My Yahtzee Challenge

[AV1611VET]

What are the odds of rolling a Yahtzee on the first roll?

Is it 1 in 7776 (6 x 6 x 6 x 6 x 6)?

Or, as I suspect, is it 1 in 1296 (6 x 6 x 6 x 6), since the first die that comes to rest sets the precedent for the other four?

 

[Endtime Survivors]

What are the odds of rolling a Yahtzee on the first roll?

Is it 1 in 7776 (6 x 6 x 6 x 6 x 6)?

Or, as I suspect, is it 1 in 1296 (6 x 6 x 6 x 6), since the first die that comes to rest sets the precedent for the other four?

Depends on who's rolling. Never discount magnetic waves. XD

 

[Bungle_Bear]

Depends how hard you pray, doesn't it? Or is it sincerity? Or both?

 

[Endtime Survivors]

Depends how hard you pray, doesn't it? Or is it sincerity? Or both?

No, my friends and I discussed that. No praying allowed. -_-

 

[Dave-W]

Don't you know that dice are evil?
They support the sin of gambling.

 

[Lulav]

What are the odds of rolling a Yahtzee on the first roll?

Is it 1 in 7776 (6 x 6 x 6 x 6 x 6)?

Or, as I suspect, is it 1 in 1296 (6 x 6 x 6 x 6), since the first die that comes to rest sets the precedent for the other four?

I would say six of one half dozen of the other.

Also you can't support the second choice as more than one can land at the same time.

 

[TLK Valentine]

What are the odds of rolling a Yahtzee on the first roll?

Is it 1 in 7776 (6 x 6 x 6 x 6 x 6)?

Or, as I suspect, is it 1 in 1296 (6 x 6 x 6 x 6), since the first die that comes to rest sets the precedent for the other four?

The second -- 1 in 1296. The first one applies if you're looking for the odds of it coming out Yahtzee on a specific number.

(Incidentally, I've had this happen once...)

 

[Endtime Survivors]

(Incidentally,

I wish an understanding of odds, numbers, and statistics were incidental for me, too.

 

[TLK Valentine]

I would say six of one half dozen of the other.

Also you can't support the second choice as more than one can land at the same time.

No, he's right. probabilities are determined by multiplying the number of possible outcomes..

for example, the odds of rolling any specific number on a die is 1 in 6. the odds of rolling doubles on a pair of dice are also 1 in 6, assuming it doesn't matter what the "first" die is, since the second one only has to match it.

The odds of rolling a specific set of doubles, however, is 1 in 6*6, or 1 in 36. There are 36 possible outcomes on a pair of dice (assuming they're not loaded ) and only one will do.

 

[TLK Valentine]

I wish an understanding of odds, numbers, and statistics were incidental for me, too.

The math isn't that hard; just repetitive... and the numbers tend to get big pretty quickly.

Once upon a time, a man walking the streets in Las Vegas found a $5 casino chip on the sidewalk. He walks into the nearest casino and heads for the Roulette table. He feels lucky, so he bets his red $5 chip on red 5.

(Now, a standard American Roulette wheel has 37 numbers: 1-35, zero, and double zero. Meaning the odds of him winning are 1 in 37... the payout on a single number bet is only 35 to one, however... the house always wins, even when you do, but I digress...)

Sure enough, Red 5 wins. the man's $5 is now $180. He wants to bet it again.... on Red 5. (he still feels lucky!)

(Odds of winning twice: 37*37 = 1 in 1369)
Well, he wins again, and his $180 is now $6480. Still feeling lucky, he lets it ride again... on Red 5.
(Odds of a roulette wheel hitting the same number three times in a row: 37*37*37 = 1 in 50653)

BOOM! Another winner! Now everyone on the Casino floor is paying attention to this high roller, he's up to $233,280... and he wants to let it ride one more time.

(Odds of a roulette wheel hitting the same number 4 times: 1 in 37*37*37*37 = 1,874,161)

Nobody say a word... if the house takes the bet and loses, our gambler wins $8,398,080...

The wheel spins one way, the ball goes the other... it slows down, hops the track, and lands right in Red 5!
... then hops out again and lands right next door in Black 22.

The house always wins.

The man walks out of the casino and pulls out his cell phone. He calls his brother.

"Hey Bro, how's Vegas?"
"It's fun! I just got back from some gambling."
"No kidding? How'd you do?"

"Meh..... I lost five bucks."

 

[KWCrazy]

What are the odds of rolling a Yahtzee on the first roll?

Is it 1 in 7776 (6 x 6 x 6 x 6 x 6)?

