Originally posted by eldragonflyOK, well you implied I was a "dummy", so I returned the favor.
My mistake, still stop being rude, it serves no purpose. 😉
Yes, I could see how their question is ambiguously offered. But hopefully you see now why they quote the answer they quote.
I think your alternative interpretation is a more interesting question, but that also becomes ambiguous. For instance, are we supposed to have knowledge of what particular face card is initially drawn and removed from the deck, or do we only know that it is a face card? The most interesting case would be where we only know that it was a face card (we don't know which particular face card or type of face card). In that case, I think mtthw and geepamoogle have already given the right answer on the previous page.
Originally posted by LemonJelloThe problem was poorly worded.
OK, well you implied I was a "dummy", so I returned the favor.
Yes, I could see how their question is ambiguously offered. But hopefully you see now why they quote the answer they quote.
I think your alternative interpretation is a more interesting question, but that also becomes ambiguous. For instance, are we supposed to have knowledge of what p ...[text shortened]... card is initially drawn and removed from the deck, or do we only know that it is a face card?
Originally posted by LemonJelloThere you go.
Yes, I would agree with you on that.
But i was trying to work up to bayes theorem, in other words i don't really fathom the solution to the baseball playoff game, other than in my mind the first answer would something involve just *crunching* the games won stats.
Originally posted by eldragonflyTry the problem on p.8 of this thread. Part 1 is easy. Part 2 will help you understand conditional probability.
There you go.
But i was trying to work up to bayes theorem, in other words i don't really fathom the solution to the baseball playoff game, other than in my mind the first answer would something involve just *crunching* the games won stats.
Originally posted by PBE6did you take into consideration the score (11-4) ?
I ran a simulation too. Each number below represents 5000 trials, and the average is the value for all 50,000 trials:
0.556419113
0.557171952
0.541043724
0.563888088
0.543978749
0.576020851
0.567821491
0.565077508
0.582945285
0.568644561
Average = 0.562301132
This is closer to mtthw's answer (56.5% ) than my answer (56.8% ), which gives me added confidence in mtthw's answer.
you could do that when you run your simulation and see that because the mets won with almoust triple points increases the odds for them to have palyed in NY to 1.66 to 1
that goes double for mtthw!
Originally posted by alexdinoGood try 🙂
did you take into consideration the score (11-4) ?
you could do that when you run your simulation and see that because the mets won with almoust triple points increases the odds for them to have palyed in NY to 1.66 to 1
that goes double for mtthw!
Originally posted by PBE6you should swich
I can help with the concept, although I'll drag in the Monty Hall problem (the bazillionth time for some of these guys).
You're playing Let's Make a Deal with Monty Hall. There are three doors on stage, and behind one there is a prize (the prize door is randomly selected before the show, 1/3 chance it could be behind any door). For simplicity's sake, there ...[text shortened]... should have a good handle on what conditional probability means and how to use it.
i assum the chances are 2 to 1
Originally posted by alexdinoThe original problem says nothing about score distributions, it only gives probabilities for win/loss.
did you try it?
what was your result?
and i say it maters because its given in the original text
It had crossed my mind that you could infer something from the actual scoreline, but we aren't given nearly enough information to do that.
The good old Monty Hall problem. The intuitive answer is that either door is equally likely to have the prize and therefore switching is an even trade of odds.
However, in the Monty Hall problem, the door being opened doesn't provide as much information as it would seem to, because the door selected is not random, but rather is intentionally a losing door. The host knows the layout of the doors, and he always has a losing door to choose, so he can do that.
Had the door that was selected been a random choice (meaning there was a chance that the revealed door held the prize), it WOULD be the case that switching would be equally likely to win or lose.
I heard a very simple way to reason this out that makes sense, but it's not coming to me now.
I think it went something like this.
When you first choose the door, you have 1-in-3 chance to be right. Since the revealed door will always be a losing door, that means the remaining door has a 2-in-3 chance of having the prize. Remember, the host always reveals a losing door, so if you chose wrongly, the remaining door will have the prize.
Here's another problem about conditional probability to keep the thread going.
Suppose one is scheduled to play a best-of-five match where every game of the match is decisive: either a win or a loss results. The probability that he will win the first game is 1/2. But in subsequent games thereafter, the probability of his winning each game depends on the match history and his related confidence level:
If he won the previous game, he is feeling heightened confidence and the probability of his winning the upcoming game is 2/3. If he lost the previous game, he is feeling somewhat demoralized and the probability of his winning the upcoming game is only 1/3. There is an exeception to this rule, however. If he has won the last two games in a row, then he is really feeling on a roll, and his probability of winning the upcoming game is 3/4. If he has lost the last two games in a row, then he is really losing his confidence and his probability of winning the upcoming game is only 1/4.
What is the probability that he wins the match, given that he wins the first game of the match?
Originally posted by LemonJelloplease refrasethe question because it has too many lacunes
Here's another problem about conditional probability to keep the thread going.
Suppose one is scheduled to play a best-of-five match where every game of the match is decisive: either a win or a loss results. The probability that he will win the first game is 1/2. But in subsequent games thereafter, the probability of his winning each game depends on ...[text shortened]... What is the probability that he wins the match, given that he wins the first game of the match?