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

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.

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)

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

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

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.


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.

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%.


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.

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.

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!