# The Price is Right (math/stat geeks here!)

I wish I remembered high school statistics.

What’s the probability of getting a \$1.00 on the big wheel?

No, I don’t mean landing on the 1.00 square. I already know the probability on that is 1/20.

For those who don’t know the “Price is Right”, to whiddle down the six contestants to two for the showoff, they have them spin a wheel with various values on them. In increments of 5 cents from 5 cents up to a dollar (spread in what appears randomly). You’re goal is to get closest to a dollar without going over. You’re allowed to spin the wheel twice.

OK, lets say the person ahead of you has \$0.95 on the board. Therefore the only way of getting to showdown is to get exactly. \$1.00

What I do know.

The first spin determines if you have spin again. If you land on \$1.00 (prob. 1/20) you win. If not (19/20) you spin again. (Your first spin could be considered your throw-a-way spin.) On the second spin, you’re chances of landing on the square that will bring you’re total to a dollar is 1/20.

What I don’t remember is how to place those facts together into the overall probability of winning the big whell and moving on to the final round.

Thanks for the help.

How’s my posting? 1-800-AM A TROL

OK, let me see if I understand the question. There is a wheel with 20 positions, labeled 5,10,15…100. You turn the wheel once. If you don’t get 100, you turn it again. If the total of the one or both turns is 100, you win.

So basically you have two ways to win: 1) get a 100 on the first try, or 2) not get 100 on the first try, but get a 100 as a sum of the two tries. Let’s look at each case.

The chance of getting 100 on the first turn is obviously 1/20.

The chance of having to turn it again is 19/20. If you did have to turn again, the chance of the total comingout to 100 is 1/20. That is, regardless of what you got on the first try, there is only one outcome that makes the sum 100. Therefore, the chance of both of these happenign (i.e. winning by the 2nd method) is 19/20 times 1/20, or 19/400.

Therefore, the chances of winning by either method is 1/20 plus 19/200, or 39/400. Normally you need to be careful adding probabilities, but here the two options are mutually exclusive so it is valid.

i’m not a math geek, but i’m willing to try anything to delay my homework.

first spin: 1 in 20, as previously stated.

odds that you won’t get \$1 of the first spin: 19/20

second spin, the odds that you will come up with the needed change to make \$1: 1/20

so maybe 19/400, a little less than 1 in 20?

it seems to me whenever i watched the show, during sick days, summer, and teacher’s institutes in the late 80’s/ early 90’s, about once every 5 to 10 shows.

so you have better odds hitting the \$1 on your first spin than hoping to make \$1 in two spins.

but this could be completely bs. i’m a grad student and haven’t taken a stats class…ever. this is just me remember freshman in high school math…

so maybe i shouldn’t have offered…

damn i didn’t think anyone up at 3:15 in the morning would be quick enough to beat me. i guess i was wrong…

ah well, it isn’t 3:15 everywhere in the world, you know…

scr4 is right, I just wanted to throw my 2 cents in.

A lot of times it’s easier to do these problems in reverse: What’s the probability you DON’T get \$1.00?

19 out of 20 chances you won’t get it on the first spin, and if you spin again (which you will, assuming you didn’t get the \$1), another 19 chances out of 20 you won’t get it on the second spin, either. So it’s (19/20)(19/20).

So the probability you DO get \$1.00 is

1-(19/20)(19/20).

## Therefore, the chances of winning by either method is 1/20 plus 19/200, or 39/400

Is this right ?

10/200+19/200= 29/200

Or am I wrong ?

funneefarmer,

I think “1/20 plus 19/200” was a typo. It should have been 1/20 plus 19/400.

Thanks.

You spin once–you get either \$1 or some other number from 5 to 95 in increments of 5.

So if you get the \$1–game over.

If not, you spin again.

Only one of those (second) numbers will give an even \$1–still 1 in 20 (for the second spin).

Does this change the odds that have been proposed so far?

Mjollnir,

Isn’t that the same as the original question?

DOH!

I’ve been up for about 30 hours, so maybe I’m not thinking straight, but it seems like you’re complicating the problem.

You get two shots at getting the dollar. The first one is 1/20. If you miss, you get a second try, which also offers odds of 1/20.

Therefore, you have a 1/10 chance of getting the \$1.00 in two spins.

Think of the problem this way: Let’s say there was a \$1.00 symbol on the wheel, and nineteen empty spots. Spin the wheel, and try and hit the dollar. If you miss, you get to try again. It’s the EXACT same problem, other than that on the ‘real’ wheel you have to hit a different symbol on the second go-round.

dhanson,

The second try isn’t quite 1/20–you have to miss the first try to even get to the second try, so the second try is (19/20)(1/20).