Or, as I suspect, is it 1 in 1296 (6 x 6 x 6 x 6), since the first die that comes to rest sets the precedent for the other four?

It's one in 7776 to roll a Yahtzee on any given number; say, I call sixes. it's much easier with 6 choices of variables. If I roll four of five dice the odds of matching the first would be 1 in 1296. However, there are 6 possible numbers to match. The probability that I roll a Yahtzee in any particular game is about 1 in 22; 4.6% .

 

[Lulav]

And what are the probabilities of each different yatzee?

You can have a Yatzee with:

1's, 2's, 3's, 4's , 5's and 6's.

Each increase in the number lightens the side, the sides with 6's being the least heavy and the sides with the 1's, the most.
This is determined by the concave 'dot' as well as the paint within it.

 

[John 1720]

What are the odds of rolling a Yahtzee on the first roll?

Is it 1 in 7776 (6 x 6 x 6 x 6 x 6)?

Or, as I suspect, is it 1 in 1296 (6 x 6 x 6 x 6), since the first die that comes to rest sets the precedent for the other four?

If it is just Yahtzee, then the answer is 1 in 1296 because the first die is independent and we only need to match it with the other 4.

• (1/6 * 1/6 * 1/6 * 1/6 )= 1/1296.

But your example specifically shows all 6's. Therefore the first die is no longer independent , since it necessarily must be six, which you have a 1/6 chance of rolling.

Therefore rolling Yahtzee with all "sixes" must necessarily be.

• (1/6 * 1/6 * 1/6 * 1/6 * 1/6)= 1/7776) in 7776.

But I believed you only were illustrating 1 example of 6 outcomes for Yahtzee so I would defer back to your original question. Therefore the correct answer is 1 chance in 1296 .

Regards, Pat

 

[Greg J.]

It is sometimes enlightening to understand the probability of an event occurring—when all possible outcomes are equal in probability—is simply the number of outcomes that are "wins" divided by the total number of possible outcomes (the result is the % chance in the range of 0 to 1 [just multiply by 100]). How to calculate that is where it gets more interesting.

The world would be a better place if the independence of events was taught in elementary school. That is, rolling a 6 on one die will not affect the other rolls. Or, more to the point, a streak of bad luck does not mean one is "due" some good luck.

 

[AV1611VET]

It is sometimes enlightening to understand the probability of an event occurring—when all possible outcomes are equal in probability—is simply the number of outcomes that are "wins" divided by the total number of possible outcomes (the result is the % chance in the range of 0 to 1 [just multiply by 100]). How to calculate that is where it gets more interesting.

The world would be a better place if the independence of events was taught in elementary school. That is, rolling a 6 on one die will not affect the other rolls. Or, more to the point, a streak of bad luck does not mean one is "due" some good luck.

Uh-huh.

 

[FrumiousBandersnatch]

The world would be a better place if the independence of events was taught in elementary school. That is, rolling a 6 on one die will not affect the other rolls. Or, more to the point, a streak of bad luck does not mean one is "due" some good luck.

Indeed. Some people seem to have trouble with the idea that if you happen to toss a coin and get a sequence of 'heads' or 'tails', not only do the odds of the next few tosses continue to be 50/50 (assuming ideal conditions), but continuing to toss the coin won't somehow 'compensate' for that sequence in the long term.

The gambler's fallacy seems to be a powerful intuition, on a par with the 'commentator's curse', where people are reluctant to make a positive assessment of some situation (e.g. sport) in case doing so somehow 'spoils it'.

 

[TLK Valentine]

It's one in 7776 to roll a Yahtzee on any given number; say, I call sixes. it's much easier with 6 choices of variables. If I roll four of five dice the odds of matching the first would be 1 in 1296. However, there are 6 possible numbers to match.

right -- you need four dice to all match 1 number... the odds of which are 1 in 1296. But what number will that be?

The probability that I roll a Yahtzee in any particular game is about 1 in 22; 4.6% .

If you roll the dice 59 times.... sounds like a long game.

 

[TLK Valentine]

It is sometimes enlightening to understand the probability of an event occurring—when all possible outcomes are equal in probability—is simply the number of outcomes that are "wins" divided by the total number of possible outcomes (the result is the % chance in the range of 0 to 1 [just multiply by 100]). How to calculate that is where it gets more interesting.

The world would be a better place if the independence of events was taught in elementary school. That is, rolling a 6 on one die will not affect the other rolls. Or, more to the point, a streak of bad luck does not mean one is "due" some good luck.

indeed -- the odds of a die roll coming up yahtzee with, lets say, all sixes (1 in 7776), is actually no different than any other possible combination...

.... of course, those combinations don't score as many points.