The Roll of the Dice

[Hobbes]

If I throw three six sided dice - what is the chance of rolling a six?

Cheers, Chris

[Gray_Lensman]

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[Hobbes]

This could be interesting - both different to the answer I saw in The Times!

42%

[DarthMath]

I agree with Hobbes.

(5/6)*(5/6)*(5/6) = 125/216

1-(125/216) = 91/216 = 42,13%

[Gray_Lensman]

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[Hobbes]

I stand corrected... For some reason (probably clutsy fingers) when I plugged 5/6 into the calculator I came up with .8888888 instead of .833333333... Ooops. 42.14% is indeed correct.

For 3 separate dice being throw simultaneously it is still 50% however.

1/6 + 1/6 + 1/6 = 3/6 or 50%

This is what is bugging me - surely it makes no difference how the dice are thrown?

[Gray_Lensman]

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[DarthMath]

The result 42% is for the case of dice rolled simultaneously. 1/6+1/6+1/6 is not good. It's a matter of conditions of experience and equiprobabilty, and it can be quite disturbing some times.

[Hobbes]

In one case you are throwing a single die 3 separate times hence a 42.13% chance of one of those 3 separate throws resulting in a 6.

In the 2nd case you are throwing 3 separate independent dice a single time.

It does make a difference.

Try both methods separately yourself and tabulate the results over several hundred die tosses. You'll find that the results do in fact differ depending on the method.

So you are saying that if we were sitting in a pub and you had one die and I three - I threw all three and you threw your one three times you would have more chance of getting a six?

If this is the case there must be a lot of rich people out there!

[Hobbes]

The result 42% is for the case of dice rolled simultaneously. 1/6+1/6+1/6 is not good. It's a matter of conditions of experience and equiprobabilty, and it can be quite disturbing some times.

You appear to have Math in your name for a reason Darth

This has been disturbing us all at work all day! I always thought I should have a 50% chance with three dice. With all the dice games I have been playing since year dot it is no wonder I have always felt that I have been unlucky with my rolls - especially when playing my wife at Yahtzee!

Cheers, Chris

[DarthMath]

Remember that I'm the Dark Lord of the Maths, thanks to ComtedeMeighan !!
I've "eaten" probabilities for quite a long time during university, almost to nausea !! Maybe that's an explanation ...

P.S. : for the pseudonym, to have Mathieu as a first name is helping, too !!

[gchristie]

it is no wonder I have always felt that I have been unlucky with my rolls - especially when playing my wife at Yahtzee!

Cheers, Chris

From The Yahtzee Page

"Any Yahtzee after 1 throw
0.077%
Any Yahtzee after 2 throws
1.3%
Any Yahtzee after 3 throws
4.6%

Adding it all up we have a 4.6% chance of a Yahtzee which is roughly 1 in 22. So, for every 27 tries you should expect to get one Yahtzee over the long run. In real play only 10 goes target five-alike (excluding full house and the two straights) so you should expect one yahtzee per 2.2 games."

This definitely made me feel a little better.

[Gray_Lensman]

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[Hobbes]

Incidentally, this subject was appropriate during some event designs for the 1861 w/Kentucky scenarios.

For example: During the design of the event(s) regarding the random entry by either the USA or CSA, I wanted an approximate 25% chance PER GAME of that happening... However, it had to be spread out over 6 turns.

The closest I could come was to use 5% (within the event syntax) for each of those 6 turns.

1 - (95/100)^6 = 26.49%

If I had instead assigned 25% for each turn (within the event syntax), the probability of a random entry PER GAME would have been 82.20%

1 - (75/100)^6 = 82.20%

It is certainly something a game designer should understand. I still don't really. I still don't see a difference with throwing three dice or throwing one three times.

I understand the 6x6x6 = 216 possible outcomes and 5x5x5 = 125 don't have a six. So 125/216 * 100 = 58% chance of not getting a six.

What about the toss of a coin? Can the same maths be applied?
2x2 = 4 possible outcomes and 1x1 = 1. So 1/4 * 100 = 25% chance of not getting a head in two tosses?

I think I will have to give this more thought in the morning

Cheers, Chris

[DarthMath]

It is certainly something a game designer should understand. I still don't really. I still don't see a difference with throwing three dice or throwing one three times.

I understand the 6x6x6 = 216 possible outcomes and 5x5x5 = 125 don't have a six. So 125/216 * 100 = 58% chance of not getting a six.

What about the toss of a coin? Can the same maths be applied?
2x2 = 4 possible outcomes and 1x1 = 1. So 1/4 * 100 = 25% chance of not getting a head in two tosses?