Think of it this way–flipping a coin two times, what’s the chance you get a tails? Not 1/2+1/2=1, but 1/2 + (1/2)(1/2) = 3/4.

I know you wanted the probability of getting \$1.00 in one or two spins, but I just had to nitpick to throw a monkey wrench in your premise.

You could also get \$0.95 in one or two spins to tie the other person. IIRC (I haven’t watched TPIR in ages) then (assuming the 3rd person didn’t get \$1.00 to beat both of you) they have a spin-off between the two of you, one spin each, with the person hitting the higher number winning.

So, what would the odds be of getting either \$0.95 to tie or \$1.00 to win?

Your Official Cat Goddess since 10/20/99.

“I get along well with everybody.” --I.M.F.

Kat, man I was hoping nobody would notice that little hole in the situation.

Thanks, folks.

How’s my posting? 1-800-AM A TROL

If you assume the person will stand on .95 if they get it on the first spin:

There’s a 2/20 chance you get .95 or 1 on the first spin.

If you don’t (an 18/20 chance) there’s a 1/20 chance you get .95, a 1/20 chance you get 1–so a 2/20 chance you get .95 or 1.

So you get 2/20 + (18/20)(2/20) = 76/400.

Of course, this all ignores the fact that you’re only out after the second (or any subsequent) spin if your total is over \$1.00.

Example 1: First spin is \$0.80. Second spin is \$0.40. Total is \$1.20; you’re out.

Example 2: First spin is \$0.25. Second spin is \$0.60. Third spin is \$0.05. You can go for spin #4.

So a full evaluation of the probability of getting \$1.00 would have to include the possibility of multiple spins, and those odds would (I think) be dependent on the actual values you spun. (My math is nowhere near good enough to tackle that, so I won’t even try.)

If you want to get into really deep water, there’s also the game theory aspects to the showdown. IIRC, there are three contestants competing, and you don’t need to spin \$1.00 to win. All you need to do is beat them. If you’re #3, and the other two have gone bust, spinning even \$0.05 will let you win.

And I think there’s another (or two?) section on the wheel, with something like a “lose immediately” or “go directly to jail, do not pass go” effect. So I think (even in my example last paragraph), you’d still have to spin and not lose, even if the two people before you crapped out.

…but when you get blue, and you’ve lost all your dreams, there’s nothing like a campfire and a can of beans!

Whoa there, Da Ace! Hold up! What version of “The Price is Right” are you playing?

With “TPiR” (the American game-show) you only get two spins. (with the exception of tie-breakers – Kat) If you don’t beat the other guy by then, you’re out. I’ve never heard of the version where you get four spins. Is this a home version or something?

There are no ‘casulity’ spots on the wheel. Just values of 0.05 to 1.00 in 0.05 increments. No “lose immediately” squares.

In my example you had to total \$1.00. Not 0.95 (let’s ignore the tie possibility), not 1.05 (you’d loose immediately, anyway.) Basically I was asking what was the probability of getting a specific total amount on that wheel. And it looks like they’ve solved it there (Jeez, after thinking about it, why couldn’t I figure it out? It wasn’t that weird statistics stuff I learned and promptly forgot last year.)

Now that I’m done beating up on ya’ man, the second part of the post is interesting. Throwing out all other qualifiers, it is theoretically possible to score only \$0.15 and still win the showdown, if the other contestants only manage to hit the nickel square twice. (You could score five cents, Ace, and win if you’re the last contestant and the other two manage to spin to total over a dollar. Of course, that would require them to have shit-for brains, but that’s another matter.)

Is there a way of actually determining the probability of winning the game

How’s my posting? 1-800-AM A TROL

Yes and no.

Technically, you “only” have to consider all possible cases of what each player gets in each spin, determine who wins and loses, and calculate the overall ratio. There’s usually a more efficient way than the brute-force method of looking at each combination of spins separately, but it’ll still be tedious and I won’t do it for you.

The real problem is that you have to assume a strategy for the participating players. If they get, say, \$.60 on the first spin, will they take a second spin? How does this depend on what the previous players did, and which position the current player has? (The first player would behave differently from the middle or last one.) The straightforward choice would be to assume perfectly rational behavior for each player, meaning that each will do what gives him the greatest probability of winning. But that in turn will depend on the other players, and – presto! – you’re in the beautiful world of game theory!

It only gets more complicated from there on.