I was doing a statistics problem about discrete random variables, and I ran into something I didn't know how to do.

I was setting up a problem that required me to show the probability of beating someone on "the price is right" showcase showdown wheel. I did that fine, and I was going to do more than the problem asked by finding out the probability if I was given 2 spins of the wheel, but I could not wrap my head around how to do it.

does anyone know how to set that up?

my first spin would have to be less than their spin, and my second spin would not be able to go over $1.

(in case you don't know, the price is right wheel consists of 20 spaces $0.05 to $1.00)

I set regular problem up by assigning in integer value to all the prices to make it easier to work with. 1 was the lowest value and 20 was the highest. the problem was simple, P[x>y]=1-y/20 (where y is some value they got and x is mine)

this is going to bother me all day if I can't figure it out =/.

(this is not homework help, I don't have to solve this, its just bugging me)

Sorry about my ignorance, but could you describe in detail the rules and procedure of the game here? Does the first person get two spins also, or just you? Do you add the amounts or just take the largest individual amount?

-Dale

the person before you will spin some value, I called 1 through 20, and you have two chances of spinning the wheel to beat their spin, without going over. if you beat it in the first spin, you can stop (that was the original problem). however, if your first spin is not greater than their value, you can spin again and the two amounts are added.

the sum of your two spins could be less than theirs (you loose), between their's and 20 (you win), or over 20 (you loose).

so basically, its the probability of x1+x2>y given that x1.<.y

I just don't know how to set it up.

So if A is the event x>y and B is the event 20>=x+z>y then the probability of winning is P(A U B). In other words, approach it from the idea that you spin the wheel twice no matter what. If your first spin, x, is greater than y then you win, even if your second spin, z, puts you over. The key then is to think of an (x,z) plane and determine the area of the region corresponding to B (remember that the line x+z=k will be a line with a slope of -1 in this plane with k=20 being the main diagonal)

-Dale

Hmm, I didn't like my last post. If it helped you then good, otherwise consider this one instead.

Think of it as a cube with three planes through it slicing it up into different regions:
x=y
x+z=20 -> x=20-z
x+z=y -> x=y-z

With x in the vertical direction you win if you are in the region above the first plane or the region between the other two planes. So if A is the region above the first plane, B is the region below the second plane, and C is the region above the third plane then the probability of winning is:
P(A U (B ∩ C))
P(A) + P(B ∩ C) - P(A ∩ B ∩ C)

Don't know if that helps either, sorry.

-Dale

Btw, if you just do it exhaustively you find out that there are 5130/8000 ways to win. So the second person has a better chance to win than the first.

Of course, this assumes that the first person only gets one spin. If the first person gets two spins then you cannot determine the odds since there is no clear decision about when to take the second spin.

-Dale

so basically, its the probability of x1+x2>y given that x1.<.y

I just don't know how to set it up.

Given x1 and y for the first spins, for a win we must have 20>=x1+x2>y, or 20-x1>=x2>y-x1. The probability of getting x2 in this range is just (20-y)/20. Sum over the x1's to give the probability of a win given y. Then sum over the y's.

This isn't taking into account what happens in the event of a tie. If your first spin is equal to y, the usual game allows them the choice of spinning again or going into a spin off? A spin off has a 50/50 for each player, so you'd go for the spinoff if your chance of totalling over 20 is .5 or more. Presumably you'd also have this spin off if x1+x2=y.

shmoe, I don't think that is right. (20-y)/20 is the answer to the original problem (i think).

could you explain that 1 more time dale? I am not quite understanding. if you could put it in a pdf (probability density function), it would be easier to understand.

maybe if you work it out with some values. I came up with most of what you had dale, but I am not sure how I can make it into something where I write it in a PDF.

shmoe, I don't think that is right. (20-y)/20 is the answer to the original problem (i think).

Yes, given that y>=x1, you have the same probability of winning as you did on your first spin, there are still (20-y) numbers on the wheel that give a win. Multiply this by the probability that y>=x1 and sum up over the y's. Then add in the probability that you win outright, that is when x1>y (summed over the y's again).

What about ties? Are you counting them as losses? And when x1=y? Spin again or ???

what about if I go over? the first probability didn't take that into account(because you can't go over with one spin), this model should. that is where I am getting hung up.

That shouldn't be a problem. After your first spin (assuming you haven't won yet) you have 20-y ways to win with your second and still y ways to lose. These aren't the same ways to lose as before, x1 of them are from going over and y-x1 are from not getting higher than y, but a loss is a loss.

Sorry cato, I can't describe my ideas any better. But shmoe is right and his way is easier and clearer anyway. If A is winning the first time then P(A) is (20-y)/20 and P(A') is y/20. If B is winning the second time given that there is a second time then as shmoe mentioned P(B|A')=P(A). So then the probability of winning overall is P(A) + P(A')P(B|A') = (20-y)/20 + y/20 (20-y)/20. Incidentally, summing over y from 1 to 20 and dividing by 20 gives 5130/8000, agreeing with my exhaustive number above.

-Dale

ok, thanks. it seems odd, but I guess its correct, thanks guys =]