# Radio Competition - HELP NEEDED!

#### ProducerGeoff

##### New Member
Hi, I'm a producer on a big (US) Radio station. I'm developing a new competition, but I need help with the probability...

The game is simple (although some details are changed here for obvious reasons!) We have an amount of cash in a case... all you have to do is tell us how much is in there to win it.

It's definitely over $20,000 and to win you must guess the EXACT amount in dollars and cents (i.e.$36,745.36) so it's seven digits. After each guess we will reveal whether the amount is HIGHER or LOWER.

What's the average amount of guesses it will take for someone to guess the amount?

Once I have the probability I can work out how big we can make the prizes! Your help is MUCH appreciated!

Thanks.

G

#### GameGod

##### New Member
Your upper bound is $99,999.99 right? Then between the$20,000 and upper bound you need to specify whether there is an even spread, i.e. is $99,999.99 as likely as$20,000, or does the probability change continuously?

#### ProducerGeoff

##### New Member
There's only one prize, and yes it's equally as possible that it's any value in the range.

#### Dason

It depends on the person's guessing strategy. A person that just starts at the lower bound and increases their guess by a penny until they hit the correct amount will take a lot longer (on average) than somebody that starts in the middle and continues to bisect the possibilities until they get it right. Is there a particular strategy you're interested in?

#### ProducerGeoff

My reason for asking is that I want the competition to last for three months. Once a prize has been guessed we'll rewrite the cheque for a new amount. If I have a budget of $100k and it's likely to be guessed once a month (approximately 100 guesses per month) then I'll need to split the prize in three. If it's likely to be guessed on average once every 50 guesses then i'll split the prize in 6. It won't be one person guessing, you'll only get one guess. Then we say whether it's higher or lower and open the phones for the next person to guess, so there would't be a single guessing 'strategy' Is there a way of solving this mathematically? Geoff #### GameGod ##### New Member Bisection is probably the best strategy in terms of probability, so if you plan for your customers to use bisection then you will at worst save some extra money Looks like there are 7,999,998 values to pick to start with. P(Correct on Try 1)=1/7,999,998. P(Correct on Try 2)=7,999,997/7,999,998*2/7,999,998. P(Correct on Try 3)=7,999,997/7,999,998*((7,999,998-2)/7,999,998)*4/7,999,998. P(Correct on Try 4)=(7,999,997/7,999,998)*((7,999,998-2)/7,999,998)*((7,999,998-4)/7,999,998)*(8/7,999,998). Those are my first thoughts. As an early stats student I will leave anything further than that to those who know better. #### spunky ##### Can't make spagetti Is there a way of solving this mathematically? there probably is, but i'm with Dason on this one. without making any assumptions on the guessing strategies people will use, there're probably a myriad of ways in which said probabilites are calculated. if you have evenly-spread values between 0.01 and 99,999.99 it makes sense that people would be more likely to say 'higher' if you're around the lower bound of 0.01 or 'lower' if you're close to the upper bound of 99,999.99.... with (this is me just talking) a 0.5 chance of saying higher/lower if they're around the middle. this is an interesting question actually... we just need someone who's either done research on the probability distribution of guesses in scenarios similar to this one OR if you're willing to give us a strategy to work with. #### Dason ##### Ambassador to the humans When in doubt - simulate! I tried three different guessing strategies and used R to simulate the competition. The first strategy was the bisection method where the next guess is just right in the middle of the remaining values. The second 'strategy' was to just randomly (uniformly) guess a value from the remaining possible values. The third was similar to the second but the guess is randomly (uniform) but only from the middle 50% of the remaining values. Code is below followed by the output Code: # Author: Dason Kurkiewicz # May 30, 2013 # http://www.talkstats.com/showthread.php/44051-Radio-Competition-HELP-NEEDED! # Function to perform the optimal guessing strategy optimalguess <- function(lower, upper){ # optimal strategy round(mean(c(lower, upper))) } # Function that guesses uniformly amoung the possible values randomguess <- function(lower, upper){ # not so great sample(seq(lower, upper), 1) } # Function that guess uniformly amoung only the middle portion # of the remaining possible values randommiddle <- function(lower, upper){ middle <- round(mean(c(lower, upper))) int <- round((upper - lower)/4) # middle half of possible values vals <- seq(middle - int, middle + int) if(length(vals) == 1){return(vals)} sample(vals, 1) } # guessmethod: function that takes two parameters (lower, upper) and returns # a next guess # lower : numeric - lower boundary for possible values (in cents) # upper : numeric - upper boundary for possible values (in cents) simulateCompetition <- function(guessmethod = optimalguess, lower = 2000000, upper = 9999999){ actual <- sample(seq(lower, upper), 1) i <- 1 guess <- guessmethod(lower, upper) while(guess != actual){ i <- i + 1 if(guess < actual){ lower <- guess }else{ upper <- guess } guess <- guessmethod(lower, upper) } i } N <- 1000 optimal <- replicate(N, simulateCompetition(optimalguess)) randommiddle <- replicate(N, simulateCompetition(randommiddle)) random <- replicate(N, simulateCompetition(randomguess )) summary(optimal) summary(randommiddle) summary(random) Code: > summary(optimal) Min. 1st Qu. Median Mean 3rd Qu. Max. 14.00 21.00 22.00 21.91 23.00 23.00 > summary(randommiddle) Min. 1st Qu. Median Mean 3rd Qu. Max. 11.00 22.00 24.00 23.76 26.00 33.00 > summary(random) Min. 1st Qu. Median Mean 3rd Qu. Max. 12.00 27.00 31.00 31.87 36.00 52.00 Obviously the optimal strategy did the best on average and we could expect that the competition will take close to 22 rounds if people guess optimally and at least 50% of the time it will last between 21 and 23 rounds. If they use the strategy of guessing randomly from the middle 50% we could expect the competition to last 24 rounds and at least 50% of the time it will last between 22 and 26 rounds. The full 'random' guessing strategy gives an average of about 31 and we could expect about 50% of competitions to last between 27 and 36 rounds. #### GameGod ##### New Member Sorry to butt in, but I have to ask - by "optimal strategy", do we mean bisection until the right value is reached? #### Dason ##### Ambassador to the humans Sorry to butt in, but I have to ask - by "optimal strategy", do we mean bisection until the right value is reached? Yes. $$\hspace{.3cm}$$ #### spunky ##### Can't make spagetti Sorry to butt in, but I have to ask - by "optimal strategy", do we mean bisection until the right value is reached? it looks like that's what it's doing. i mean, just look at the function of optimalguess. it's taking the middle of the mid-point of the upper and lower bounds. Code: optimalguess <- function(lower, upper){ # optimal strategy round(mean(c(lower, upper))) } Last edited: #### Dason ##### Ambassador to the humans My reason for asking is that I want the competition to last for three months. Once a prize has been guessed we'll rewrite the cheque for a new amount. If I have a budget of$100k and it's likely to be guessed once a month (approximately 100 guesses per month) then I'll need to split the prize in three. If it's likely to be guessed on average once every 50 guesses then i'll split the prize in 6.

It won't be one person guessing, you'll only get one guess. Then we say whether it's higher or lower and open the phones for the next person to guess, so there would't be a single guessing 'strategy'

Is there a way of solving this mathematically?

Geoff
I didn't notice this part before. How much of this type of information will the people calling in have available to them? Will they know the overall budget and all that? Because if you've already given out 70000 in prizes and are running another round then if they knew the budget they could use that info and say "we know that the amount is between 20000 and 30000" (as opposed to 20000 and 99999.99)