I have a question. Let's say that there are 27 players involved in a tournament. Each one is ranked according to his/her skill. Player 1 can beat all others. Player 2 can beat all except player 1. Player 3 can beat anyone except players 1 and 2, etc.
In the first round, the players form 9 groups of 3 players each. When it's over, the 9 winners of each group form 3 groups of 3 players and compete again.
What is the probability that player 3 will make it out of the first round as one of the 9 winners. What is the probability that he will make it out of the second round as one of the 3 finalists?
> I have a question. Let's say that there are 27 players involved in a > tournament. Each one > is ranked according to his/her skill. Player 1 can beat all others. > Player 2 can beat all > except player 1. Player 3 can beat anyone except players 1 and 2, etc.
> In the first round, the players form 9 groups of 3 players each. When > it's over, the 9 winners > of each group form 3 groups of 3 players and compete again.
> What is the probability that player 3 will make it out of the first > round as one of the 9 > winners. What is the probability that he will make it out of the > second round as one of > the 3 finalists?
The question can't be answered without making some assumptions. I'll assume that all the groups are chosen at random at every stage.
The probability that Player 3 will make it out of the first round is equal to the number of ways in which he can survive, divided by the number of outcomes. For this purpose we can actually ignore the other eight groups, since their outcomes have no bearing on his survival or otherwise.
Whatever group he's in, there are only two people who can beat him, and 24 who can't. There are two slots to fill in his group (other than his own). He survives if the first slot is filled by any of the 24 people, and the second by any other of the 24: 24 * 23. There are 26 * 25 possible ways to fill the slots, so his survival probability is (24 * 23) / (26 * 25) which is 552/650 which is 0.84923+. It's surprisingly low, isn't it? I mean, okay, it's third-highest - but it's still lower than I thought it would be.
Conditional probabilities can get murky (can't they, Monty?), so the following may be flawed. To make my working more obvious to the probability-minded, I haven't bothered cancelling common factors.
Let us look at ways in which #2 and #3 can both survive Round 1. We have already filled the first group (in one of 24 * 23 ways). We place #2 firmly in the second group, so we've now filled four places, leaving 23 players, but #2 can't face #1, so #3 *and* #2 survive with probability 552/650 * (22*21)/(23*22) = 552/650 * 462/506 = 255024/328900 = 0.77538+.
We deduce that player #3 survives without #2 with probability 552/650 * 44/506 = 24288/328900 = 0.07384+.
Assuming #2 did NOT survive, player #3 *will* survive to the final unless he meets #1 in the semis. He has seven suitable opponents out of eight, and we need to draw two of them, so that's (7*6)/(8*7) = 42/56. We need to multiply this by the probability that #3 survived but #2 did not, which gives us 42/56 * 44/506 = 1848/28336 = 0.06521+
Assuming that #2 DID survive, player #3 must avoid not only #1 but also #2. So he has only six suitable opponents out of eight. We need two, so that's (6 * 5) / (8 * 7) = 30/56. We need to multiply this by the probability that both #2 and #3 survived, which gives us 30/56 * 255024/328900 = 7650720/18418400 = 0.41538+.
#3 survives to the final if either of these mutually exclusive events occurs, so we add:
1848/28336 + 7650720/18418400 = (34037203200 + 216790801920) / 521903782400 which is 250828005120 / 521903782400 or 1437 / 2990 or 0.480602+.
> > I have a question. Let's say that there are 27 players involved in a > > tournament. Each one > > is ranked according to his/her skill. Player 1 can beat all others. > > Player 2 can beat all > > except player 1. Player 3 can beat anyone except players 1 and 2, etc.
> > In the first round, the players form 9 groups of 3 players each. When > > it's over, the 9 winners > > of each group form 3 groups of 3 players and compete again.
> > What is the probability that player 3 will make it out of the first > > round as one of the 9 > > winners. What is the probability that he will make it out of the > > second round as one of > > the 3 finalists?
> The question can't be answered without making some assumptions. I'll assume > that all the groups are chosen at random at every stage.
> The probability that Player 3 will make it out of the first round is equal > to the number of ways in which he can survive, divided by the number of > outcomes. For this purpose we can actually ignore the other eight groups, > since their outcomes have no bearing on his survival or otherwise.
