[bushindo]

This problem is inspired by a recent viewing of the famous Press Your Luck show where Michael Larson memorized the lights patterns and broke CBS's bank.

Assume that a contestant on a game show has 10 spins. Each spin of the wheel might land on a one of these outcomes with equal probability. The possible outcomes on the spins are

(Whammy, 1000 dollars, 3000 dollars, 500 dollars, 2000 dollars )

Each contestant starts with 0 dollars, and accumulates money depending on what outcome they land on. Landing on a Whammy means that the current money is wiped out, and the amount is reset to 0. Landing on a Whammy does not affect the number of spins remaining for the player.

1) What is the expected winnings after 10 spins ?

2) In this scenario, the wheel is modified to contain 1 additional outcome. The chance of landing on any outcome is 1/6

(Whammy, 1000 dollars, 3000 dollars, 500 dollars, 2000, 1000 dollars + 1 extra spin )

Landing on the last outcome to add 1000 dollars and 1 extra spin to the player's account. If a player starts with 10 spins, and continue until all spins are gone, what is the expected earnings?

[Guest]

Is the answer to the first one...

4057.237 ?

Edited May 24, 2009 by psychic_mind

[bushindo]

Is the answer to the first one...

4057.237 ?

Simulations says so. Upon closer inspection, this problem is harder than i thought. I thought i had an analytical solution, but now it's back to the drawing board.

[Guest]

Thats alot to calculate, because of where and when you land on the whammy, and wipes you out. I would like to see how that is calculated.

[Guest]

OK I'll explain what i did.

If you win money on a spin then on average you'll win (500+1000+2000+3000)/4 dollars or 1625.

If you are unlucky enough to get the whammy on spin 10 (1/5 chance) you'll win 0.

The chance that you'll get a whammy on spin 9 and not 10 is: 1/5 * 4/5

The chance taht you'll get a whammy on spin 8 and not 9 and not 10 is: 1/5 * 4/5 * 4/5

...and so on...

The chance that you wont get a whammy is (4/5)^10.

Now, if you got your last whammy on say spin 6, then you'll have 4 spins of money coming which would be 1625*4. So now we multipy probability by money won and:

(1/5 * 0) + (4/5 * 1/5 * 1625) + ((4/5)^2 *1/5 * 2 * 1625) + ... + (4/5)^10 * 10 * 1625

i think I missed the very last part of that calculation before. SO my revised answer is: 5802.067814

Even if my method is correct (which it may well not be) I may have done the calculation wrong so it would be good if someone could check.

[Guest]

nevermind

Edited May 24, 2009 by James8421

[Guest]

nevermind

There are 5^10 outcomes if im not mistaken. Which is about 10 million. You can try that method if you like .

[Guest]

There are 5^10 outcomes if im not mistaken. Which is about 10 million. You can try that method if you like .

I realized that almost immediatly after I posted it , thats why I edited it

[Guest]

I tried to write an equation to get the answer..

outcomes: 3000, 2000, 1000, 500 and whammy

on the first spin on average you will get $1300

now if the whammy was a just $0 and not a wipe out then every spin you will get $1300, every spin the prob. of a whammy is 1/5, but your average decreases in larger increments every spin.

1300

+ 1300*(1 - 1/5)

+ 1300*(1 - 1/5 - (1/5)^2)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4 - (1/5)^5)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4 - (1/5)^5 - (1/5)^6)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4 - (1/5)^5 - (1/5)^6 - (1/5)^7)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4 - (1/5)^5 - (1/5)^6 - (1/5)^7 - (1/5)^8)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4 - (1/5)^5 - (1/5)^6 - (1/5)^7 - (1/5)^8 - (1/5)^9)

= $10,156.25

[bushindo]

OK I'll explain what i did.

If you win money on a spin then on average you'll win (500+1000+2000+3000)/4 dollars or 1625.

If you are unlucky enough to get the whammy on spin 10 (1/5 chance) you'll win 0.

The chance that you'll get a whammy on spin 9 and not 10 is: 1/5 * 4/5

The chance taht you'll get a whammy on spin 8 and not 9 and not 10 is: 1/5 * 4/5 * 4/5

...and so on...

