|
#1
|
|||
|
|||
Dice Game
This could probably go in the probability forum, but theres way too much philosophy around here and not enough math so lets give it a shot.
A friend of mine posed this to me after he was asked it on a job interview, heres the game, its pretty simple. You roll a fair 6 sided die and are given an amount of money equal to the number you roll, 1=$1 2=$2...etc. The caveat is that if you wish to reroll you may, but would then be stuck with whatever you get the second time. The question is, how much would you be willing to pay to play this game? read: what is the EV (this is what he was asked to figure out on the spot) A slightly less trival extension, what if you get up to 6 rolls, meaning you can choose to reroll 5 times before you are stuck with the 6th. (he was given this as 'homework') Finally and most interestingly, how about generalizing it for n rolls...just for kicks |
#2
|
|||
|
|||
Re: Dice Game
Clearly if you get a 1-3 you should reroll and if you get a 4-6 you should keep. If the game was played with only one roll the $EV is 3.5. Thus the EV of this game is:
1/6 * (4) + 1/6 * (5) + 1/6 * (6) + 1/2 * (3.5) = $4.25 |
#3
|
|||
|
|||
Re: Dice Game
[ QUOTE ]
Clearly if you get a 1-3 you should reroll and if you get a 4-6 you should keep. If the game was played with only one roll the $EV is 3.5. Thus the EV of this game is: 1/6 * (4) + 1/6 * (5) + 1/6 * (6) + 1/2 * (3.5) = $4.25 [/ QUOTE ] Very nice, but thats the easy part. I'm guessing most people who hang out around here could get that pretty quick, I'm more interested in the n-generalization since I have an answer but its not all that elegant, wondering if someone can come up with a better approach. |
#4
|
|||
|
|||
Re: Dice Game
Well, I thought about it for awhile. The answer is definitely 5 - 3/(2^n), though I'm having trouble proving it. I'll think about it some more.
|
#5
|
|||
|
|||
Re: Dice Game
If you had infinite rolls, wouldn't you expect to end up with a 6?
|
#6
|
|||
|
|||
Re: Dice Game
[ QUOTE ]
If you had infinite rolls, wouldn't you expect to end up with a 6? [/ QUOTE ] But you don't have infinite rolls. You have a certain number of rolls and then you stop. Hm. Maybe my optimal strategy is not, in fact, optimal. If you have 20 rolls left and you roll a 5, my strategy says stop. That is clearly wrong, but it's where my answer comes from. |
#7
|
|||
|
|||
Re: Dice Game
[ QUOTE ]
You roll a fair 6 sided die and are given an amount of money equal to the number you roll, 1=$1 2=$2...etc. The caveat is that if you wish to reroll you may, but would then be stuck with whatever you get the second time. The question is, how much would you be willing to pay to play this game? read: what is the EV [/ QUOTE ] The average for 1 die is 3.5, so keep 4,5,6 (avg = 5) and roll 1,2,3 again. The EV is (1/2)*5 + (1/2)*3.5 = $4.25 for 2 dice. [ QUOTE ] A slightly less trival extension, what if you get up to 6 rolls, meaning you can choose to reroll 5 times before you are stuck with the 6th. [/ QUOTE ] EV(n) = EV for n rolls. Work backward from 2 remaining rolls. Since EV(2) = 4.25, keep 5+ (avg = 5.5) and roll 1,2,3,4: 2 rolls left, keep 5+, EV(3) = (1/3)*5.5 + (2/3)*4.25 = 4 + 2/3 3 rolls left, keep 5+, EV(4) = (1/3)*5.5 + (2/3)*(4 2/3) = 4 + 17/18 4 rolls left, keep 5+, EV(5) = (1/3)*5.5 + (2/3)*(4 + 17/18) = 5 + 7/54 5 rolls left, keep 6, EV(6) = (1/6)*6 + (5/6)*(5 + 7/54) = 5 + 84/325 =~ $5.27 for 6 dice. [ QUOTE ] Finally and most interestingly, how about generalizing it for n rolls...just for kicks [/ QUOTE ] For n > 6, EV(n) = (1/6)*6 + (5/6)*EV(n-1) recursion: EV(n) = 1 + (5/6)*EV(n-1) EV(6) = 5.27 EV(7) = 1 + (5/6)*5.27 EV(8) = 1 + 5/6 + (5/6)^2 * 5.27 EV(9) = 1 + 5/6 + (5/6)^2 + (5/6)^3 * 5.27 ... EV(n) = sum [i = 0 to n-7](5/6)^i + (5/6)^(n-6) * 5.27 from finite sum geometric series: EV(n) = [1 - (5/6)^(n-6)]/(1 - 5/6) + (5/6)^(n-6) * 5.27 EV(n) = 6 - 6*(5/6)^(n-6) + 5.27 * (5/6)^(n-6) EV(n) = 6 - 0.73*(5/6)^(n-6), for n > 6 dice. |
#8
|
|||
|
|||
Re: Dice Game
Looks like you worked through this a lot like I did, but I like your presentation here, makes it nice and easy to follow. I had it down as a messy recursion type deal, but that last line there
[ QUOTE ] EV(n) = 6 - 0.73*(5/6)^(n-6), for n > 6. [/ QUOTE ] seems to sum it up nicely in the closed form I was looking for. |
#9
|
|||
|
|||
Re: Dice Game
[ QUOTE ]
Looks like you worked through this a lot like I did, but I like your presentation here, makes it nice and easy to follow. I had it down as a messy recursion type deal, but that last line there [ QUOTE ] EV(n) = 6 - 0.73*(5/6)^(n-6), for n > 6. [/ QUOTE ] seems to sum it up nicely in the closed form I was looking for. [/ QUOTE ] So I get the job? |
#10
|
|||
|
|||
Re: Dice Game
[ QUOTE ]
So I get the job? [/ QUOTE ] Hired! |
Thread Tools | |
Display Modes | |
|
|