It makes no difference whether the spots are chosen one at a time or simultaneously. With the same number of candidates for the same number of spots the odds (calculated before the process starts, of course) of you winning a spot are the same. Odds calculated along the way change with each pick of course - either to 1/1 if you are chosen, otherwise they increase to 1/2 as you outlined. But that’s not relevant to the OP.
Your two methods of calculation end up with the same odds, 100/101, when done before any results come in.
In that second line, you’re leaving out one term. 1/6 + 1/5 = 11/30, not 1/3. There is a 1/6 chance of getting the first slot and a 1/5 chance of getting the second, but you must subtract 1/5*1/6=1/30, because you can’t get both slots.
Another way the OP can think of this is that there are 15 (unordered) pairs of people that can be chosen, 5 of which contain the OP, so there’s a 5/15=1/3 chance of being selected.
No, that’s not quite the right way of looking at it. The 1/5 probability is a conditional probability: you have a 1 in 5 chance of being chosen on the second round provided you have no been chosen on the first. So the right equation is:
1/6 + (5/6 * 1/5) = 1/3
The 5/6 term is the probability that you will not be chosen on the first round.
The odds as calculated before anything happens are the same regardless of whether the spots are going to be filled all at once or one at a time. If it does end up being done one at a time, the odds change (and so must be re-calculated) after each pick, as with each pick one possibility has collapsed to 1/1 and others to 0/1.
It’s a perfectly fine way of looking at it. You’re looking at filling the slots sequentially and I’m looking at filling them simultaneously. Both are valid ways of determining the probability.
Your approach actually does work, but it’s convoluted enough that I needed a couple minutes with scratch paper to convince myself that it’s correct, and based on your explanation, I’m not sure that you understand it. It has nothing to do with filling the slots sequentially or simultaneously; instead, Giles is using a standard approach that will work for any two events, and you’re using a formula I’ve never seen before that only works for mutually exclusive events.
In more formal probabilistic terms, if A is the event that a player is picked for the first slot, and B is the event that a player is picked for the second slot, what Giles is computing is P(A) + P(A[sup]c[/sup])P(B|A[sup]c[/sup]), and what you’re computing is P(A) + P(B|A[sup]c[/sup]) - P(B|A[sup]c[/sup])P(A). Those are both equal if A and B are mutually exclusive (as they are here), but as I mentioned above, I’ve never seen the second approach before. If I ever teach an undergrad probability class, I’ll definitely use this as the basis for a fairly difficult test question, so I owe you one.
I’m using the standard method of calculating probabilities using the principle of inclusion exclusion. In this case there are two events, and the P(A or B) = P(A) + P(B) - P(A and B). It doesn’t matter if the events are mutually exclusive.
I’m thinking of this as two separate events, selecting one of six and selecting one of five. It’s possible for the person to be selected either time, but not both (since the OP is only one person), so the events are A=selected from the six and B=selected from the five. What we want is P(A) + P(B) - P(A and B). You can think of this as two selections happening simultaneously. (It’s basically picking apart the idea of counting the total number of pairs of people and the number of pairs that include the OP).
Using conditional probabilities is sequential in that you calculate the probability of getting the first slot, and then the probability of getting the second provided that you didn’t get the first.
Two comments:
[ol]
[li]What exactly are your events A and B here? If you define them the way you want to, the formula I wrote down is the correct representation of what you’re calculating.[/li][li]P(A and B) = P(A)P(B) only if A and B are independent. Any two events that are mutually exclusive are not independent, so you can’t use that equality in this case.[/li][/ol]
A is the event that the person is selected from a set of six.
B is that the person is selected from a set of five.
I’m considering these as two independent selections (which I’ll call drawings) and determining the probability of being selected exactly once. This is equivalent to considering the two drawings as you did, with the second being dependent on the first. The event (A and B) is the event that the person is selected both times, so we must subtract the probability that it occurs. So what we want is P(A) + P(B) - P(A and B).
Saying that two events are independent means that the outcome of one has no influence on the other. Since the probability that you’re selected in the draw of five depends on whether you were selected in the draw of six, they are not independent, and so you can’t write P(A and B) = P(A)P(B).
I think that you are missing the fact that my A and B aren’t the same as your A and B. I’m modeling this situation as two independent (and I’m well aware of what that means) drawings–one of five and one of six–and looking at the probability of being selected in exactly one of the drawings. To do that, P(A xor B) = P(A) + P(B) - P(A AND B), which in this situation is 1/6 + 1/5 - (1/6)(1/5).
Using conditional probabilities is equivalent. If you factor out the 1/5 above, you get 1/6 + 1/5 (1 - 1/6) = 1/6 + (1/5)(5/6).
In both we have to eliminate the possibility of being selected for both slots. One method adds 1/5 then subtracts (1/6)(1/5), the other multiplies (1/5) by (5/6) before adding it.
If you claim that both draws are independent, you’re asserting that there’s a 1/30 chance that you’ll be selected for both slots. Is that really what you mean?
To be more clear, I’m modeling this the same way you’d model the following situation. One bag contains 6 balls, another contains 5. Each bag contains exactly one black ball. One ball is selected from each bag. What is the probability of selecting exactly one black ball.
You’re modeling this the same as using one bag containing 6 balls, exactly one of which is black, and asking if two balls are selected without replacement, what is the probability of selecting the black ball in the first or the second draw.
In the first setup the two draws can happen simultaneously and in the second one they are sequential (which I mentioned in my first post). These are two different models, but the particular question we are asking is equivalent.
This is the right track. It’s not a probability question. Straight probability assumes that all those competing have an equal chance, but obviously, they would not.
A pro golfer is going to have a better chance than a weekend duffer. While the latter may play the round of his life and make the team, it’s impossible to determine the probability of that.
That might be true of a sport like baseball, or maybe basketball, where there are defined positions with different skill sets. Golf doesn’t work that way. Golf “teams” are just individual players that pool their individual scores for a total. It’s not like there’s Position 1 that handles all the tee-offs and Position 2 that handles the putting.
It’s amazing how much probability is influenced by who is your friend. Do you know these guys? Do any of the other “possibilities” hang out with these guys? If your competition is a drinking buddy of two or more choosers, I would estimate your odds of being chosen as about .000000025%.