Brain Teasers
Chess
Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.Probability
X was starring in the Super Chess Challenge. He played 8 people at once! What are the odds of X being victorious every game? It is assumed that everyone is on the same playing level.
Answer
.0000595%In a chess game, one can win, lose, draw, stalemate, resign, or opponent resigns. That is a total of 6 possibilities for one game! Odds of winning one game are 1 in 6. Odds of winning all 8 games are (1/6)^8. So,
(1/6)^8=.000000595, or .0000595% !
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Comments
I'm afraid this is a poorly defined problem with no good answer.
There are certainly not six different outcomes.
Resign is a subcase of lose, opponent resign is a subcase of win,
and stalemate is a subcase of tie. There are only
three relevant outcomes of a game of chess: win, lose, or tie.
(Actually, there are a few really oddball cases like a double-forfeit,
but we should ignore those for a problem like this).
If everyone is on the same playing level, then the probability of
a win should be equal to the probability of a loss, but we
know nothing about the probability of a tie relative to these.
Even this is not fully accurate, because the probability of
winning with white may well be different from the probability of
winning with black.
There are certainly not six different outcomes.
Resign is a subcase of lose, opponent resign is a subcase of win,
and stalemate is a subcase of tie. There are only
three relevant outcomes of a game of chess: win, lose, or tie.
(Actually, there are a few really oddball cases like a double-forfeit,
but we should ignore those for a problem like this).
If everyone is on the same playing level, then the probability of
a win should be equal to the probability of a loss, but we
know nothing about the probability of a tie relative to these.
Even this is not fully accurate, because the probability of
winning with white may well be different from the probability of
winning with black.
Also, the probabilities of the six outcomes, if they were prevalent, would definitely not be 16.6666% each.
is there any kind of draw other than a stalemate?
do u know how to play chess?
i like chess!
Oct 07, 2002
News flash! If your opponent resigns, you don't really actually win - the game just really actually ends that way. I'd say the next time CheeseEater means to say "Although this teaser is ligitimate" there is a 67% chance that he will again say "Though this teaser has ligitimacy" and a 29% chance that he will say "Yo, teaser, legit". Simply disricpiptcable, or whatever.
I would suggest that your chances of winning are much less if your opponent is Bobby Fisher, Gary Kasparov, etc.
Not if you are Kasparov, Fischer, etc. It did say "same playing level". Probability is a mathematical construct expressing the ratio of equally likely events to one another. Whenever the outcomes are not equally likely, Mathematical probability has nothing to say. Even if the players are equally skilled, it has more to do with the nature and rules of chess whether a stalemate or a win are equally likely. If you know anything about real chess competition, you will know that they are not equally likely. How many matches between Kasparov, Fischer etc. end in a stalemate? Not many!
This question is nonsense, if all the players played at a very high standard, more draws could be expected.
Whereas if it were a low standard, less draws would be expected.
Basically there are 3 options, win, lose, or draw.
We can say the odds of winning and losing are the same.
Letting D(S) be the probability of drawing a game where both players have the skill S, then we see that the chance of winning one game would be (1-D(S))/2, so the chance of winning 8 games in a row would be (1-D(S))^8/256.
Whereas if it were a low standard, less draws would be expected.
Basically there are 3 options, win, lose, or draw.
We can say the odds of winning and losing are the same.
Letting D(S) be the probability of drawing a game where both players have the skill S, then we see that the chance of winning one game would be (1-D(S))/2, so the chance of winning 8 games in a row would be (1-D(S))^8/256.
Just one note not really related to the teasor. I don't think Garry Kasparov ever played against Bobby Fischer, and in that sense I don't think Fischer played against any champions other than Spassky.
Some of the comments already expressed how poorly this teaser was written. However, there's another incorrect assumption. If 'Bob' plays 8 players at his skill level at the same time than he is at a GREAT disadvantage. Bob is playing eight games while his opponents only have to focus on their one game. That's why chess grandmasters play simul. games against inferior opponents.
If X was in the Super Chess Tournament, do you really think he'd be resigning his games?
Respectfully, the answer and explanation are silly on several levels.
You distinguish "win" and "lose" from "resign" and "opponent resigns". Yet the vast majority of non-drawn games are ended in a resignation. You rarely see checkmate at moderately competitive levels and even less so, as you go up from there.
Next, a stalemate is an extremely rare event, at any level. I have played hundreds of tournament games (not for a few years, admittedly) and never once was involved in a stalemate in any of them.
So the main point is: the "six outcomes" you defined are not independent, and are far from equal probability - at any level of play.
One poster stated that Fischer didn't play any "champions" other than Spassky - very far from the truth. Fischer had to play for years at various qualifying levels, including matches against former American champions and many other national champions from around the world, just to qualify to play Spassky. The ladder to the world championship is, and was, a very steep climb.
You distinguish "win" and "lose" from "resign" and "opponent resigns". Yet the vast majority of non-drawn games are ended in a resignation. You rarely see checkmate at moderately competitive levels and even less so, as you go up from there.
Next, a stalemate is an extremely rare event, at any level. I have played hundreds of tournament games (not for a few years, admittedly) and never once was involved in a stalemate in any of them.
So the main point is: the "six outcomes" you defined are not independent, and are far from equal probability - at any level of play.
One poster stated that Fischer didn't play any "champions" other than Spassky - very far from the truth. Fischer had to play for years at various qualifying levels, including matches against former American champions and many other national champions from around the world, just to qualify to play Spassky. The ladder to the world championship is, and was, a very steep climb.
May 03, 2008
There are only three outcomes: win, lose, draw.
P(win) = P (lose)
P(win) != P(draw)
So chance of winning is
P(win) = 1-P(lose)-P(draw)
P(win) = (1-P(draw))/2
P(win) = P (lose)
P(win) != P(draw)
So chance of winning is
P(win) = 1-P(lose)-P(draw)
P(win) = (1-P(draw))/2
How did this puzzle ever get past the judging process?!? This is stupid and horribly, horribly wrong.
This could be a good puzzle but it needs to specify things more. The writer was I think aiming at a puzzle based on the idea that for each game win, lose or draw are equally likely. The probability of winning all 8 games would then be (1/3)^8 I think.
However the assumptions need to be set out. For example, there is nothing in the puzzle that states that particular outcomes are as likely as each other. The players are equal but that just means the probability of a particular outcome is the same for each game. If the players are precisely equal then you could argue that every game will be a draw, so the chance of winning them all (or even one of them) is zero.
Maybe the best place for this puzzle would be as one of those semantic puzzles which all depend on things which aren't given - you give them to groups of people to start them thinking. E.g. with children.
However the assumptions need to be set out. For example, there is nothing in the puzzle that states that particular outcomes are as likely as each other. The players are equal but that just means the probability of a particular outcome is the same for each game. If the players are precisely equal then you could argue that every game will be a draw, so the chance of winning them all (or even one of them) is zero.
Maybe the best place for this puzzle would be as one of those semantic puzzles which all depend on things which aren't given - you give them to groups of people to start them thinking. E.g. with children.
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