The chance that you wont get a whammy is (4/5)^10.

Now, if you got your last whammy on say spin 6, then you'll have 4 spins of money coming which would be 1625*4. So now we multipy probability by money won and:

(1/5 * 0) + (4/5 * 1/5 * 1625) + ((4/5)^2 *1/5 * 2 * 1625) + ... + (4/5)^10 * 10 * 1625

i think I missed the very last part of that calculation before. SO my revised answer is: 5802.067814

Even if my method is correct (which it may well not be) I may have done the calculation wrong so it would be good if someone could check.

psychic_mind got problem 1. Onto problem 2.

[Guest]

I tried to write an equation to get the answer..

outcomes: 3000, 2000, 1000, 500 and whammy

on the first spin on average you will get $1300

now if the whammy was a just $0 and not a wipe out then every spin you will get $1300, every spin the prob. of a whammy is 1/5, but your average decreases in larger increments every spin.

1300

+ 1300*(1 - 1/5)

+ 1300*(1 - 1/5 - (1/5)^2)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4 - (1/5)^5)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4 - (1/5)^5 - (1/5)^6)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4 - (1/5)^5 - (1/5)^6 - (1/5)^7)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4 - (1/5)^5 - (1/5)^6 - (1/5)^7 - (1/5)^8)

+ 1300*(1 - 1/5 - (1/5)^2 - (1/5)^3 - (1/5)^4 - (1/5)^5 - (1/5)^6 - (1/5)^7 - (1/5)^8 - (1/5)^9)

= $10,156.25

Hey, I think my previous answer was wrong...

this logic gets me to the same ans as psychic_mind.

i tried to extrapolate the equation form a coin toss.. if H then $10, if T then whammy...

so

Toss Outcome 1 Outcome 2 Avg Equation

1 10 0 5 10/2

2 5+10 0 7.5 10/2 + 10/4

3 7.5+10 0 8.75 10/2 + 10/4 + 10/8

similarly

1300*[ (4/5)^0 + (4/5)^1 + (4/5)^2 + (4/5)^3 + (4/5)^4 + (4/5)^5 + (4/5)^6 + (4/5)^7 + (4/5)^8 + (4/5)^9]

= $5,802.06

Now on the first try you will get an avg. of $1300

on the second try you will get $3000/5, 2000/5, 1000/5, 500/5 or a wipe out which is -1300/5

i.e 1300*(1-1/5)

now the next spin, will get $3000/5, 2000/5, 1000/5, 500/5 or a wipe out which is -(1300 + 1300*(1-1/5))/5

the equation now is:

1300*( 1

+ (1-1/5) //.8

+ (1- (1+0.8)/5) //0.64

+ (1- [1+0.8+0.64]/5) //0.512

+ (1- [1+0.8+0.64+0.512]/5) //0.4096

+ (1- [1+0.8+0.64+0.512+0.4096]/5) //0.32768

+ (1- [1+0.8+0.64+0.512+0.4096+0.32768]/5) //0.262144

+ (1- [1+0.8+0.64+0.512+0.4096+0.32768+0.262144]/5) //0.2097152

+ (1- [1+0.8+0.64+0.512+0.4096+0.32768+0.262144+0.2097152]/5) //0.16777216

+ (1- [1+0.8+0.64+0.512+0.4096+0.32768+0.262144+0.2097152+0.16777216]/5) //0.134217728

)

=1300(1+0.8+0.64+0.512+0.4096+0.32768+0.262144+0.2097152+0.16777216+0.134217728)

=$5,802.06

nevermind, psychic_mind 's ans has been confirmed

Edited May 24, 2009 by adiace

[Guest]

ingore this im redoing this now i forgot i deleted a part for the initial test and didnt undo that one sec

ok my math ill post in the bottom but i used java code for it its a little confusing but it gives an answer also i couldn't remember how to do powers so those are the separate loops the code is embarrassing but it works in a rush

for the first loop (extra space but dont land on it) i got 3200.... but the final answer after 100 outer loops is 3840.9006567847555

the copyright symbol is a c in parenthesis and the wink guy is a b in parenthesis

and the system.out.println are for debugging and spot checking

counter2=-1;

answer=0;

counter=0;

while(counter2<100)

