Let 'Em Roll odds

[clemon79]

(I hope David still reads here, at the very least.)

I'm trying, mathematically, to figure out the overall odds of winning at Let 'Em Roll, assuming the player has all 3 rolls and sets cars aside as they come up.

Here's the information I have so far (Check me. If I missed something somewhere, prove me wrong):

Total permeutations of 5 dice: 6^5, or 7776.

Since each die has three faces with cars on them, the chance of an individual die coming up with a "good" result (an upturned car) is 1 in 2. Therefore the chances of all five coming up "good" is 2^5, or 1 in 32.

Which means 1/32nd of those 7776 permeutations are winners, or 243
That also means that 1/16th of those have at least 4 "good" faces, or 486. But only half of those have ONLY four "good" faces, so there are 243 of those as well.
Also, 1/8 have at least three good faces. 972. Minus the 486 with four or more, for 486. I'm noticing a pattern here.
972 discrete 2-face rolls. 1944 with but a single car.

So, recap:

Number of rolls with ONLY 5 cars: 243 (1/32)
Number of rolls with ONLY 4 cars: 243 (1/32)
Number of rolls with ONLY 3 cars: 486 (1/16)
Number of rolls with ONLY 2 cars: 972 (1/8)
Number of rolls with ONLY 1 cars: 1944 (1/4)
...and the number with ZERO cars should be the same as with five: 243.

Something is clearly wrong here, as this does not add up to 7776. What happened?

[parliboy]

Chris, let's simplify things a bit.  We know that any cube has a 50/50 chance of being a car.  So instead of dice, we'll use everyone's other favorite statistical prop, coins.

What are the odds of 5 coins coming up heads within three total flips, assuming any heads flipped are banked between each flip.

There are 2^5 or 32 possible combinations.  (Edit: these odds are after one flip)

5h, 0t: 1/32
4h, 1t: 5/32
3h, 2t: 10/32
2h, 3t: 10/32
1h, 4t: 5/32 (here you made an error)
0h, 5t: 1/32

I could build a tree to get the right answer.  But I'd rather restate things to make the problem solvable for mere mortals like myself:

Each coin is flipped invididually three times.  What are the odds that each coin will be a head at least one of the three flips?

The odds of one coin being a head one of the three times is .875.  If it lands on a head the first time, the other two are moot.  So the odds of having all five coins be a head at least once is .875 ^ 5, which is about .5129, or 51.29%.  That should be right, but those who have had more than three credit hours in statistics can probably give you a more traditional approach to the answer.

[clemon79]

I bow to your simplification skills. That all makes 100% perfect sense and avoids a lot of messy math. My initial suspicion was that the odds were better than 50/50, but I wasn't sure by how much. Thanks!

[willmorris]

Let's see how the math should work.

Five dice, all coming up as cars, happens every one every thirty-two times.  So do five dice, none coming up as cars.

Five dice with four cars can happen five different ways.  XCCCC, CXCCC, CCXCC, CCCXC, and CCCCX.  Five rolls every thirty-two times.  The chances are the same for just one car.

Five dice, with three cars, can happen ten different ways.  XXCCC, XCXCC, XCCXC, XCCCX, CXXCC, CXCXC, CXCCX, CCXXC, CCXCX, CCCXX.  Ten rolls out of thirty-two And of course, the chances for two cars are the same.

In short:

Five cars - 1/32
Four cars - 5/32
Three cars - 10/32
Two cars - 10/32
One car - 5/32
Zero cars - 1/32

EDIT - parliboy, you just beat me to this one.

[parliboy]

[quote name=\'clemon79\' date=\'Jun 16 2004, 01:00 AM\'] I bow to your simplification skills. That all makes 100% perfect sense and avoids a lot of messy math. My initial suspicion was that the odds were better than 50/50, but I wasn't sure by how much. Thanks! [/quote]
 Thanks.  This method makes it easy to calculate odds from any position in the game without the "messy math" as you called it.  After one roll, you use .750 instead of .875; and after two rolls, you use .500.

Example: If you have two cars after one roll, there are three cubes in play.  So the odds of winning in that position are .750 ^ 3 or just over 42%

[tvrandywest]

You've simplified the "come out roll". I get it. Thanks.

But on the tape day of Mandel's "Half-Off" debut David Hammett and I were in the audience discussing just this subject as it related to the "Let 'Em Roll" game on that episode. We never came to agreement on the odds of a win for that day's player. It was a discussion of the odds of that player winning the car SUBSEQUENT to his first roll.

As I remember the specifics, on that first roll the player got 3 cars and 2 high-dollar amounts on the other 2 cubes. With 2 more rolls available to that player, what are the odds that he nails the car? Keep in mind that the player can roll the additional 2 cars on his second roll (game over) OR add 1 more car on the second roll (game continues) and the fifth car on the final roll when he rolls only the final cube OR get no help on the second roll (game continues) but adds the other 2 cars on the third and final roll.

