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

[SterlingNorth]

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.

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How's my posting? 1-800-AM A TROL

[scr4]

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.

[ubermensch]

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....

[ubermensch]

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...

[scr4]

ah well, it isn't 3:15 everywhere in the world, you know...

[Cabbage]

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).

[funneefarmer]

Therefore, the chances of winning by either method is 1/20 plus 19/200, or 39/400
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Is this right ?

10/200+19/200= 29/200

Or am I wrong ?

[Cabbage]

funneefarmer,

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

[funneefarmer]

Thanks.

[Earl Snake-Hips Tucker]

How about this:

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?

[Cabbage]

Mjollnir,

Isn't that the same as the original question?

[Earl Snake-Hips Tucker]

DOH!

[Sam Stone]

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.

[Cabbage]

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.

[Kat]

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

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?

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Your Official Cat Goddess since 10/20/99.

"I get along well with everybody." --I.M.F.

[SterlingNorth]

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

Thanks, folks.

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How's my posting? 1-800-AM A TROL

[Cabbage]

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.

[G.B.H. Hornswoggler]

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.



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

[SterlingNorth]

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

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How's my posting? 1-800-AM A TROL

[TheIncredibleHolg]

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

[Harvey The Heavy]

And don't forget that the wheel isn't as random as it looks. I've seen players do a "finesse" spin and get the amount that they were aiming for.

[G.B.H. Hornswoggler]

Sorry about that Sterling; I guess my last jury duty must have been even more boring than I'd thought, since I clearly fell asleep and dreamed my own version of TPIR.

But I could have sworn you could keep spinning as long as you didn't break $1.00...

(walks away slowly, shaking head)

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

[Kat]

With "TPiR" (the American game-show) you only get two spins. (with the exception of tie-breakers -- Kat)

*wonders if she should be a smart-ass and bring up the Bonus Spin for cash if the player gets exactly $1.00, figures SterlingNorth will argue (correctly) that the Bonus Spin isn't really a part of the Showcase Showdown and decides against it*

[Earl Snake-Hips Tucker]

I wanna try!! I wanna try!!

First, my answer: 20 in 381.

Here's the math behind it:

There are 20 numbers on the wheel.
So, with two spins 20 * 20 = 400.

However, if you get $1 on the first try, it's over.

That leaves 19 * 20 + 380.
Add one for the $1 on the first spin, and you have 381 total possible spin results.

There are only 20 possibly $1 combinations.

So, 20 in 381.

I'm not a sadistics (uh, statistics) expert, so help me out if I'm wrong.

[Cabbage]

Mjollnir,

You're right, there are 381 different spin combinations--the one where you get $1 on the first spin, then all the 19*20 other combinations.

The problem with concluding the probability to be 20/381 is this: The probability of getting a $1 on the first spin is 20 times more likely than any other given combination. For example, there's a 1 in 400 chance of getting .05 first, then .95, but there's a 1 in 20 chance of getting $1 on the first spin.

So instead of adding the number of different combinations as 380+1, we have to count the $1 spin 20 times--380+20.

[Sam Stone]

Now that I've had some sleep, let me chime in that the easiest way to do these problems is often to calculate the odds of NOT getting the result.

In this case, your chance of missing on the first spin is 19/20. If you get that result, you get a second spin, and the chance of missing is also 19/20. Thus, the chance of missing both spins is 19/20 X 19/20, or 361/400. So, your chance of winning is 1-361/400, or 39/400.

If you do it the other way, by calculating the chance of making the spin (as I did in a drunken stupor), you get an incorrect result because the two variables are not independent.

When the odds are long like this, the answers come out close (if you do it the 'win' way, you get 1/10, or 40/400). But when the odds are better, the error gets to be large.

For instance, if you had a 25% chance of winning on each spin, the correct calculation would be to take the chance of losing, at .75 X .75, or .5625.. and subtract one, for a chance of winning with two spins of just over 43%. If you do it the 'win' way, by taking your .25 chance and adding it together, you get a chance of 50%. In the first case, the difference between the right and wrong answer was only 1/400. But in the second case, the difference was over 6%.

[hardcore]

dhanson:

I think there is a correct way to calculate it using the "win" method, you just have to be careful. Using your 25% case it goes like this: .25 + (.75)(.25)=0.4375

You have a 1 in 4 chance on the first turn, and the the remaining 3 out of 4 cases have a 25% chance.

scr4 calculated the OP case by this same "win" method.

[Sam Stone]

Sure. There are several different ways of calculating it. I was just pointing out that in cases like this, the simplest way is often to calculate the odds of NOT winning.

These calculations come up in poker all the time. Let's say you have a 4-flush in Texas Holdem. A player bets into you. How much money does there have to be in the pot for you to call with two cards to come?

In Holdem, there are 47 unseen cards after the 'flop'. Nine of them make your flush. If you don't make it on the next card, there are 46 unseen cards, and nine of them make your flush.

A common mistake in poker books is to say that your odds of making your flush are 9/47 + 9/46, or 38.7%. The correct way, by calculating the odds of losing, yields the correct answer of 35%.

For relative longshots like flush draws, either answer is probably close enough. But for hands with more 'outs' (like straight flush draws), the numbers are quite different.

[hardcore]

The point is that you get the same result whether you calculate the odds for winning or the odds for losing and subtract from 1. The simpler way is usually a matter of personal perspective. I do it both ways to check my answer.

If a poker book shows the calculation you posted, then it made the mistake of not multiplying the 2nd factor (9/46) by the remaining cases (38/47). This makes it come out correctly to 35%.

Then again, it has been a loooong time since my college statistics class!