Silly dicey problem

[science_man_88]

Originally Posted by TheMawn

Here's an interesting monkey wrench...

If you're not limiting yourself to 2's in order to make the game go faster then there's some human non-randomness that gets thrown into the mix, too. For example, if the caller says "One of the dice is a 3" the player wins if they're both 3. It would be like this as opposed to waiting for specifically a 2.

How does the caller decide which die to consider? For example, if you get a 2 and a 5, the caller could say "One of the dice is a 2" or "One of the dice is a 5," right? If he always goes with the left-most one, for example, then the odds become 1 in 6.

my point in the spoilers was permutations versus combinations because:

(1,2)
(2,1)
(2,2)
(2,2)
(2,3)
(3,2)
(2,4)
(4,2)
(2,5)
(5,2)
(2,6)
(6,2)

there are twelve if we don't care about order ( aka don't treat them as n-tuples) only 6 are distinct so the odds become 1 in 6 if order matters then technically it's 1 in 12 if we delete the one duplicate as though order doesn't matter for that one ( as my attempt at a general formula would do ) we get 1 in 11.

[retina]

Originally Posted by TheMawn

Here's an interesting monkey wrench...

If you're not limiting yourself to 2's in order to make the game go faster then there's some human non-randomness that gets thrown into the mix, too. For example, if the caller says "One of the dice is a 3" the player wins if they're both 3. It would be like this as opposed to waiting for specifically a 2.

How does the caller decide which die to consider? For example, if you get a 2 and a 5, the caller could say "One of the dice is a 2" or "One of the dice is a 5," right? If he always goes with the left-most one, for example, then the odds become 1 in 6.

If you give the caller free choice then it makes no difference how the caller chooses the dice to call, the odds are always 1 in 6.

[sdbardwick]

Originally Posted by retina

If you give the caller free choice then it makes no difference how the caller chooses the dice to call, the odds are always 1 in 6.

I've had that concern nibbling at the back of my mind...I didn't do the math before posting my idea. Didn't offer the game up yet, but I would have done the math before putting $$ on the line :)

[TheMawn]

Originally Posted by retina

If you give the caller free choice then it makes no difference how the caller chooses the dice to call, the odds are always 1 in 6.

I just ran some simulations on excel and this statement is entirely correct. Even using a random selection process the odds are 1 in 6.

This is very similar to the boy / girl thing LaurV mentioned: if there is any way at all to distinguish one die from the other the odds become 1 in 6.

For the "casino game" to actually work, you have to commit to a number i.e. 2. You have to choose 2, then look at the dice and then say whether or not at least one of the dice is 2. Now of course this number can be randomly chosen but for the sake of speeding up the game, you can't avoid the 25/36 cases where there are no 2's at all.

[Paul]

Originally Posted by sdbardwick

Against my better judgment, I've become involved in a discussion about probability elsewhere (no, it is not the Monty Hall problem) but it might be too counterintuitive for some to accept.

Scenario:
No trickery - each die is a standard six-sided cube, with each face numbered uniquely 1 to 6, and fair (not altered or loaded).
You need to determine the probability of each die in a pair of dice showing a 2, given certain information.


You throw two fair six-sided dice simultaneously so that each lands by itself in one of two separate boxes, labelled L and R. You cannot see into the boxes, but your partner can. Your partner then truthfully tells you that at least one of the die is showing a 2.

What are the odds that both of the cubes show a 2?

Please explain the process used to arrive at your answer.


I'll post my original response later, if required. I'm more interested in explaining why the incorrect answer is incorrect than anything else.

I come up with 1/11, others are vehemently supporting the more intuitive 1/6. I've done 3.6MM simulated rolls in Excel; the results support my answer.

Hi,
can anybody tell me how to work out the odds of throwing 5 dice and getting 5 of the same number twice in two consecutive throws?
Thanks,
P

[wblipp]

Originally Posted by Paul

can anybody tell me how to work out the odds of throwing 5 dice and getting 5 of the same number twice in two consecutive throws?

This is sounding like a homework problem. I'll give most of it away, but leave some calculation.

First, the problem is ambiguous. Does 5 ones followed by five sixes qualify? If no, then it is the same as getting 10 dice. If yes, then it is probability of doing one multiplied the probability of doing it on the second roll - the probability of A and B for independent events is the product.

Next, figure out the probability on the first roll. Think in terms of rolling (or perhaps revealing) the dice one at a time. The first roll can be anything - then each additional is ...

[Uncwilly]

...

[Xyzzy]

Originally Posted by Paul

can anybody tell me how to work out the odds of throwing 5 dice and getting 5 of the same number twice in two consecutive throws?

Here is a start:

[Paul]

Hi Xyzzy,
thank you for the link, I watched it and got the basic idea.
As Wblipp said, the statement was a bit a ambiguous, just to clarify, what I was trying to say was that I had thrown 5x5's and on the next throw threw 5x5's again. I am not really into exponential math but I would imagine that the odds would be heavily weighted against it occurring. Would there be a algebraic formula on could employ to work it out?
Thank you for your help,
Regards,
Paul.

[wblipp]

Originally Posted by Paul

what I was trying to say was that I had thrown 5x5's and on the next throw threw 5x5's again

Let's talk about this some more to be sure I understand. Because you say "had," I think this means that you are not asking about the probability of throwing the first 5x5's. You are imagining yourself in the situation of that having already happened. Mathematicians would probably phrase this as "Given that I have just thrown 5x5's."

You then want to know, given that, what's the probability the very next throw will do it again.

If that's what your asking, try this thought experiment. Imagine I am standing next to you throwing dice, and I have just thrown 2-2-3-4-6. Do you believe the probability that I will next throw 5x5 is different than the probability you will? If so, what is the method the dice use to know whether they are in your hand or mine?

Most people conclude the probability is the same, so the history has no effect on the probability. Depending on the point we are trying to make, we might call this "memorylessness" or we might say the two rolls are independent.

[retina]

Originally Posted by wblipp

... I have just thrown 2-2-3-4-6.

I think a more interesting problem would be to ask what are the chances you roll the combination 2-2-3-4-6 again given that you've just rolled the combination 2-2-3-4-6 now?

And an even trickier problem. With only one trial what are the chances you roll the same 5-die combination twice?

Note to those not paying 100% attention: There is a difference between a combination and a permutation.