Two Dice Puzzle

I think the best thing to do now is just to do it.

Everybody, believers and non believers alike, shake two dice in a cup, slam them on a table, and record the number of times you have just one two, and record the number of times you have double twos.

I've got a few minutes now, I'll give it a try and report back.

Quote: Dalex64

I think the best thing to do now is just to do it.

Everybody, believers and non believers alike, shake two dice in a cup, slam them on a table, and record the number of times you have just one two, and record the number of times you have double twos.

I've got a few minutes now, I'll give it a try and report back.

I did this last night. I got it every 10.8 rolls

I just rolled the dice for 10 minutes.

I got just one 2 : 102 times
I got double twos : 7 times
That is about 1 in just over 14.
Not very lucky, about 22% fewer than I was expecting.

To speed things up, I didn't use a cup. I grabbed the first dice I found, which were about 10 cheap parcheesi style dice. I rolled them quckly, sometimes changing dice, did not keep track of the total number of rolls, counted single two appearances in my head and double twos on my fingers. When I got to 80:5 I tried to pick the dice that looked like they rolled more twos. Didn't help much, as like I said I ended up at 102:7. So even the "I think they are rolling a lot of twos" dice came in at 22:2

If I had to do it with a cup and with peeking, it would have taken a lot longer, or I wouldn't have been able to throw the dice as often in the same amount of time.

Quote: AlanMendelson

I think this is his strategy. To confuse, mislead, deny, attack and then run away.

He's already linked this dice bet to a video poker bet -- did you see that?

I just posted on my forum conflicting statements that he has made -- first supporting 1/6 then saying it's 1/11.

It's amazing.

Anyway... I can be in Vegas to do the bet and do lunch on Saturday.

Alan, you are a good egg. I for one could and would hang with you in Vegas if I can ever get back there. Singer on the other hand is poison But you're a free speech and a stand up guy. Good for you!

Quote: Dalex64

I got just one 2 : 102 times
I got double twos : 7 times
That is about 1 in just over 14.
Not very lucky, about 22% fewer than I was expecting.

Thank you.

Maybe my simulation got overlooked, but here are my results of a billion bets resolved:

Quote: wizard

Wins=90912246 (9.0912%)
Losses=909087754 (90.9088%)

Quote: Wizard

but <snip>

paying 9 to 1 on the 2-2
i gets (calculate)
a player probability of a net loss over (flat bet of course)
360 dice rolls (proposed)
to be abouts 63.8%


paying 9 for 1 i gets a player probability of a net loss over 360 dice rolls to be abouts 76.9%

a sure bet contest for sure

btw,
Alan M is a lucky man
except maybe at marriages at a craps table

Go Alan Go!

Sally

Quote: Wizard

Thank you.

Maybe my simulation got overlooked, but here are my results of a billion bets resolved:

Yes, I saw your code and your results. I think it is a nice, clear, and concise program. I did a similar one in perl (and with fewer iterations since perl is slower) hoping others might run it.
So, people can run a sim in excel, perl, and C. Anyone want one written in Java or Python?

But, seeing is believing, and I thought I better do what I was asking others to do, in an actual experiment.

Three more days to MonkeyFest. I'm hoping Alan will confirm showing up at the SLS Friday evening to settle this.

Oops. I missed the "monkeyfest."

Yes, I can drive up Friday but I am not sure on the time. What time is the event and if I arrive late is there an alternative plan? My departure time on Friday will depend on two things:

1. what time we deliver the new TV show (it's a weekly show)
2. what excuse I give my girlfriend and if it works

would someone please PM me the "meet up" info and is there a cell number for someone so I can coordinate? thanks

Quote: AlanMendelson

Oops. I missed the "monkeyfest."

Yes, I can drive up Friday but I am not sure on the time. What time is the event and if I arrive late is there an alternative plan? My departure time on Friday will depend on two things:

1. what time we deliver the new TV show (it's a weekly show)
2. what excuse I give my girlfriend and if it works

Thanks for asking. The tentative plan is dinner at the SLS and then some gambling. I'll push for a late dinner. We should be around until 10PM or so.

Thanks but now I am unlikely to arrive in time. My girlfriend is coming and we won't be able to leave LA till 6pm, so arrival before 10 is not possible.

If there is a way to do this Saturday morning -- prior to LUNCH -- please let me know. PM would be best. And if anyone does PM me please leave a cell number, thanks.

By the way, I blocked Singer from my forum again. I gave him another chance to "behave" but he just can't and once again returned to his practice of attacks.

Quote: AlanMendelson

Thanks but now I am unlikely to arrive in time. My girlfriend is coming and we won't be able to leave LA till 6pm, so arrival before 10 is not possible.