I think I will have to give this more thought in the morning

Cheers, Chris

You have 4 faces on a coin ?
Strange England ...
In fact, all depends of what kind of probability you're searching. Write correctly the wording in term of numbers and events is the key in any probability exercise.

[Hobbes]

You have 4 faces on a coin ?
Strange England ...

We do if we toss it twice quickly

[Gray_Lensman]

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[DarthMath]

For the coin, events are completely independant and it's not the same law which is used ( it's a simple binomial case with a probability of 0.5 ). So each time, you have 1 chance to have one face, and 1 chance to have the other.
So if your probability is "to obtain a head", it's 50% whatever the number of tries.

Edit : Quite fast, Mr Lensman !!

[Hobbes]

Because throwing 3 dice at once is a single event

but

throwing 1 die 3 times is 3 separate sequential but dependent events.

It's a brain f*** is what it is!

[Hobbes]

For the coin, events are completely independant. So each time, you have 1 chance to have one face, and 1 chance to have the other.
So if your probability is "to obtain a head", it's 50% whatever the number of tries.

Edit : Quite fast, Mr Lensman !!

But if I throw 2 coins?

[DarthMath]

But if I throw 2 coins?

Simultaneously ?
In this case, you have a similarity with the dices' case. 75% to obtain at least one head, for example.

Edit : Sorry, not "at least". It's a different case. So : 75% to obtain one head.

[Hobbes]

Gray :-

In one case you are throwing a single die 3 separate times hence a 42.13% chance of one of those 3 separate throws resulting in a 6.


Darth :-

The result 42% is for the case of dice rolled simultaneously.


Are you chaps in agreement? This seems to be the opposite view?
But I know I'm struggling here

[Gray_Lensman]

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[DarthMath]

Gray :-

In one case you are throwing a single die 3 separate times hence a 42.13% chance of one of those 3 separate throws resulting in a 6.


Darth :-

The result 42% is for the case of dice rolled simultaneously.


Are you chaps in agreement? This seems to be the opposite view?
But I know I'm struggling here

No problem.
No we're not. 42% is the result for 3 simultaneous dices to obtain one six on one of them. Equiprobability, no interdependance.

[Gray_Lensman]

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[dooya]

To clarify things.

If you throw 3 dice (all at once) (a single event):

The odds are 1/6 + 1/6 + 1/6 = 3/6 = 50% chance of at least one of the 3 dice being a 6.

If you throw a single die (3 separate times) (3 separate sequential but dependent events):

The odds are 1 - (5/6)^3 = 42.13% chance of one of the 3 separate die rolls being a 6.

Throwing a single dice three times constitutes 3 dependent events? If so, the chance of throwing a 6 must change for every single throw, because otherwise the events would be independent by definition. Does throwing a 6 (or whatever) change the odds of the dice for following throws?

Furthermore, if you have dependent events, you can not calculate probabilities by multiplying the probability of the single events, because P(A u B) = P(A) * P(B) iff A and B are independent.

Actually, I see no difference between throwing 3 dices at once or one dice three times - the dice throws are three independent events in both cases.

[DarthMath]

To clarify things.

If you throw 3 dice (all at once) (a single event):

The odds are 1/6 + 1/6 + 1/6 = 3/6 = 50% chance of at least one of the 3 dice being a 6.

If you throw a single die (3 separate times) (3 separate sequential but dependent events):

The odds are 1 - (5/6)^3 = 42.13% chance of one of the 3 separate die rolls being a 6.

Sorry I disagree on the wording. I think just quite the opposite.
If I follow you, if I throw 7 dices all at once, I have a 7/6 (:bonk probability to obtain at least one 6 !!
"at least", "dependent", "sequential", "simultaneous" etc ... means something in proba. in term of mathematical representation ( and we disagree on this representation, as it seems ). That's the difficulty in probabilities.

[Hobbes]

Are there any lurkers out there struggling with this?

It's too late for me tonight, but I will get my head around this by the weekend!

Thanks, Chris

[DarthMath]

Throwing a single dice three times constitutes 3 dependent events? If so, the chance of throwing a 6 must change for every single throw, because otherwise the events would be independent by definition. Does throwing a 6 (or whatever) change the odds of the dice for following throws?

Furthermore, if you have dependent events, you can not calculate probabilities by multiplying the probability of the single events, because P(A u B) = P(A) * P(B) iff A and B are independent.

Actually, I see no difference between throwing 3 dices at once or one dice three times - the dice throws are three independent events in both cases.

Exactly !!
In our case, simultaneous means independent. And that's why we have this result of 42% for having a 6.

[Gray_Lensman]

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