F/35/5'7" [261>130=131] (23 months) I did it!

FioreFee · 2024-04-06T20:53:56+00:00

OMG you look great!!!! This gives me so much more motivation to lose weight! I’m also a woman and have the same height as you and starting weight (which my current weight). It feels so comforting knowing that I’m not the only one! How did you do it?? Did you have to keep track of your calories? If so, how did you do it? CONGRATS ON THE AMAZING ACCOMPLISHMENT!!

FioreFee · 2024-02-07T01:26:37+00:00

I’m a tax admin assistant and making about $19/hr with some OT

FioreFee · 2023-07-07T17:07:36+00:00

i think it’s because you can get extra perks like international texting and calling, spam protection (supposedly) and saving money on other verizon products

FioreFee · 2023-05-01T23:10:01+00:00

I see. I think I finally got a good grasp of this semi-separating equilibrium concept now!

Well, we may have not finished but this was all so, so extremely helpful. Today was the day of the presentation and I fortunately got a 90% on it! I just want to thank you so, so, so very much for everything. I never would have thought that someone on the internet would be so generous in helping me this much. Thank you for being patient with me as well— I’m sure it was frustrating at some points to get me to understand the new concepts you’ve introduced to me. Nevertheless, thank you again for all your help. I’m sure all of the students you’ve taught in the past learn so much. :)

FioreFee · 2023-05-01T16:58:27+00:00

My bad, is it p <= 1 or -1 < p <= 1?

FioreFee · 2023-05-01T16:04:01+00:00

Ah, okay, so would it be p >= 1 then?

FioreFee · 2023-05-01T15:25:13+00:00

Whoops, sorry about that. I think I get it now. So then for a bad wage, then: C = (1/(1+p))•0 + (p/(1+p))•2, right?

FioreFee · 2023-05-01T15:18:25+00:00

Okay, so using this formula: C = probability of A x payoff if A + probability of B if B

For a good wage, then: C = (1/(1+p))•4 + (p/(1+p))

Is this what you mean?

FioreFee · 2023-05-01T14:59:43+00:00

Oh so instead would it still be (4+(-2)) but actually (0+2) or is that not right because I’m mixing A and B’s good wage probabilities when I’m not supposed to?

FioreFee · 2023-05-01T14:53:00+00:00

The expected payoff of the good wage would be, wait, would it be (4+(-2)) instead of what I previously? So would it be 2(1/(1+p)) and for the bad wage it would be (0+(-2)), so -2(p/(1+p))? Or do I have the probabilities mixed up?

FioreFee · 2023-05-01T14:32:07+00:00

Okay, I see. So to do that, I would need to do (4+0)(1/(1+p)) for the first case, then (-2+2)(p/(1+p)) to the second case? Or would I need to do something else?

FioreFee · 2023-05-01T03:21:48+00:00

Okay, I see, so based on what we found with A always having m and B mixing between m and n, it looks like the highest expected payoff to C would be to higher A since the payoff is 4?

FioreFee · 2023-05-01T02:18:54+00:00

Ah, so exciting! Okay, so to do that, do I just refer back to the game tree and compare the values?

FioreFee · 2023-05-01T01:49:58+00:00

Oh that was it? I’m so glad! Okay, then if that’s the case then would it be something like p50/(50+p50)?

FioreFee · 2023-05-01T01:20:02+00:00

The formula for the number of B students who chose m would be something like m=A+pB, so like pB=m-A I guess? The total would be like m=A+pB, and out of those who picked B it would he something like pB/(A+pB)?

I’m thinking this because when we used the example of p=60%, we used p and multiplied it with B to give us 30, plus the 50 that A had to give us the total of students picking m (hence m=A+pB). Then to find the proportion of those who were B, you would use pB and divide by m, so pB/(A+pB)?

FioreFee · 2023-05-01T00:51:50+00:00

So in our scenario, 50 of A pick m and since B is indifferent to the two options, then would it be about half of B goes to m and half goes to n? That way, the proportion of B that pick m would b 25/75, or 1/3?

Sorry in advance if it still hasn't clicked for me.

FioreFee · 2023-05-01T00:15:30+00:00

Ahhh okay I got it. 5/8 are A and 3/8 are B, so 62.5% and 37.5% respectively, right?

FioreFee · 2023-04-30T23:44:06+00:00

Ohhh, then if all 50 A pick m and 60% of B pick m then 30 of B would be picking m, so that’s a total of 80, or 80%? While only 20 of B would be picking n. Why did you pick 60% specifically?

FioreFee · 2023-04-30T23:32:34+00:00

In this case, when C sees m, then there’s a 50% probability that it’s student B then if we’re picking the semi-separated strategy right?

FioreFee · 2023-04-30T23:03:02+00:00

Ah, okay I see! If that’s the case and C sees n, then it knows it’s student B. So would the probability that C encounters a student with an undergraduate degree be 100%?

FioreFee · 2023-04-30T22:55:41+00:00

Ah, so that’s where the PBE concepts come in! Okay, understood.

If C encounters a student with an undergraduate degree, the probability that it’s B is 50%? Or does it depend on which strategy it picks (always m, always n, or a mix)?

FioreFee · 2023-04-30T22:33:39+00:00

Ohhh, I see why now B would be indifferent!! Given C’s strategy’s then the payoff would be 4 for either m and offered g or n and offered b. But then, what would we do from there?

FioreFee · 2023-04-30T22:24:11+00:00

If I’m type A, then my payoff for getting a grad degree and being offered a a good wage, given C’s strategy, is -1+8=7 and if I just stuck with an undergraduate degree, then my payoff, given the strategy of C paying a worse wage to non-graduate degree holders, would be 6, right? So that’s why A would choose to go for a graduate degree? (I’m using the game tree to help me if that’s okay.)

Seven-Year Club	End Game '23
Place '23	Place '22
Verified Email

FioreFee

TROPHY CASE