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October 28

probability question

I'm terrible at probability, this is not homework though i admit it sounds like one. I'm trying to work out something for an argument. Say you have two groups of 10 people, one is a control group and the other is your study group; you fed that group a drug, or gave them special coaching or something. So after a test, 8 people in the effect group pass and 2 people in the control group pass. What are the odds that this occurred entirely by chance? I'm thinking you start by saying the total chance of passing is 10/20 = 50% so it's sort of a coin toss right? I'm thinking you need to distribute 10 "passes" between the two groups where the chance is 50%, so each "pass" is represented by the toss of a coin and you call the groups "heads" and "tails". Then all I need is the odds of getting (at least) 8 "passes" (coin tosses) in ONE of those groups (heads). Does that sound right? So what's the odds of tossing at least 8 heads out of 10 coins? Vespine (talk) 00:04, 28 October 2011 (UTC)[reply]

Ok, i found this pretty cool website and I think I've got the answer of about 5%. About 4% for 8 heads, about 1% for 9 heads and 0.1% for 10 heads. That sounds about right to me, can anyone see any glaring errors? Vespine (talk) 01:09, 28 October 2011 (UTC)[reply]

If I didn't specify why I added 8,9 and 10 heads it's because I want the chance of AT LEAST that result. Vespine (talk) 03:55, 28 October 2011 (UTC)[reply]

See our article Bayesian_inference#Posterior_distribution_of_the_binomial_parameter. See also Binomial_proportion_confidence_interval. Bo Jacoby (talk) 06:47, 28 October 2011 (UTC).[reply]

The point is that you do not know the exact probability $P_{1}$ of a person passing having being given the drug, nor the exact probability $P_{2}$ of a person passing having not being given the drug. You know that $P_{1}$ is about $\mu _{1}={\frac {8+1}{10+2}}=0.75$ give or take $\sigma _{1}={\sqrt {\frac {\mu _{1}(1-\mu _{1})}{10+3}}}=0.12$ . And you know that $P_{2}$ is about $\mu _{2}={\frac {2+1}{10+2}}=0.25$ give or take $\sigma _{2}={\sqrt {\frac {\mu _{2}(1-\mu _{2})}{10+3}}}=0.12$ . Then you know that $P_{2}-P_{1}$ is about $\mu =\mu _{2}-\mu _{1}=-0.50$ give or take $\sigma ={\sqrt {\sigma _{1}^{2}+\sigma _{2}^{2}}}=0.17$ . The probability that $P_{1}>P_{2}$ is approximately CDF[NormalDistribution[-0.5, 0.17],0]=0.998359. (See[1]) Bo Jacoby (talk) 07:47, 28 October 2011 (UTC).[reply]

Seems to me you could be after Fisher's exact test for a 2 × 2 contingency table. Standard statistical software gives a two-sided p-value of 0.02301, i.e. for two groups of 10 people, given that 10 people in total have passed, the probability of 8 or more of them being in the same group (either the control group or the study group) by chance alone is 2.3%. Qwfp (talk) 14:55, 28 October 2011 (UTC)[reply]

Right, Fisher's exact test is what a well-trained scientist would use to analyze these statistics. If the numbers were larger, a chi-square test might be used instead. Looie496 (talk) 15:58, 28 October 2011 (UTC)[reply]

The well-trained scientist is a frequentist rather than a Bayesian. Bo Jacoby (talk) 20:29, 28 October 2011 (UTC).[reply]

Yowzer! So looks like I was a fair way off (still much closer then the person I'm arguing with). Thanks for the replies and those links too. Vespine (talk) 21:45, 1 November 2011 (UTC)[reply]

Puzzling junior high maths question

I know you guys can't do homework for people, but please indulge me just this once as this is making me (and a few other people) tear out their hair. This is not for me (a long way from junior high!!) but for a daughter of a friend of my wife. The friend has put the question up on Facebook and the consensus is that it is not solvable. 90 [ ] 10 [ ] 7 = 17 replace the [ ] with either + - * or / in order to get the answer (17). I think that the question must be a misprint, and that the answer should be 16 since 90 / 10 + 7 = 16. Am I correct and is this a misprint or is there really an answer which I can't see? Thanks guys 121.44.65.209 (talk) 06:39, 28 October 2011 (UTC)[reply]

