Calculate the odds of that!!!
http://xfinitytv.comcast.net/tv/The-...d=FCST_hero_tv
By the way - before Denis jumps on me - it is not homework - so I ain't sho'in nutting.........
Calculate the odds of that!!!
http://xfinitytv.comcast.net/tv/The-...d=FCST_hero_tv
By the way - before Denis jumps on me - it is not homework - so I ain't sho'in nutting.........
“... mathematics is only the art of saying the same thing in different words” - B. Russell
Would that be the odds with the slight circular movement of the container (as shown) before dumping or without?
Btw, Drew is wrong! Jolly Ranchers do not last forever. They begin to dissolve (within their unopened wrappers) several months after they've lodged between cushions or fallen behind headboards, et cetera.
Last edited by mmm4444bot; 07-15-2012 at 08:50 PM. Reason: Grammar
Last edited by mmm4444bot; 07-16-2012 at 02:07 PM. Reason: Restored quote pointer for system testing
I think there is only one car in each die - and by rolling five together you are supposed to get total three cars (I think!). You can get three tries by guessing prices of three products.
“... mathematics is only the art of saying the same thing in different words” - B. Russell
“... mathematics is only the art of saying the same thing in different words” - B. Russell
This game is called Let ’Em Roll. The contestant starts the game with 1 free roll, and by pricing products correctly can win up to 2 more for a total of 3 rolls max. The contestant rolls five dice a maximum of three times in an attempt to win a new car. The five dice are exactly the same: Each has a car pictured on three sides and dollar amounts ($500, $1000, and $1500) on the other three sides. A contestant who rolls cars on all five dice wins the car. If a rolled die does not show a car, the contestant can either take the money shown on the die and leave the game or “freeze” the die or dice showing the car and roll the remaining dice, should the contestant have rolls remaining. If the contestant obtains five dice with cars by the end of three rolls, she or he wins the car; if not, she or he wins the total dollar amount shown on the dice on the last turn.
If we assume the contestant priced correctly and has 3 total rolls:
P(win)=P(5 cars 1st roll)+P(4 cars 1st roll then win)+P(3 cars 1st roll then win)+P(2 cars 1st roll then win)+P(1 car 1st roll then win)+P(0 cars first roll then win)
There is probably a more elegant way but it is small enough for brute force to do the job with some help from pascal's triangle.
P(5 cars 1st roll) = 1/32
P(4 cars 1st roll then win)=(5/32)[1/2+1/4]
P(3 cars 1st roll then win)=(10/32)[1/4+1/2(1/2)+1/4(1/4)]
P(2 cars 1st roll then win)=(10/32)[1/8+3/8(1/2)+3/8(1/4)+1/8(1/8)]
P(1 cars 1st roll then win)=(5/32)[1/16+4/16(1/2)+6/16(1/4)+4/16(1/8)+1/16(1/16)]
P(0 cars 1st roll then win)=(1/32)[1/32+5/32(1/2)+10/32(1/4)+10/32(1/8)+5/32(1/16)+1/32(1/32)]
P[win)=.031+.117+.176+.132+.049+.007=51.2%
My gut feeling is that these are pretty good odds compared to some of the other games.
Of course this is under the assumptions that the contestant priced 2 prizes correctly and goes for the car no matter what. Since the contestants do not win 3 rolls 100% of the time and might take the money on a particularly bad first roll the actual result will be significantly lower. Say they only price 1 prize correctly and get 2 rolls total, the chance to win the car drops to 1/32+5/32(1/2)+10/32(1/4)+10/32(1/8)+5/32(1/16)+1/32(1/32) = 23.7%, and if they get both wrong 3%.
Last edited by Yogi; 07-31-2012 at 04:31 AM.
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