The Wizard will bank this bet: 1/6 vs 1/11

[Alan Mendelson]

Will you allow me to set the dice?

Have you seen my video about dice sliding?

Have you seen my quotes in the Las Vegas Review Journal about the Wynn dice sliding case? http://www.reviewjournal.com/news/cr...wynn-las-vegas

You know that I was stopped shooting at three Vegas casinos: Bellagio, MGM and NYNY ?

[kewl]

Will you allow me to set the dice?

No, you use a cup as per the OQ, and you don't cheat.

[Alan Mendelson]

No, you use a cup as per the OQ, and you don't cheat.

As I said before, at a craps table when throwing two dice, then the 2-2 will appear 1/36 times.

But everytime you tell me that two dice were thrown and at least one is showing a 2, the chance that the other die shows a 2 is always 1/6.

think about it.

[kewl]

As I said before, at a craps table when throwing two dice, then the 2-2 will appear 1/36 times.

But everytime you tell me that two dice were thrown and at least one is showing a 2, the chance that the other die shows a 2 is always 1/6.

think about it.

And that's exactly what I'm challenging you to do. Prove that when at least one is showing a 2, the chance that the other die shows a 2 is always 1/6.
We are not interested in all other, non-2 throws, not counting them.

So since I'm saying your methodology is flawed in your previous video and you are not eligible to come after the establishment throw and pick the other die and show it has 6 faces, therefore 1/6, I'm saying you prove it to us, the ignorant, with real physical dice in a real throws experiment. That's how it works in the real life doesn't it? No jumping two's from one die to the other and no setting of one die and concluding "the other" is 1/6.

The real deal.

[Alan Mendelson]

And that's exactly what I'm challenging you to do. Prove that when at least one is showing a 2, the chance that the other die shows a 2 is always 1/6.

Now you are being childish. Even if I am allowed to slide the dice there is no guarantee that when at least one die is showing a two "that the other die shows a 2 is always 1/6."

And the problem with actually rolling dice is that anything can happen -- which is why some people win at craps even though the house has an advantage on all bets except the odds.

Rolling dice will not prove anything and I am the first to admit it. In fact this is why I never expected to win my "lunch bet" with the Wizard especially since the number of rolls would be so few.

[kewl]

There are two ways to have our proof.
Mathematical , which you are rejecting.
And empirical, which you are uncertain about.

Will you accept computer simulation then? No, right?

Why? It is the same thing and has the benefits of saving time and ability to generate huge number of trials.

[Alan Mendelson]

I would accept computer simulation if you asked the question correctly. Here is the correct question:

Two dice were rolled and at least one of the two dice was identified as showing a "2." What are the odds that the other die is also showing a "2"?

Go ahead. Simulate that for me.

[kewl]

I would accept computer simulation if you asked the question correctly. Here is the correct question:

Two dice were rolled and at least one of the two dice was identified as showing a "2." What are the odds that the other die is also showing a "2"?

Go ahead. Simulate that for me.

Cool. The more correct question is though:

Two dice were rolled and at least one of the two dice was identified as showing a "2." What are the odds that both dice show "2"?

as per the OQ and per the OP in this thread. Agreed?

[indignant99]

junk deleted

[kewl]

No, it isn't. You fail to acknowledge that the 3 could have been either die.
Consequently, the 5 could have been on either (other) die. The answer is
also 1/11, however ludicrous this might sound to you.

Yeah this one is worded unclear. The way I read it, it implies the peeker announces random value no matter what the outcome is. Check my second response where I clarify the two scenarios.

[Alan Mendelson]

Cool. The more correct question is though:

Two dice were rolled and at least one of the two dice was identified as showing a "2." What are the odds that both dice show "2"?

as per the OQ and per the OP in this thread. Agreed?

I'm not sure how you use language in a computer simulation but the two key phrases from the original question are these:

"At least one of the dice is a 2."

What is the probability that both dice are showing a 2?

If you can use those phrases in your simulation it would be okay with me. But again, how do you translate English into a simulation for a computer? I am unfamiliar with it.

[kewl]

I'm not sure how you use language in a computer simulation but the two key phrases from the original question are these:

"At least one of the dice is a 2."

What is the probability that both dice are showing a 2?

