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# chess

Sat 30 Jan, 2016 01:19 pm
Two chess grandmasters, Andrew and Jacob, are having an epic chess showdown to determine which of them is the uncontested best player in the world!

The showdown consists of N games. In each game, one player plays as White and the other plays as Black. In the first game, Andrew plays as White. After each game, the player who loses it chooses which color they'll play as in the following game. However, the victor of the final game wins the entire showdown, regardless of the results of the previous games!

In each game, each player may decide to attempt to win or attempt to lose:

If both players play to win, then Andrew wins with probability Ww if he plays as White (and loses with probability 1 - Ww, as there are no draws at this high level of play). Similarly, he wins with probability Wb if he plays as Black.
If both players play to lose (achieved by tipping over their own king as quickly as possible), then Andrew loses with probability Lw if he plays as White, and loses with probability Lb if he plays as Black.
If exactly one player wants to win a game, then he's guaranteed to win it.
Assuming both players play optimally in an attempt to win the showdown, what is Andrew's probability of besting Jacob?
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Type: Discussion • Score: 2 • Views: 4,281 • Replies: 14
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maxdancona

1
Sat 30 Jan, 2016 01:27 pm
Quote:
If both players play to win, then Andrew wins with probability Ww if he plays as White (and loses with probability 1 - Ww, as there are no draws at this high level of play).

I don't think this is a realistic depiction of chess. There is a real possibility that with perfect play from both players every game will be a draw.

If you look at the play between top international grandmasters, I believe that more than half of the games end in draws (this is certainly true for the last few championships).

merry619

1
Sat 30 Jan, 2016 01:39 pm
@maxdancona,
so what do you think with what prob Andrew will win if the values of Ww, Wb, Lw, Lb are respectively 0.9 0.8
0.0 1.0 ?
maxdancona

1
Sat 30 Jan, 2016 01:42 pm
@merry619,
I believe that if we are talking about chess, that every game where they both want to win will result in a draw..

So the question is how often does either player decide that they want to lose? If you tell me how to decide this, I can give you a mathematical answer.

I see no reason that either player wouldn't want to win a game.
merry619

1
Sat 30 Jan, 2016 01:47 pm
@maxdancona,
because the wining a game depends on last round. so if there is 2 round then andrew will try to loose in first so that he can choose his move (white or black) in 2nd means last game.
maxdancona

1
Sat 30 Jan, 2016 01:52 pm
@merry619,
If every game that they both try to win ends in a draw. Then this strategy doesn't make sense.

The game theory square would be

Code:``` W L W D 1 L -1 ? ```

If you play me in this game, the best you can hope for is to draw every game (and unless you decide to lose the tournament, that is the best I can hope for as well).
maxdancona

1
Sat 30 Jan, 2016 01:56 pm
@merry619,
Let's figure out a proper strategy then... (although this now has nothing to do with chess and can be substituted with dice rolls).

merry619

1
Sat 30 Jan, 2016 01:56 pm
@maxdancona,
Can you help to solve this problem?

Uh oh, the weather forecast predicts that a rainstorm will soon break out over the infinite, 1-dimensional number line you call home!

The forecast is remarkably precise. In fact, it's known that exactly N raindrops will fall, with the ith drop striking the number line at position Xi exactly Ki seconds after the start of the storm. No two drops will strike the number line at exactly the same position and time.

You'd like to stop as many of the drops as possible, but you don't exactly have an umbrella... so your boomerang will have to do. Your plan is as follows:

Stand at some (possibly non-integral) position A on the number line, and choose some other position B (A ≠ B).
At some point in time, throw your boomerang from A to B. This point in time can be arbitrarily long before or after the start of the storm (the storm won't start for a while, so you have time to prepare). The boomerang will travel along your chosen line segment at a constant speed of S units per second.
Make your boomerang spin in place at position B for some non-negative amount of time (which can be arbitrarily large).
Have your boomerang travel back along the line segment from B to A at the same speed, and catch it as soon as it gets back to position A.
During the inclusive time interval from when you throw the boomerang to when you catch it again, if the boomerang is ever at exactly the same position as a raindrop at exactly the time that it strikes the number line, it will intercept it... just like an umbrella! What's the maximum number of raindrops you can intercept in this way with a single throw?

merry619

1
Sat 30 Jan, 2016 01:58 pm
@merry619,
ya, you are right this just a mathematical problem
0 Replies

maxdancona

1
Sat 30 Jan, 2016 02:02 pm
@maxdancona,
Easy, with the numbers you have given, when they both want to lose Black always suceeds losing and White never succeeds.

In games 1 through N-1 both players will always try to lose (so they can pick the color of game N).

If Andrew starts out as Black on game 1, he will lose every game until N-1. He will then choose to be White for Game N (which he will try to win). His chance of winning the whole shebang will be 0.9

If Andrew starts as White on game one, Jacob will lose every game until N-1 and will choose to be White, meaning that Andrew will play as black. In that case his chance of winning will be 0.8.

merry619

1
Sat 30 Jan, 2016 02:07 pm
@maxdancona,
so, how can we generalize this answer? Can we have a general formula?
suppose rounds N is very large like 1,000,000,000
maxdancona

2
Sat 30 Jan, 2016 02:11 pm
@merry619,
I gave you a generalized formula.

If N = 1,000,000,000 then N-1 = 999,999,999. This works for any positive integer.
merry619

1
Sat 30 Jan, 2016 02:14 pm
@maxdancona,
lets say N, Ww, Wb, Lw, Lb are respectively
1000000000
0.5 0.5
0.12345 0.56789
then what will be the prob that andrew win?
maxdancona

1
Sat 30 Jan, 2016 02:22 pm
@merry619,
Now we are getting into the binomial theorem.

You can read a pretty good explanation of that here...

https://en.wikipedia.org/wiki/Binomial_theorem
merry619

1
Sat 30 Jan, 2016 02:23 pm
@maxdancona,
i tried but not able to apply binomial on this. can you solve this
0 Replies

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