{

counter2++;

counter=0;

while(counter<10)

{

a=1;

b=1;

c=1;

counter3=0;

while(counter3<counter)

{

c=c*4;

counter3++;

}

counter3=0;

while(counter3<counter+1)

{

b=b*6;

counter3++;

}

counter3=0;

while(counter3<counter2)

{

a=a*6;

counter3++;

}

answer=answer+((1/a)*(©/(*(1625*counter));

counter++;

System.out.println(" a " + a+ " b " + b + " c " + c + " answer " + answer);

}

answer=answer+1/a*1744.830464;

System.out.println("answer " +counter2+ " " + answer);

}

System.out.println("answer " + answer);

}

}

the start is the exact same except instead of 4/5's its 4/6 thats the first round

the secondround you multiply every cell by 1/6 and add 1000 like so

4/6*1/5*1/6(1625+1000)+(4^2/6^3)(1625**2+1000)+...

third round multiply bu 1/6^2 and add 2000 so on so forth this is what i got

also i used this exact program for the first answer and it worked i just set the outer loop to 0 loops and set a =1 and b to b=b*6 anyway my last program had a small error in it but im hopeful

Edited May 25, 2009 by final

[Guest]

ok wouldnt let me edit again for some reason so here is everything same as above except small change in java and answer and everything else

im almost positive the program is right and pretty sure the math is too but i wont be surprised either way

i got first loop still 3200... but 100th outer loop 4198.819673194855

b in parenthesis was for some reason changed to wink guy

c in parenthesis was changed to copyright

system.out.println for debug purposes but i kept it in case anyone was actually going to try this code

counter2=-1;

answer=0;

counter=0;

while(counter2<100)

{

counter2++;

counter=0;

while(counter<10)

{

a=1;

b=1;

c=1;

counter3=0;

while(counter3<counter)

{

c=c*4;

counter3++;

}

counter3=0;

while(counter3<counter+1)

{

b=b*6;

counter3++;

}

counter3=0;

while(counter3<counter2)

{

a=a*6;

counter3++;

}

answer=answer+((1/a)*(©/(*(1625*counter+counter2*1000));

counter++;

System.out.println(" a " + a+ " b " + b + " c " + c + " answer " + answer);

}

answer=answer+1/a*(1744.830464+counter2*1000);

System.out.println("answer " +counter2+ " " + answer);

}

System.out.println("answer " + answer);

}

math is the same and i still no my programming style is sloppy but i was rushing and if you want me to comment it send me a self addressed envelope and ill mail it to you i swear

[Guest]

this is annoying as hell it wont let me edit my own post. thats messed up now i have three post in a row and look like almost as much of an idiot as i am...oh well

had to edit again forgot to change something else from the test case now im not too sure the answer because its so low but i cant find an error maybe you guys can

ok wouldnt let me edit again for some reason so here is everything same as above except small change in java and answer and everything else

im almost maybe kinda positive the program is right and knda pretty sure the math is too

but 100th outer loop 2443.1819163948576

b in parenthesis was for some reason changed to wink guy

c in parenthesis was changed to copyright

system.out.println for debug purposes but i kept it in case anyone was actually going to try this code

counter2=-1;

answer=0;

counter=0;

while(counter2<100)

{

counter2++;

counter=0;

while(counter<10)

{

a=1;

b=1;

c=1;

counter3=0;

while(counter3<counter)

{

c=c*4;

counter3++;

}

counter3=0;

while(counter3<counter+1)

{

b=b*6;

counter3++;

}

counter3=0;

while(counter3<counter2)

{

a=a*6;

counter3++;

}

answer=answer+((1/a)*(©/(*(1625*counter+counter2*1000));

counter++;

System.out.println(" a " + a+ " b " + b + " c " + c + " answer " + answer);

}

//below is the change had forgot to change it from 4/5^10 to 4/6^10

answer=answer+1/a*(281.7998611+counter2*1000);