If you can use those phrases in your simulation it would be okay with me. But again, how do you translate English into a simulation for a computer? I am unfamiliar with it.

I don't . I'm unfamiliar with simulations of this sort too. So I am now calling for a volunteer to do the sim. I think OnceDear has prepared spread sheet in excel some time ago.

[indignant99]

. The answer to that one is 1/6
That's a different question - one that assumes the peeker will announce the value of a random die even if there is no 2 under the cup. The answer to that one is 1/6.
This is not the question asked here though.

No it isn't. Since one die is a three (it could be on either die), you get the eleven possibilities in the "3" criss-cross, and you have two chances to hit in that criss-cross. The answer is 2/11, no matter how ludicrous that sounds.

[Alan Mendelson]

OnceDear's spread sheet and all those other spreadsheets looked at the combinations of 2 dice that contained at least 1 two, and from that they said the answer is 1/11. They did not respond to the question.

And when I said they didn't respond to the question that's when they started with "you don't know which die is a 2."

And to that I answered does it make a difference?

So the simulation comes down to how the question is translated into math and whether the simulation looks at one die or two dice or combinations of two dice... and that takes us back to start of this entire discussion.

I won't accept your simulation unless it looks at one die. Guess why I said that!

[kewl]

No it isn't.

Thanks, nice analysis. But I already responded to your previous post, where I say what the question implies at first read (not worded clearly enough) and how I made a subsequent response, clarifying both scenarios.

Edit:

Oh, you are correct of course, my bad.
The answer is 2/11. Totally blacked out on this one, thanks.

[kewl]

I won't accept your simulation unless it looks at one die.

What do you mean?
It has to look at the result of fair throw of a pair of dice, does it not??

[kewl]

Let me rephrase.

If the question was:

You roll a pair of dice and no matter what the outcome is the peeker will announce the value of one of the dice (at random). What are the odds that there will be a: pair, a (announced value) and 1, a (announced value) and 2, a (announced value) and 3...whatever),
then the answer will be 1/6.

And if the question was (in your case with 5 and 3):

You roll a pair of dice. The peeker makes announcement when at least one 3 or one 5 is under the cup. What are the odds that there are exactly 5 and 3 under the cup?
then the answer is 1/11.

Now, compare the second question with the one asked here. Keep in mind that there was discussion about whether the peeker announces a value every time or only if there is a 2 under the cup and it was agreed upon, and Alan concurred, that the case in discussion is the one where the peeker only announces if there is a 2. Otherwise he says nothing.

The second part of this answer is incorrect. The correct answer is 2/11.
Because we have two 3's and two 5's vs only two 2's in OQ.
Still not 1/6. But twice as much as 1/11. Guess why this is?

[indignant99]

I have 6 men wearing 6 different colored shirts. Across the street are 6 women wearing 6 different colored shirts. Each group has one red shirt. If I select the guy with the red shirt from group 1, what are the odds of blindly picking the matching shirt from group 2.

Hint--it ain't 1 of 11.

Wrong proposal, s-i-n-g-e-r backwards. You need to choose a shirt, happens to be red, but you don't know whether it was a man's or a woman's. Whichever (mens or womens), now pick again off the "other" group. If you are eyeballing a man's shirt, you have 5 ways to mismatch picking a woman (red). If you are eyeballing a woman's shirt, you have 5 ways to mismatch picking a man (red). You have 10 ways to mismatch, only 1 way to match.

The problem ain't just - forget the men altogether - pick a red woman. 1/6. That's a complete and utter castration of the question.

[Alan Mendelson]

The second part of this answer is incorrect. The correct answer is 2/11.
Because we have two 3's and two 5's vs only two 2's in OQ.
Still not 1/6. But twice as much as 1/11. Guess why this is?

Are you asking me? Yes, there are two choices for each die. Much different than the question about being a 2 and what are the odds for another 2.

[OnceDear]

I don't . I'm unfamiliar with simulations of this sort too. So I am now calling for a volunteer to do the sim. I think OnceDear has prepared spread sheet in excel some time ago.

He didn't understand it then. He won''t understand it now.
Sim rolling 50000 times, showing all random outcomes and indicating the times where at least one die was a deuice.. And no. I did not set one of the fcuking dice.