Not exactly sure about this one so I need a Math whiz to my rescue. How do I calculate the probability of an event occuring at least 1 time out of 8 if its single event probability is 0.06? Do I add it up (0.06*8) and arrive at 0.48? For example, my probability of winning a race is 0.06. I am forced to race everyday until I win, up to a maximum of 8 races a day. So does that make my probability of winning at least 1 race out 8 0.48? doesn't seem right if I put it this way. Or is the probability still 0.06? Or, if I get 8 coin tosses. What is the probability of a head occurring at least 1 time? Okay I think this is a much better example. Yup, I'm an idiot. Pls help.
It's been many years since I did probablity at college. So I might be off. Well all thing being equal your probability of winning a race is 1 in x (where x is the number of peopel taking part) The fact that there are eight races in a day doesn't change the probablity of you winning a race unless another factor changes. I..e. someone drops out or more people race. As for a coin toss it's a 50/50 chance that a head would come up at least once. As each coin toss has a 50/50 chance of coming up heads. But are you actually meaning the chances of 7 tails and only 1 heads coming up
Hey thanks. I figured out my problem after I posted. Let me know if this makes sense. The probability of at least one head occurring in 8 tosses is 1 - (0.5^8)=0.99609. 0.5^8 is the probability of 8 tails in 8 tosses. So if my probability of winning a race is 0.06 (I'm not very good), the probability of me losing 8 in 8 is 0.94^8=0.6. Therefore my winning at least 1 is 0.4.
Yes, you're correct. Each race is an independent event, with each result not affecting the next. The chances of losing 8 in row is therefore (1 - 0.06)^8, so the chances of winning at least one is 1- [1 - 0.06)^8]. Which, as you say, comes out to just under 0.4. (note, this is the chance of winning at least 1 race i.e. the set includes the chances that you might win 2 races, or 6, or even all 8, though each of those outcomes has a much lower probability of occuring. It also doesn't tell you anything about which races you win, or what order you win them in. Different calculations are required for those, and all have a much lower probability than the 0.4 above.)
0.4 is far too high as it'd mean 40/60 chance. It's what we Germans call a Tree-event: First try: ........x happens ......./.............\ yes (0.06) ..... no (1-0.06 = 0.94) second try: x happens /.............\ yes (0.06) no (1-0.06 = 0.94) and so on Have a look at the diagrams on this site: http://www.austromath.at/medienvielfalt/materialien/wkeit/lernpfad/ They give examples for the likeliness of throwing dice 3 times and each time throwing a 6. (and for throwing only one 6 out of 3 tries) In your case the likeliness of winning 8 out of 8 would be (0.06)^8 The likeliness of winning one out of 8 would be 0.06 * (0.94)^7 = 0.06 * 0.65 = 0.04 (Hmm, I believe you simply made an error with the decimal point)
My maths is rusty, but I don't think this is totally right. The OP asked for the chances of winning at least 1, not the chances of winning 1 AND losing the other 7 (which is what your calculation works out, I think). For example, he could win two but lose 6 (chances = 0.06^2*0.94^6) = 0.0025. And so on, for 3, 4, 5 wins, etc. So what I and Witterquick did need is to subtract the chances of losing every single race from 1. The trouble is, that generates a figure of about 0.4, which I certainly agree intuitively seems way too high. And adding up all the chances when working it out "forwards" (i.e. summming "win one, lose 7", "win 2, lose 6", etc) will only give a figure a fairly small amount above 0.04. Why don't the two methods match up? What am I missing? I'm off out to dinner; I expect an answer on my return.
That makes the most sense. It gets easier now as I think about it. All I have to do is calculate the odds for each scenario, like: 1 win out of 8 (0.94^7*0.06) 2 wins out of 8 (0.94^6*0.06*0.06) Etc. I did think 0.4 was way too high. Really forgot the basics of probability there. Another question, so the odds of my winning at least one race would be the addition of all the possible scenarios? That is, if prob of 1/8=a, 2/8=b, 3/8=c, 4/8=d, etc., prob of at least one win would be a+b+c+d+........? Edit: Hey Holdfast, posted before I saw your post. Yup 0.4 is definitely off. It means I, who suck, could win 4 days out of 10. I'm puzzled too. The prob of losing 8 out of 8 is 0.94^8 which is only 0.6.