I'll give you my number just in case we linger until 11 or so. Otherwise, if you want action with me, I'd have to ask you to come out to Summerlin. Axel or Ace may be willing to meet more centrally to do it.

Quote: Wizard

I'll give you my number just in case we linger until 11 or so. Otherwise, if you want action with me, I'd have to ask you to come out to Summerlin. Axel or Ace may be willing to meet more centrally to do it.

I'm still happy to set this up as a UK hosted, video demo or even a live stream, at the weekend. I can promise that the whole film set would be ultra shoddy with monopoly dice and a cup or mug from my kitchen cupboard and a little hardboard slamming area. Two piles of pennies (say 500p for banker and I'll lend 500p to Alan). No apologies for using UK funny money, you'll need to use google to get the current exchange rate.

All a total time-wasting session of course. I actually trust Alan to set up his own playing environment, but I don't trust him not to play his own game. Setting aside one dice and tossing the other is absolutely not permitted in my casino.

Nonetheless, my set up would be quite quite fairly done. If anyone wants to make a professional movie production copying my scenery, they can contact me to arrange royalties.

As Sally rightly points out, with just a few hundred resolved bets, Alan would still have about a 20% chance of coming out ahead.

On a much lighter note: Can anyone see the irony in the quotes below?
It looks like Alan can't make 6, so Wizard is offering 11.

OK. I used selective quoting. So bite me.

Quote: AlanMendelson

... we won't be able to leave LA till 6pm, so arrival before 10 is not possible.

Quote: Wizard

I'll give you my number just in case we linger until 11 or so...

The problem provides insufficient information to deduce an answer. The additional information needed is what possible things your partner might say upon peeking under the cup, and what his strategy is to select what to say.

For reference, I restate the problem here: "You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, 'At least one of the dice is a 2.' What is the probability that both dice are showing a 2?"

If the partner must truthfully say either "At least one of the dice is a 2," or "Neither die is a two," then if he says the former, the conditional probability that both dice are showing a 2, given his statement, is 1/11.

But suppose he may truthfully make either of those statements or may also say, if the statement is true, "Both dice are showing a 2." Now he has two possible statements he may make if both dice are showing a 2. If he decides always to say "Both dice are showing a 2," when that is true, then if he says, "At least one of the dice is a 2," we know the conditional probability that both dice are showing a 2, given that statement, is zero!

For another example, suppose he may only truthfully say, "At least one of the dice is an N," where N is an integer between 1 and 6, inclusive. If both dice are the same, his statement is forced, but if they are different, he has a choice. Say he always decides to comment using the higher number of the two dice. Thus if the dice show 2 and 3, he says "At least one of the dice is a 3." Now the conditional probability that both dice are showing a 2, given that his statement is, "At least one of the dice is a 2," is 1/3.

Things are even worse if he is allowed to lie on occasion but happens to be telling the truth this time.

I believe the problem was really meant to be written as follows:

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and you ask him, "Is at least one of the dice is a 2?" You know he always tells the truth. He replies, "Yes." What is the conditional probability that both dice are showing a 2, given his statement?

For this problem, the answer is 1/11.

I'm sure you're right, but please explain to me what is wrong with my reasoning:

The two die rolls, despite having happened at the same time, are independent events. The unknown die has no knowledge that the other die is a two. Thus, the unknown would be just as likely to have been a two as any other number; thus, 1 in 6.

Quote: OrkFromPluto

I'm sure you're right, but please explain to me what is wrong with my reasoning:

The two die rolls, despite having happened at the same time, are independent events. The unknown die has no knowledge that the other die is a two. Thus, the unknown would be just as likely to have been a two as any other number; thus, 1 in 6.

This thread has many detailed and colorful analyses, but I'll try a quick response.

The only knowledge you have is that the result of the the roll contains at least one 2.
When calculating probabilities, you can only exclude possible results if the particular result is impossible.
Here we know that there is at least one 2, so we know the outcome must contain at least one 2.
So how many of the 36 possible results contain at least one 2? (11) The other results are impossible, since they don't contain at least one 2.
How many of those 11 results contain two 2s? (1)
So 1 out of 11.

Done beating the dead horse...

Quote: OrkFromPluto

I'm sure you're right, but please explain to me what is wrong with my reasoning:

The two die rolls, despite having happened at the same time, are independent events. The unknown die has no knowledge that the other die is a two. Thus, the unknown would be just as likely to have been a two as any other number; thus, 1 in 6.

It's in your phrasing. Which one is "the other die"? What if they're both "the other die" because they're both 2? If you count them twice, you end up with 1 in 6 because you're really counting 2 in 12. But if you *don't* count them twice, you end up with 1 in 11.