> Whatever group he's in, there are only two people who can beat him, and 24 > who can't. There are two slots to fill in his group (other than his own). > He survives if the first slot is filled by any of the 24 people, and the > second by any other of the 24: 24 * 23. There are 26 * 25 possible ways to > fill the slots, so his survival probability is (24 * 23) / (26 * 25) which > is 552/650 which is 0.84923+. It's surprisingly low, isn't it? I mean, > okay, it's third-highest - but it's still lower than I thought it would > be.
> Conditional probabilities can get murky (can't they, Monty?), so the > following may be flawed. To make my working more obvious to the > probability-minded, I haven't bothered cancelling common factors.
> Let us look at ways in which #2 and #3 can both survive Round 1. We have > already filled the first group (in one of 24 * 23 ways). We place #2 > firmly in the second group, so we've now filled four places, leaving 23 > players, but #2 can't face #1, so #3 *and* #2 survive with probability > 552/650 * (22*21)/(23*22) = 552/650 * 462/506 = 255024/328900 = 0.77538+.
> We deduce that player #3 survives without #2 with probability 552/650 * > 44/506 = 24288/328900 = 0.07384+.
> Assuming #2 did NOT survive, player #3 *will* survive to the final unless > he meets #1 in the semis. He has seven suitable opponents out of eight, > and we need to draw two of them, so that's (7*6)/(8*7) = 42/56. We need to > multiply this by the probability that #3 survived but #2 did not, which > gives us 42/56 * 44/506 = 1848/28336 = 0.06521+
> Assuming that #2 DID survive, player #3 must avoid not only #1 but also #2. > So he has only six suitable opponents out of eight. We need two, so that's > (6 * 5) / (8 * 7) = 30/56. We need to multiply this by the probability > that both #2 and #3 survived, which gives us 30/56 * 255024/328900 = > 7650720/18418400 = 0.41538+.
> #3 survives to the final if either of these mutually exclusive events > occurs, so we add:
> 1848/28336 + 7650720/18418400 = (34037203200 + 216790801920) / 521903782400 > which is 250828005120 / 521903782400 or 1437 / 2990 or 0.480602+.
> On May 14, 5:16 pm, Richard Heathfield <r...@see.sig.invalid> wrote:
> > [Problem text is spoiler space]
> > Patrick D. Rockwell said:
> > > I have a question. Let's say that there are 27 players involved in a > > > tournament. Each one > > > is ranked according to his/her skill. Player 1 can beat all others. > > > Player 2 can beat all > > > except player 1. Player 3 can beat anyone except players 1 and 2, etc.
> > > In the first round, the players form 9 groups of 3 players each. When > > > it's over, the 9 winners > > > of each group form 3 groups of 3 players and compete again.
> > > What is the probability that player 3 will make it out of the first > > > round as one of the 9 > > > winners. What is the probability that he will make it out of the > > > second round as one of > > > the 3 finalists?
> > The question can't be answered without making some assumptions. I'll assume > > that all the groups are chosen at random at every stage.
> > The probability that Player 3 will make it out of the first round is equal > > to the number of ways in which he can survive, divided by the number of > > outcomes. For this purpose we can actually ignore the other eight groups, > > since their outcomes have no bearing on his survival or otherwise.
> > Whatever group he's in, there are only two people who can beat him, and 24 > > who can't. There are two slots to fill in his group (other than his own). > > He survives if the first slot is filled by any of the 24 people, and the > > second by any other of the 24: 24 * 23. There are 26 * 25 possible ways to > > fill the slots, so his survival probability is (24 * 23) / (26 * 25) which > > is 552/650 which is 0.84923+. It's surprisingly low, isn't it? I mean, > > okay, it's third-highest - but it's still lower than I thought it would > > be.
> > Conditional probabilities can get murky (can't they, Monty?), so the > > following may be flawed. To make my working more obvious to the > > probability-minded, I haven't bothered cancelling common factors.
> > Let us look at ways in which #2 and #3 can both survive Round 1. We have > > already filled the first group (in one of 24 * 23 ways). We place #2 > > firmly in the second group, so we've now filled four places, leaving 23 > > players, but #2 can't face #1, so #3 *and* #2 survive with probability > > 552/650 * (22*21)/(23*22) = 552/650 * 462/506 = 255024/328900 = 0.77538+.
> > We deduce that player #3 survives without #2 with probability 552/650 * > > 44/506 = 24288/328900 = 0.07384+.
