Fantasy football fans will like this one; Ticket with two games, Game 1 and Game 2. Game 1: teams A vs. B with 60/40 slant of team A being the favorite over team B (underdog) Game 2: teams C vs. D with 50/50 even split to win between C & D Standard rules, the winning bettors split the total pot. 100 players, each one wagering $1. If you know in advance that 99 other players are betting on team A to win, which game (or team) do you bet on to maximize your returns? Bon appetite a tout!
Geez, I don't know if it's old age, fear of Alzheimers, or insanity, but for the past year I've been enjoying questions like these. I remember(guess it can't be Alzheimers then) in school getting sick of this stuff real quick.There must be a new fad for probability like questions, as I keep running into them in magazines, here and other places. Now to sound off my rocker, it'd be cool to have a thread with these and the poster posts the answer a week later. Possibly interview questions, or just problems people have come across over the years.
Neither- trading isnt like a wagetr in a tru sense- I cant increase or decrease betting size real time- its a win lose.
40% chance to win $99 vs 60% to win $1. How does the second game fit in? Something seems to be missing............
For example lets say you boil it dow and "odds" are that it should be choice one- Your trading and seeing that many people are taking choice one and its going against them= I sell you can make real time choice while its unfolding- and cover at break even if it turns back to the odds, probably not go long an reverse but cover- get a free look at a trend change if you will
like corbetcps said, its not like trading in the true sense since you cant increase or decrease your wager, and in real trading you would probably consider the fact that everyones betting on A as a major indicator to buy A and sell B, however the possibility of maximizing your bet if I remember correctly: $99 are being voted on Team A. Probability of Team B beating team A and team C winning: P(B+C) = (0.4) X (0.5) = (0.2) = 20% chance that you will win the entire pot. My answer would be to bet on the second game, on either team. ^ I wrote this, thought about it, and came back. If no one is betting on the second game, you're probably better off betting on team B, with a 40% chance of winning the whole pot. If you bet on A and they win, all you're getting back is your dollar... Now I'm friggin confused. Post the dam answer soon!
betting game 2 is a non win bet as its probability is 50/50 and its assumed the bettors will split the bet between teams C and D. betting team A doesnt seem worthwhile as even if you win you'll wind up splitting with 99 other winners,ergo team B is the bet i think
squirly309 - too funny - I agree post dam answer. In trading - WE BOTH can take the different team and win ! Its tru.......
I'd say: Risk vs Reward scenario does not make the trade worth taking. Stay in cash until a better opportunity comes along
This might be one of those questions that might be easier than everyone thinks it is. 40% chance of winning the 100 on the first game. And there is a 20% chance of winning it using second game. This stupid thing has been driving me nuts. What am I missing?