-52.9756135))/2 = 57.27624395%

So, not as large as my guess of 60%. I don't know whether my simplifaction would magnify or decrease the advantage, but I would now guess that the correct answer within a factor of 2, i.e. between 54% and 64%.

@Bill, nice!

Could you explain your formula?
If only P1 has this skill, I would have gone with (P future-skill + (1 - P normal))/2 or (67.53+(100-59.25))/2=54.14

"one of them always knows what will be the outcome of the next dice roll." What is exactly meant by this? Does he know the next dice roll of his opponent? Or of his own next dice roll?
Anyway this should give him the win in games with more points.

@Carroll:

I was thinking that the player with the skill will win 67.528% of the games going first and 47.024% of the games going second, or 57.276% of all games.

@Paul: I took it to mean "When I make my move, I know what number my opponent is about to roll". If I also knew what number I am going to roll next turn, things get weird (I could bluff, for example).

@Bill, yes that is right :) because You are then the opponent with skill when you go second.
This game should be fun, you roll your dice and your opponents dice.
The variant where both players roll two dice in a row should be fun too.