> > Assuming #2 did NOT survive, player #3 *will* survive to the final unless > > he meets #1 in the semis. He has seven suitable opponents out of eight, > > and we need to draw two of them, so that's (7*6)/(8*7) = 42/56. We need to > > multiply this by the probability that #3 survived but #2 did not, which > > gives us 42/56 * 44/506 = 1848/28336 = 0.06521+
> > Assuming that #2 DID survive, player #3 must avoid not only #1 but also #2. > > So he has only six suitable opponents out of eight. We need two, so that's > > (6 * 5) / (8 * 7) = 30/56. We need to multiply this by the probability > > that both #2 and #3 survived, which gives us 30/56 * 255024/328900 = > > 7650720/18418400 = 0.41538+.
> > #3 survives to the final if either of these mutually exclusive events > > occurs, so we add:
> > 1848/28336 + 7650720/18418400 = (34037203200 + 216790801920) / 521903782400 > > which is 250828005120 / 521903782400 or 1437 / 2990 or 0.480602+.
> On May 14, 5:16 pm, Richard Heathfield <r...@see.sig.invalid> wrote: > > [Problem text is spoiler space]
> > Patrick D. Rockwell said:
> > > I have a question. Let's say that there are 27 players involved in a > > > tournament. Each one > > > is ranked according to his/her skill. Player 1 can beat all others. > > > Player 2 can beat all > > > except player 1. Player 3 can beat anyone except players 1 and 2, etc.
> > > In the first round, the players form 9 groups of 3 players each. When > > > it's over, the 9 winners > > > of each group form 3 groups of 3 players and compete again.
> > > What is the probability that player 3 will make it out of the first > > > round as one of the 9 > > > winners. What is the probability that he will make it out of the > > > second round as one of > > > the 3 finalists?
> > The question can't be answered without making some assumptions. I'll assume > > that all the groups are chosen at random at every stage.
> > The probability that Player 3 will make it out of the first round is equal > > to the number of ways in which he can survive, divided by the number of > > outcomes. For this purpose we can actually ignore the other eight groups, > > since their outcomes have no bearing on his survival or otherwise.
> > Whatever group he's in, there are only two people who can beat him, and 24 > > who can't. There are two slots to fill in his group (other than his own). > > He survives if the first slot is filled by any of the 24 people, and the > > second by any other of the 24: 24 * 23. There are 26 * 25 possible ways to > > fill the slots, so his survival probability is (24 * 23) / (26 * 25) which > > is 552/650 which is 0.84923+. It's surprisingly low, isn't it? I mean, > > okay, it's third-highest - but it's still lower than I thought it would > > be.
> > Conditional probabilities can get murky (can't they, Monty?), so the > > following may be flawed. To make my working more obvious to the > > probability-minded, I haven't bothered cancelling common factors.
> > Let us look at ways in which #2 and #3 can both survive Round 1. We have > > already filled the first group (in one of 24 * 23 ways). We place #2 > > firmly in the second group, so we've now filled four places, leaving 23 > > players, but #2 can't face #1, so #3 *and* #2 survive with probability > > 552/650 * (22*21)/(23*22) = 552/650 * 462/506 = 255024/328900 = 0.77538+.
> > We deduce that player #3 survives without #2 with probability 552/650 * > > 44/506 = 24288/328900 = 0.07384+.
> > Assuming #2 did NOT survive, player #3 *will* survive to the final unless > > he meets #1 in the semis. He has seven suitable opponents out of eight, > > and we need to draw two of them, so that's (7*6)/(8*7) = 42/56. We need to > > multiply this by the probability that #3 survived but #2 did not, which > > gives us 42/56 * 44/506 = 1848/28336 = 0.06521+
> > Assuming that #2 DID survive, player #3 must avoid not only #1 but also #2. > > So he has only six suitable opponents out of eight. We need two, so that's > > (6 * 5) / (8 * 7) = 30/56. We need to multiply this by the probability > > that both #2 and #3 survived, which gives us 30/56 * 255024/328900 = > > 7650720/18418400 = 0.41538+.
> > #3 survives to the final if either of these mutually exclusive events > > occurs, so we add:
> > 1848/28336 + 7650720/18418400 = (34037203200 + 216790801920) / 521903782400 > > which is 250828005120 / 521903782400 or 1437 / 2990 or 0.480602+.