System.out.println("answer " +counter2+ " " + answer);

}

System.out.println("answer " + answer);

}

math is the same and i still know my programming style is sloppy but i was rushing and if you want me to comment it send me a self addressed envelope and ill mail it to you i swear

[Guest]

i hate when i do this but ignor most of what i say the essence is there but i forgot some a choose b stuff so its wrong for now

i think its close in theory but probably pretty far away in number working on it now but i wouldnt hold your breath (unless u want to pass out)

[Guest]

modifying my method for the 1st part for the 2nd part

1st spin: 1250 + nextspinavg./6

second spin: 1250 -sumofallpreviouspinavg./6+ nextspinavg./6

so the nth spin is requires the result of (1) sumofallpreviouspinavg. and (2) nextspinavg

writing a formula for the nth term

n0=0

and n = number of the spin

nthterm = 1250 - [summation n=1 to n {n-1thterm}]/6 + n+1thterm/6

the answer is summation n=0 to n=infinity { nthterm }

i.e.:

n0+n1+n2+...

= 0 + (1250 + (n2)/6) + [1250 - (n0+n1)/6 + n3/6] + ....

[Guest]

The equation i posted in my previous post seems correct.. but is grossly impractical. But you never know, there is a lot of math specifically designed to deal with the impractical

[Guest]

hey sorry about all my crappy post before but im so close need help if any is out there

so i used a computer program to do the math since there are 100 equations so what i did was what psychic i think first posted for the first part but then you need to do all of those plain (but with 6ths of course). then same formula times 1/6 and add 1000 for +spin to 1625 then times 1/36 +2000 anyway the numbers die at about 1/6^9 but anyway thats were i was stuck for a while. but then i realized you needed to multiply by how many normal money you got + how many (+spin) you got choose +spins

then i was stuck again untill i realized you couldnt end on a +spin so i subtracted one from my chooses

stuck again

then i figured it out (kinda) you can now get to the whammy with any number of +spins which is 1/6+1/6^2...=6/5 atleast for the 9 spins left case i took this as fact for all and ran with it and got the answer im about to give but i cant seem to prove it for the rest of them but i ran a simulation and it is pretty gosh darn by golly close

oh and i didnt multiply the no whammy case by 6/5ths for obvious reasons. my code is below

6516.052877185571

so itso facto im done and am 98.9% positive but it is really late so you never know

wham is the name of class all of this is obviously in a main except for factorial method and variables are declared and initialized (as specified or just make them double never mind i just through them in)

public static double answer;

public static int counter, counter2,counter3;

public static double a,f;

public static long fact;

public static double b,c,e;

counter2=-1;

answer=0;

counter=1;

fact=0;

wham r= new wham();

while(counter2<11)

{

counter2++;

counter=1;

while(counter<10)

{

a=1;

b=1;

c=1;

counter3=0;

while(counter3<counter)

{

c=c*4;

counter3++;

}

counter3=0;

while(counter3<counter+1)

{

b=b*6;

counter3++;

}

counter3=0;

while(counter3<counter2)

{

a=a*6;

counter3++;

}

fact= ((r.factorial(counter+counter2-1))/((r.factorial(counter2))*(r.factorial(counter-1))));

e=((1/a)*(©/(*(1625*counter+counter2*1000)*fact);

answer=answer+1.2*((1/a)*(©/(*(1625*counter+counter2*1000)*fact);

System.out.println(" "+counter2+". " +counter + " + change " + e);

counter++;

}

fact= ((r.factorial(10+counter2-1))/((r.factorial(counter2))*(r.factorial(10-1))));

e=(1/a*c*4/b*(1625*10+counter2*1000)*fact);

answer=answer+(1/a*c*4/b*(1625*10+counter2*1000)*fact);

System.out.println("change " +counter2+ " " + e);

System.out.println("answer " +counter2+ " " + answer);

}

System.out.println("answer " + answer);

}

public long factorial(double number2)

{

long number= (long)number2;

if ( number <= 1 ) // base case

return 1;

else

return number * factorial( number - 1 );

}

}

so i would love if someone could tell if im right and why that 6/5 thing worked on all of it

[Guest]

None of these answers make any sense.