OK, I'm back from dinner, and feeling well-fed, but I still don't have a bulletproof answer to this problem. Look, the trouble with adding up all the individual chances to yield an answer of "a little bit above 0.04" is that gives the (surely nonsensical?) answer that winning at least 1 out of 8 races run in a day is actually LESS likely than winning a single one-off race (which we know from the premise is 0.06). Therefore I can't see how it can possibly be the correct answer. Why would it be less likely to win at least one out of eight consecutive races, compared to winning one out of one? I can't wrap my head around that. There is some logical fallacy being made here, somewhere. But I can't work out what it is, though it must be there, in one of the methods. There must be a set of logical possibilities that aren't being accounted for somewhere, either in the additive or subtractive ways of getting to an answer. I'm actually leaning towards the 0.4 being correct, despite it intuitively seeming ridiculously high. But I can't figure out why the additive method doesn't match it, and why it yields a lower probability than the starting, one-off, race probability. Each race is independent; running more than one race shouldn't reduce the odds of winning at least one race below the starting level; it should increase it from that starting level, because you have eight independent chances and only need to get lucky at least once. Someone sort this out conclusively, please! EDIT - fuck it, I've now cheated and googled up a binomial probability calculator. Enter in the starting variables described in the premise, and it too gives an answer of 0.4 (actually 0.39etc). I still can't explain why the additive method building on the individual odds in Rhub's method demonstrates is wrong, though. EDIT 2 - I think it going to revolve around a missing set of possibilities I'm not quite seeing yet... give me a moment more to think... EDIT 3 - yep, got it. It's to do with combinations. For example, from 8 races, there are multiple ways of getting 1W7L from your day, and even more ways of 2W6L. Note, this isn't permutations; the individual order doesn't matter, only the total number of possible ways you can reach each potential set of results. The number of combinations will equal the number of permutations in the 1W7L scenario, but be increasingly different - smaller - with the larger race win scenarios. For example, for 4W4L, you have only 70 combinations but 1680 permutations. It's only the combinations that matter in this scenario. I think that's the key to the problem. I think. I'm no maths whizz though, so will be interested to hear from one whether I'm right or what the logical mistake I'm making is!
Don't know why but I'm still leaning towards 0.06. As you could win the first race and therefore stop racing. Each race would have to be considered a seperate entity. Sure maybe if you raced in every single race the chances of you winning one or more races is 0.4 but that's not the question at hand.
0.06 is definitely wrong. It's the probability of winning a single race. If you took just race 8 of the day, and asked me the probability of winning it, you're right, the previous races don't affect its result and your chances of winning it are still just 0.06. No matter how many days & races you ran, the chances of winning a single, given race remains stuck at just 0.06. But that's a different question to what the OP is asking. He needs to know the chance of winning at least one race (any race) out of a larger number of races (in this case, 8). So it's going to be greater than 0.06 as each individual race gives you a chance of 0.06 and you have multiple chances to getting it. Take the coin toss example. What are the chances of getting heads? 50%, right? Toss it a billion times. What are the chances of the billion-and-oneth toss being heads? Still just 50%, yes. But here's the thing: what are the chances every single coin toss in the preceding billion was NOT heads? Virtually zero (and not 50%). Even though each event is independent, there's still a culmulative effect on the distribution of events over a sample. Moreover, it's a binomial distribution rather than a normal distribution because the variables are discrete rather than continuous i.e. "heads or tails", "win or lose". You can't get a little heads or a partial win. So you need to calculate the cumulative binomial probability over the sample size.
You're right about combinations. For example, 2 coin tosses. Results would be the folllowing: TT, TH, HT, HH Each possibility has a 0.5^2 chance which is 0.25. The probability of one tail and and one head is 0.5^2 + 0.5^2 instead of just 0.5^2. Therefore the probabilty of winning 1 race out of 8 should not be just 0.06*0.94^7 but 8(0.06*0.94^7), taking into account WLLLLLLL, LWLLLLLL, LLWLLLLL, LLLWLLLL, LLLLWLLL, LLLLLWLL, LLLLLLWL, LLLLLLLW. After adding up all the different combinations, including losing all 8, we will get 1. The probability of winning at least one race will be just under 0.4. Which still is an odd answer if you ask me. I'm skeptical that out of 30 days, there will be about 12 I would win at least once.
I did fine at obscure, theoretical grad level math. But a basic probability and stats class kicked my butt. I just couldn't wrap my head around the concepts.