Factoring in all 3 of those potential scenarios, what were the total odds of the player getting those 2 more cars in his remaining 2 rolls? It makes my head explode!


Randy
tvrandywest.com

[parliboy]

[quote name=\'tvrandywest\' date=\'Jun 16 2004, 01:36 AM\'] As I remember the specifics, on that first roll the player got 3 cars and 2 high-dollar amounts on the other 2 cubes. With 2 more rolls available to that player, what are the odds that he nails the car? Keep in mind that the player can roll the additional 2 cars on his second roll (game over) OR add 1 more car on the second roll (game continues) and the fifth car on the final roll when he rolls only the final cube OR get no help on the second roll (game continues) but adds the other 2 cars on the third and final roll.

Factoring in all 3 of those potential scenarios, what were the total odds of the player getting those 2 more cars in his remaining 2 rolls? It makes my head explode!
 [/quote]
Using the numbers I just gave in my post just before yours:

If he has three cars on the first roll, the odds of getting two cars within the next two rolls are .750 ^ 2 or 56.25%

EDIT: Trying to think of a mathematical formula to express this...  Something relatively pocket sized so you can impress and/or scare people at parties:

(1 - (.5 ^ n)) ^ c, where n is the number of rolls remaining, and c is the number of cubes to be rolled.

This should work for all situations of the game.

(Personal note: I wonder if Mandel had to do this kind of number crunching when he submitted his game)

[tvrandywest]

[quote name=\'parliboy\' date=\'Jun 15 2004, 10:50 PM\'] If he has three cars on the first roll, the odds of getting two cars within the next two rolls are .750 ^ 2 or 56.25% [/quote]
Thanks. Intuitively it felt close to a 50-50. Makes sense.

But COUNTER-intuitively I'm baffled now that I got to see your previous post. With a 56.25% chance of getting 2 more cars on 2 rolls, I'm surprised that the odds don't drop drastically when attempting to get 3 more cars on 2 rolls. It sounds a lot tougher than a 42% chance of success you cite.

I bow to your mathematical expertise, but I must say it doesn't "feel" logical.


Randy
tvrandywest

[parliboy]

If you've got a headache now, think about this: of that 14% difference, 12.5% comes from the odds of rolling the cars on the very next roll instead of needing both rolls.  Take that away, and the odds are virtually even.

[Craig Karlberg]

That formula (1-(5^n)^c is a good formula for any roll in any situation & thus I shall call it the Let'Em Roll Probability Principle.  There.  That's my mathematical statment for today.

[aaron sica]

[quote name=\'Craig Karlberg\' date=\'Jun 16 2004, 04:46 AM\'] That formula (1-(5^n)^c is a good formula for any roll in any situation & thus I shall call it the Let'Em Roll Probability Principle.  There.  That's my mathematical statment for today. [/quote]
 Albert Einstein, you're definitely not.

[dzinkin]

[quote name=\'aaron sica\' date=\'Jun 16 2004, 07:55 AM\'] [quote name=\'Craig Karlberg\' date=\'Jun 16 2004, 04:46 AM\'] That formula (1-(5^n)^c is a good formula for any roll in any situation & thus I shall call it the Let'Em Roll Probability Principle.  There.  That's my mathematical statment for today. [/quote]
Albert Einstein, you're definitely not. [/quote]
 Maybe he's (channeling Gene Rayburn)... Alfred Einstein?

"WHEN DID YOU CHANGE YOUR NAME?" ;-)

[parliboy]

[quote name=\'Craig Karlberg\' date=\'Jun 16 2004, 03:46 AM\'] That formula (1-(5^n)^c is a good formula for any roll in any situation & thus I shall call it the Let'Em Roll Probability Principle.  There.  That's my mathematical statment for today. [/quote]
 Rather than be smarmy (not that smarmy is unwarranted) I'll play it straight.

It's .5, not 5.

[clemon79]

[quote name=\'tvrandywest\' date=\'Jun 16 2004, 12:07 AM\'] I bow to your mathematical expertise, but I must say it doesn't "feel" logical.
 [/quote]
 It does, tho. Think of the latter situation as broken down into two isolated events:

Situation (a): 2 cars on 2 dice in 2 rolls, and

Situation (b): 2 cars on 2 dice in 2 rolls, PLUS a seperate car on 1 die in 2 rolls.

So the only difference now is that extra die in (b). Since half the die is cars, and you get two chances, most of the time, this should happen. So the odds of getting 3 cars on three dice in two rolls shouldn't be all that far removed from those of getting 2 on two dice in 2 rolls.

[parliboy]

Exactly.  That extra cube will be a car 75% of the time across two rolls, meaning that a player with three cubes will have a winning result 75% as often as a player with two cubes.  And you'll note that 56% * .750 = 42%.  So it all works out.