Hi, This question can easily be answered using simple methods of combinatorics. Since the there are 2 "Gaps" with 4 possibilities ( +, -, x, / ) each, this leaves us with 4^2 = 16 possible solutions. Therefore we can brute force the answer. And you guessed it, none of the 16 possibilities work. Even though the one you have mentioned ( 90 / 10 + 7 = 16) is quite close. Greetings from Germany --Freakschwimmer (talk) 08:00, 28 October 2011 (UTC)[reply]

Actually there are 32 possibilities if you allow for different order of operations. I get the solution giving 16 is closest with the next closest being 90 − 10 × 7 = 20. If you allow the operations to go in other places than the brackets then − 90 + 107 = 17 works.--RDBury (talk) 09:10, 28 October 2011 (UTC)[reply]

-90 * 10 + 7 ≠ 17 — Preceding unsigned comment added by 203.112.82.128 (talk) 17:24, 28 October 2011 (UTC)[reply]

inspired! surely this deserved full marks? 109.150.107.49 (talk) 22:12, 28 October 2011 (UTC)[reply]

Difference in chances?

Scenario. A horse race is being held on which a sweep is being held, and we have a dispute regarding the fairness of the allocation of horses to the participants. There are 24 starters, and 24 people put in $10 each and are allocated a horse. After the race the holder of the winning horse wins $120, the holder of the second placed horse horse wins $80 and the holder of the second placed horse horse wins $40. Before the race we obviously know the names of the horses participating, and their names listed in the racebook, numbered 1 to 24.

Option 1. We have a hat (A) in which we put slips of paper, each with the name of one of the participants. We draw a name from the hat, and the first name out is allocated horse number 1, and the second name drawn is allocated horse number 2, and so on until all participants have been allocated a horse.

Option 2. We have a hat (A) in which we put slips of paper, each with the name of one of the participants. Also, we have a second hat (B) in which we have numbers 1 to 24 representing the numbers of the horses. In this scenario, a participant's name is drawn from A and then a horse number is drawn from B and is allocated to the participant. And so on until all 24 participants/horses have been matched.

Seems Option 2 is fairer because it gives the first participant drawn the chance of any of 24 horses, whereas in Option 1 the first particpant drawn has no choice but to accept the number 1 horse on the list.

Is there some sort of mathematical formula which might show which option is the most equitable? Also, is the chance of a participant drawing horse number 13 in Option 1 different from his chance in Option 2 ? Moriori (talk) 08:36, 28 October 2011 (UTC)[reply]

The seeming "extra fairness" of option 2 is an illusion with no mathematical basis. In both cases the allocation of people to horses is uniformly random. The chances for every person to get every horse is the same in both cases (and equal to 1/24 for every combination).

Put another way: Hat A randomizes the allocation. Hat B randomizes the allocation. With either one you have a random allocation, you can't have "more random than random" by using both.

The only difference is in the order in which information about the allocation is given. If, say, someone really cares about who gets horse 24 and would be really stressed out by having to wait to find out, then option 1 might not be fair for him.

Also, if the slips paper aren't shuffled properly and only make for partial randomness, then having two hats can make it closer to truly random. -- Meni Rosenfeld (talk) 10:01, 28 October 2011 (UTC)[reply]

Many thanks Meni. Forgetting lack of mathematical basis, I am puzzled by you saying the seeming "extra fairness" of option 2 is an illusion. How can that be? In Option 1, the first drawn participant must have number 1 and therefore has no opportunity to be randomly allocated any of the other 23 horses. In Option 2, the first drawn participant has the opportunity of randomly drawing any of the 24 runners. Illusion? Moriori (talk) 21:33, 28 October 2011 (UTC)[reply]

Would it help to consider a smaller field, say of two? I have two pieces of candy: #1 is chocolate and #2 is coconut. I write your and my names on two slips of paper and put them in a hat; the first name drawn out of the hat gets the chocolate and the second gets the coconut. It is true that the first drawn participant must have the chocolate, but there is a random, 50:50 chance of who that first drawn participant will be -- you or me. The same idea holds for your 24 slips. Yes, the first drawn participant must have number 1, but it is equally likely that any one of the 24 slips will be drawn, so everyone's chance of being assigned number 1 is 1/24. What might seem confusing is the drawing of the second slip, where there are only 23 slips remaining in the hat, so there is an equal 1/23 chance of the remaining 23 participants being assigned number 2. This is true, give that the first drawing has already occurred and those 23 participants have been excluded from being assigned number 1, but they all originally had a 1/24 chance of getting number 1, so their chance of making it to the pool of 23 was 1 - 1/24 = 23/24 and their overall chance of being assigned number 2 is 23/24 * 1/23 = 1/24, the same as their original chance of being assigned number 1. Likewise their chance of making it to the pool of 22 is 1 - 1/24 - 1/24 = 22/24 because they would have had to have missed out on the first two draws, and their chance of being assigned number 3 is 22/24 * 1/22 = 1/24. Cool, huh? -- 110.49.227.229 (talk) 23:35, 28 October 2011 (UTC)[reply]

