Need help with probability q's

[jammol7]

Please help me w these questions! I suck at probability

In a game a coin is tossed 5 times. If all 5 are heads, you get $36. Otherwise, you lose the money you bid. What's the highest bid you would make? What's your expected profit?

What I have so far

Probability of winning is 1/32. So expected payoff is 1/32 x 36 which is 1.125. So would the highest bid just be $<1.125?? Like I bid $1.12 and my expected profit is half a cent?? It doesnt sound right to me

What if the same game is played 10,000 times? What's your highest bid?

What I have so far

Ok I don't understand this question at all. All the games are independent so why should my bid be different?

What kind of distribution is this?

What I have so far

Idk whether the answer should be binomial or normal distribution. Binomial bc there's only either heads or tails. But normal bc it's being played 10,000 times so would it be normal?

pls help!! Thank you!

[jammol7]

Bump!

I'm especially confused about question 2, can anyone pls explain? Thank you!!

[1729]

Please help me w these questions! I suck at probability

Don't worry, probability is also my weakest in math! All I am certain of is the probability of me telling probability to kiss my ass is 100%.

This is probably too late, but I thought I give a crack at your problem, hope this helps!! (:

In a game a coin is tossed 5 times. If all 5 are heads, you get $36. Otherwise, you lose the money you bid. What's the highest bid you would make? What's your expected profit?

What I have so far

Probability of winning is 1/32. So expected payoff is 1/32 x 36 which is 1.125. So would the highest bid just be $<1.125?? Like I bid $1.12 and my expected profit is half a cent?? It doesnt sound right to me

Yep that sounds correct, great job. Alternatively you can do your answer * 10000 and you'd get the same answer.

What if the same game is played 10,000 times? What's your highest bid?

Do you know the expected number for a binomial distribution? Note that the bid doesn’t change but your expected profit does.
Essentially for this one you want the expected number of wins from 10000 trials and a probability of 1/32
Furthermore if you calculated the expected number of wins you'd know the expected number of money you'd get from the game so you can bid any amount less than this value, and you'd be golden, since probabilistically you'd get all that money back anyway.
And you're looking for the zero-sum value. Aka the maximum bid before you start statistically losing money, which should be should go like

This

10000/32*36=11250

. Given that you should get the answer.

What kind of distribution is this?

It remains a binomial distribution, but if you increase the number of trials that high, it can be approximated further accurately by a normal distribution. This part however, remains ambiguous, the distribution for one game is binomial but the distribution of outcomes for 10,000 games is normal.

[colline]

Knowing this information the answer should go like 10000/32*36=11250

Yes that's correct! But don't forget $11,250 is your expected payoff, so your actual bid would be less than that.

It remains a binomial distribution, but if you increase the number of trials that high, it can be approximated further accurately by a normal distribution. This part however, remains ambiguous, the distribution for one game is binomial but the distribution of outcomes for 10,000 games is normal.

Yep it's binomial. Just fyi, CLT isn't actually on the methods study design.

[keltingmeith]

Just fyi, CLT isn't actually on the methods study design.

This isn't actually an application of CLT*, this uses the fact that the normal distribution approximates a Binomial distribution. The larger the number of trials, the better the approximation. Is required for methods - but if I'm reading the study design properly, not for this year. So maybe I'm unnecessarily cutting hairs? Eh

*=I mean, yes, CLT proves this fact, but when has methods cared about methods of proof before?

[colline]

This isn't actually an application of CLT*, this uses the fact that the normal distribution approximates a Binomial distribution.

Sorry my bad! I was more referring to the OP, since he was asking would the distribution be normal when n is large.

The larger the number of trials, the better the approximation.

I may be wrong as I haven't really touched probability past VCE, but is this LLN?

[keltingmeith]

I may be wrong as I haven't really touched probability past VCE, but is this LLN?

Oh boy, I haven't gotten to talk probability like this since undergrad

But no, not quite. The law says that given enough trials, the sample and true mean will be the same thing with a probability of 1. Exactly how they're the same thing (that is, the type of convergence) depends on if you're applying the weak or strong law. The weak law says that the distance between the true mean and sample mean will always be greater than some epsilon with probability 0, the strong law says that the two are equal with probability 1. They sound like the same thing, but they're not - we say that the weak law converges in probability, whereas the strong law converges almost surely. The difference between these two is a question for an analysis class

Okay, so why is it that more trials = better approximation? Because CLT. As I said, CLT isn't required for methods, but it does prove this one result that they require 😅

But also, this is getting off topic really quickly and into second and third year uni maths territory. Feel free to take further questions to my inbox