Since the possibilities are 500, 1000, etc, the amount won has to be a multiple of 500.

[Guest]

were not actually looking for the amount you will win but the expectance or expected value. this is a mathematical property that equals the probability of each outcome multiplied by its individual value which can also be found by averageing all your outcomes in a simulation.

[Guest]

Another possible way to solve without using computer programs:

Each turn either gives a positive amount or a "negative" (whammy means negative of the amount bagged until the start of current turn)

So, the expected positive amount per turn is (1000 + 500 + 2000 + 3000)/4 = 1625

Now, simulate 10 scenarios and each time, place the whammy at a different turn. So, once case is where there is no whammy in all 10 turns, second whammy on 9th turn (T9), third whammy on T8.... whammy on T1 and all others positive. For each time, assign probability of 1/5 for whammy and 4/5 for a positive outcome. Sum up the expected winning in all scenarios and you have the total expected winning at the end of 10 turns (5802).

See attached picture below.

[Guest]

So if I tell you that there are 10 apples, and you can carry 3 at once, and you have to take them across the room, you'll tell me it would take you 3.33 trips? Logic is a large factor in puzzles... but I see what you're saying.

[Guest]

So if I tell you that there are 10 apples, and you can carry 3 at once, and you have to take them across the room, you'll tell me it would take you 3.33 trips? Logic is a large factor in puzzles... but I see what you're saying.

No this is different from the example you gave. Unlike the mode and median, the mean never has to be a value which is in the set of data.

Anyway. I ran a simulation written in C...

about 6500

I think final got that.

[Guest]

first 10 spins

1250*[(5/6)^0 + (5/6)^1 + (5/6)^2 + (5/6)^3 + (5/6)^4 + (5/6)^5 + (5/6)^6 + (5/6)^7 + (5/6)^8 + (5/6)^9]

11th spin: 1250*(5/6)^10 * the prob of getting atleat one extra spin in 10 spins, this can be calculated using binomial distrituion

1250*(5/6)^10 * (.8385)

12th spin: 1250*(5/6)^11 * the prob of getting atleat two extra spin in 11 spins

1250*(5/6)^11* (.5693)

.

.

.

therefore the answer is:

1250*[(5/6)^0 + (5/6)^1 + (5/6)^2 + (5/6)^3 + (5/6)^4 + (5/6)^5 + (5/6)^6 + (5/6)^7 + (5/6)^8 + (5/6)^9

+(5/6)^10 * (.8385)

+(5/6)^11* (.5693)

+(5/6)^12* (.3226)

+(5/6)^13* (.1581)

+(5/6)^14* (.0690)

+(5/6)^15* (.0274)

]

(limiting to 16 tries as the prob. of spinning after that is less than 1%)

= $6,626.85

[bushindo]

first 10 spins

1250*[(5/6)^0 + (5/6)^1 + (5/6)^2 + (5/6)^3 + (5/6)^4 + (5/6)^5 + (5/6)^6 + (5/6)^7 + (5/6)^8 + (5/6)^9]

11th spin: 1250*(5/6)^10 * the prob of getting atleat one extra spin in 10 spins, this can be calculated using binomial distrituion

1250*(5/6)^10 * (.8385)

12th spin: 1250*(5/6)^11 * the prob of getting atleat two extra spin in 11 spins

1250*(5/6)^11* (.5693)

.

.

.

therefore the answer is:

1250*[(5/6)^0 + (5/6)^1 + (5/6)^2 + (5/6)^3 + (5/6)^4 + (5/6)^5 + (5/6)^6 + (5/6)^7 + (5/6)^8 + (5/6)^9

+(5/6)^10 * (.8385)

+(5/6)^11* (.5693)

+(5/6)^12* (.3226)

+(5/6)^13* (.1581)

+(5/6)^14* (.0690)

+(5/6)^15* (.0274)

]

(limiting to 16 tries as the prob. of spinning after that is less than 1%)

= $6,626.85

adiace and final got it. Well done.