As Meni pointed out, there may be psychological effects involved. If would be equally fair to use hat B with the slips numbered 1 through 24, and go down the list of participants alphabetically, letting them draw their numbers. Even though poor old Wulfovna Zeigarnik will never get a chance to choose her own horse, always being given the last remaining slip in the hat, she has the same chances of being assigned any one horse as any other participant. She not may feel in control of her own fate, but that fate is illusory. -- 110.49.227.229 (talk) 23:48, 28 October 2011 (UTC)[reply]

The first drawn participant must have horse 1. But each person has the same chance to be the first drawn participant. Or any other nth drawn participant. -- Meni Rosenfeld (talk) 17:20, 29 October 2011 (UTC)[reply]

But, but, but, but, but! Please indulge me. Maybe I didn't make it clear in my two options above that I am thinking opportunity and fairness rather than chance in the sense of odds/probability whatever (despite the section header). So can I try to better explain? Horse Number 1 is the critter that every participant wants to draw because they think it is unbeatable (the fools).

In my Option 1, Wulfovna Zeigarnik is the first name drawn out from the list of participants. At that moment the other 23 gamblers have lost the opportunity to be assigned Horse Number 1. (It's gone baby, gone, because WF has it).

In my Option 2, Wulfovna Zeigarnik is the first name drawn out from the list of participants. But at that moment everybody still has the opportunity to be assigned Horse Number 1. The next step is that a name is drawn from the list of horses, to be allocated to Wulfovna Zeigarnik. Sure, she may fluke Horse Number 1. But if she draws Horse Number 7 (say), then the remaining 23 gamblers still have the opportunity to draw their favoured Horse Number 1.

So. Is not Option 2 a fairer way of allocating the horses? Moriori (talk) 23:53, 29 October 2011 (UTC)[reply]

No there is no extra fairness. And having hats C, D, E, F, G... all messing around with who gets the horse assigned makes it no fairer. One person will be assigned the horse at random and that';s that. Dmcq (talk) 00:02, 30 October 2011 (UTC)[reply]

When you say "I am thinking opportunity and fairness rather than chance in the sense of odds/probability", you need to better explain what you mean by fairness if you mean something other than "equal chance for all". There certainly are other psychological factors, such as suspense, at play here. That is why when prizes are awarded they are typically given in order of increasing value, with the grand prize awarded last. Were the grand prize to be awarded first you might argue that it is not "fair" to make both the audience and the disappointed participants sit through the distribution of the remaining prizes, but that is about showmanship, not fairness in the sense of "equal chance for all". You may well successfully argue that your two hat solution offers better showmanship, but that has nothing to do with "fairness" as it is typically understood when discussing the probabilities behind games of chance, and your question did specifically ask for a mathematical analysis, not a psychological one. -- 110.49.224.102 (talk) 03:31, 30 October 2011 (UTC)[reply]

Fair enough. Are the following the same?

Option 1 = Random name drawn and allocated specific horse

Option 2 = Random name drawn then random horse drawn

Moriori (talk) 02:11, 1 November 2011 (UTC)[reply]

Yes, they are equally fair (in the sense that, going into the drawing, every participant has equal chance of being awarded any horse). (By "specific horse" in option 1, I assume you mean a horse chosen in advance of the name drawing, possibly numerically, or possibly in increasing odds of winning the race. Clearly, if "specific horse" means saving #13 for when Wulfovna's name is drawn, then it would not be fair.) -- 110.49.227.102 (talk) 07:09, 1 November 2011 (UTC)[reply]

General Topology

I consider the spaces of continuous functions from R to R and my qiestion is: is this space Frechet Urysohn? and, Is it Sequential? Can somone help with that? 89.139.242.220 (talk) 16:05, 28 October 2011 (UTC) Forgot to mention, The topology I'm using on this space is the point open topology... 89.139.242.220 (talk) —Preceding undated comment added 17:42, 28 October 2011 (UTC).[reply]

The link Fréchet-Urysohn space takes you to a subsection of the Sequential space article. — Fly by Night (talk) 20:02, 29 October 2011 (UTC)[reply]