New Pricing Game idea: Two and Two

[tpir04]

Four prizes are shown and described to the contestant. All that must be done is to choose two prizes and have the actual retail price be LESS than the actual retail price of the other two prizes. Then the prices are revealed and the outcome determined.

I know the show needs more quickie games which is why I tried to keep the rules as simple as possible. Based on the way I structured the game, I believe there are multiple ways for the contestant to win. For example, there are four prizes: $1000, $1500, $2000, $2500. The contestant could select $1000 and $2000 or $1000 and $1500. Of course, the number of winning combinations varies on the prices.

So what do you think?

[Teddy]

Sounds like a more complicated version of Take Two and Poker Game.

[gamesurf]

I like the simplicity, I like how it's easy to understand.  It sounds like Hi-Lo, but in practice it's closer to Easy as 123 with more room for error.

The difficulty is in setting it up--unless you're using four prizes that are REALLY close to each other, this would probably be very easy for most contestants.

Just for the sake of argument let's pretend that the $1,000 prize is just a dollar cheaper to avoid ties:

AB ($2499) vs CD ($4500) - easy win
AC ($2999) vs BD ($4000) - win
AD ($3499) vs BC ($3500) - close win
BC ($3500) vs AD ($3499) - close loss
BD ($4000) vs AC ($2999) - loss
CD ($4500) vs AB ($2499) - loss

On paper it SEEMS like the contestant has a 50/50 chance of winning, and they would, if they were picking randomly. But they're not--to win the game all the contestant REALLY has to do in this scenario is identify one prize.

If they can identify the highest priced prize and avoid it, they win. If they can pick the lowest priced price, they win. Even if they pick both, they still win in this scenario.

The only way to lose would be to put the cheapest prize on the high end (which is hard to do--how often does the cheapest prize get picked in Most Expensive?). Even though these are made up-numbers, they illustrate that there's a LOT of winning combinations that most contestants will be able to identify.

On the other hand, if you try to lower the number of winning combinations, you have to make the prices more extreme--say. $500, $600, $1200, $1250. But then it becomes even more obvious which ones are the "high" prizes and which are the "low" ones.

[b_masters8]

I don't know if Chuck Woolery would take offense to you using his famous commercial break throw line-- remember "Two and two, be right back at you" (2:02, standard length of a commercial break then)?

[tpir04]

I don't know if Chuck Woolery would take offense to you using his famous commercial break throw line-- remember "Two and two, be right back at you" (2:02, standard length of a commercial break then)?

That's right! I totally forgot about that! Well, there a dozens of names I could choose for the game if the conflict arises.

If they can identify the highest priced prize and avoid it, they win. If they can pick the lowest priced price, they win. Even if they pick both, they still win in this scenario.

The only way to lose would be to put the cheapest prize on the high end (which is hard to do--how often does the cheapest prize get picked in Most Expensive?). Even though these are made up-numbers, they illustrate that there's a LOT of winning combinations that most contestants will be able to identify.

Still, I have seen a lot of times where the contestant (and I must admit, I have done so too) picked the cheapest prize in Most Expensive. Even so, perhaps the price range could be reduced to make it slightly harder? Then again, it does not matter how the game is set up as long as there is one way the contestant could lose.