[unreality]

In continuation of my other magic trick riddle, I have invented a new one...

* I have 5 dice- 3 red and 2 black. They are normal, untampered 6-sided dice

* roll the 5 dice without me looking and sum up the faces

* flip all of the red dice over so that their bottom sides are showing face up

* add the new red dice values to the total and this is your FINAL TOTAL

* remove the red dice from the playing area and hide them so that only the two black dice are on the table now

* choose EITHER of the black dice (I have no idea which) and flip around so that it's bottom face is showing up

* tell me to turn around

I have a 2/3 chance of guessing the correct FINAL TOTAL: how? and why only 2/3?

Edited August 5, 2008 by unreality

[Guest]

to start off

and their flip-sides will always be 21, so you're left figuring out which black die has been turned over. gotta think about that one some....

Edited August 5, 2008 by Cherry Lane

[Guest]

ok, I can get 1/2, but not 2/3

21 plus die A plus the flip side of die B (the flip side number will be 7 minus the number showing)

or

21 plus die B plu the flip side of die A

(if A and B are the black dice)

I guess to improve your odds we have to get into probabilities that the black dice either showed the same number or the flip sides of each other (so once one is flipped they are now the same) In either of these cases you would be sure of knowing the sum.

Edited August 5, 2008 by Cherry Lane

[unreality]

you're close ;D

btw nice sig... I'm just that good at starting trends

[Guest]

you're close ;D

btw nice sig... I'm just that good at starting trends

Thanks .. reread it. I was editing as you were answering.

And yes, I copied you on the sig. Lets see if it works...I don't know why no one has solved MY riddle yet!

[unreality]

ok, I can get 1/2, but not 2/3

21 plus die A plus the flip side of die B (the flip side number will be 7 minus the number showing)

or

21 plus die B plu the flip side of die A

(if A and B are the black dice)

I guess to improve your odds we have to get into probabilities that the black dice either showed the same number or the flip sides of each other (so once one is flipped they are now the same) In either of these cases you would be sure of knowing the sum.

what you're right about:

the flip sides of a dice always total to 7, so the three red dice contribute 21 to the final total- that is the rule I used which is hard to spot if you're having the magic trick performed to you but easy once you think about it. So what the red dice were is meaningless, the total is just 21 plus the two black dice. However one of the black dice has been flipped over... which is it? It might seem like 1/2 chances, but in your edit you realized that's not the case............ How does it have a 2/3 chance of success instead of 1/2?

[Guest]

there is a 1/6 chance that the 2 black dice rolls resulted in the same number and a 1/6 chance they rolled the reverse of each other, for a 1/3 chance you know the sum.

For the other 2/3 chance possibilities, you have a 1/2 chance of correctly guessing which die was turned over, for a total probability of 1/3 + (2/3)*(1/2) = 2/3

[unreality]

CL is right ;D And before anyone else had a crack, too! nice I'll wait for some more people to solve it

[unreality]

a further problem:

What would your chances of guessing the right total be if the person had the option of flipping over neither, just one die, or both?

[Guest]

a further problem:

What would your chances of guessing the right total be if the person had the option of flipping over neither, just one die, or both?

I'll leave this one alone. Probabilities are not my forte (that's why it took 3 posts and edits to answer your first question).

[unreality]

maybe I'll think of another magic trick today ;D

[Guest]

the pips on opposit sides of any standard dice add to 7, so the total of the red dice is ALWAYS 21.

Here's where I disagree with your math...

The total of the black dice is either...

Dice A +(7 - Dice B) OR

Dice B +(7 - Dice A)

So you have a 50/50 shot at guessing the answer (unless the 2 black dice are the same number then the answer is always 28)

The only way I can think of having a 3rd possibility is to concider the possibility that they didn't actully flip a die

[unreality]

mojojojo = nope there is a way to conquer the 50-50 odds to give you a 2/3 chance of performing it successfully instead of a 1/2

[Prime]

I see how your probability is 7/12, but not 2/3.

Sum of any 2 opposite sides of dice is 7. So your manipulations with red dice added 21 to the Final Total. You may subtract the same number and disregard red dice altogether.

There are 36 ways to roll 2 dices. When you see two dice showing the same number like (5,5), then it does not matter which one was turned -- their sum was 7 before turning one over. In all other variations you have a 1/2 chance. For example, if you see (5,2) then the sum could have been either 10, or 4.

(6*1 + 30*1/2)/36 = 7/12

I'm ready to give up on 2/3. What's the trick?

[unreality]

dang, you're right. The chances are 7/12 not 8/12

if the two black dice are the same, the final total is always 28 (1/6 of this happening)

if the two black dice are different, you have a 1/2 chance of guessing the correct original values of the black dice, so there is a 2.5/6 of getting it wrong and 2.5/6 to getting it right, therefore a 3.5/6 or 7/12 of guessing the final total correctly

I don't know what I was thinking.. 1 & 6 could be 11 or 66 of course

[Prime]

there is a 1/6 chance that the 2 black dice rolls resulted in the same number and a 1/6 chance they rolled the reverse of each other, for a 1/3 chance you know the sum.

For the other 2/3 chance possibilities, you have a 1/2 chance of correctly guessing which die was turned over, for a total probability of 1/3 + (2/3)*(1/2) = 2/3

When you see two dice showing the reverse of each other -- you still have only 1/2 chance of guessing. E.g. (5,2)-- their sum before flipping could have been either 10, or 4.

[Guest]

When you see two dice showing the reverse of each other -- you still have only 1/2 chance of guessing. E.g. (5,2)-- their sum before flipping could have been either 10, or 4.

yes, I see that now. In my haste to come up with unreality's answer, I neglected to stop and think about whether it actually made sense!

[Prime]

In continuation of my other magic trick riddle, I have invented a new one...

* I have 5 dice- 3 red and 2 black. They are normal, untampered 6-sided dice

* roll the 5 dice without me looking and sum up the faces

* flip all of the red dice over so that their bottom sides are showing face up

* add the new red dice values to the total and this is your FINAL TOTAL

* remove the red dice from the playing area and hide them so that only the two black dice are on the table now

* choose EITHER of the black dice (I have no idea which) and flip around so that it's bottom face is showing up

* tell me to turn around

I have a 2/3 chance of guessing the correct FINAL TOTAL: how? and why only 2/3?

I would rephraze the question in your problem as following: "Will you bet an even money, that I will not guess the Final Total?"

[unreality]

good idea, since it's 7/12 over 6/12 so there's a slight house-advantage for the magician Hehe, you phrased it better than I could've