It was two years ago today...

[Alan Mendelson]

This is not a question about rolling dice and seeing results.

This is a simple question: You have two dice one of which is a 2. What are the odds that the other die is also showing a two.

You can moan and groan all you want about conditional bullshit and look at graphs all you want showing the number of dice combinations displaying a 2...

But the simple fucking answer is 1/6.

Now fuck off.

[RS__]

This is not a question about rolling dice and seeing results.

This is a simple question: You have two dice one of which is a 2. What are the odds that the other die is also showing a two.

You can moan and groan all you want about conditional bullshit and look at graphs all you want showing the number of dice combinations displaying a 2...

But the simple fucking answer is 1/6.

Now fuck off.

The simple yet wrong answer is 1/6.


The question that was asked actually has entirely to do with rolling the dice and the (probability of the) result.

Conditional probability isn't the easiest thing to figure out. But at this point, Alan, you're just showing your ignorance. Try not to get so mad, it won't help you figure out the answer.


Mickey, you're wasting your time rolling the dice, if you think it's going to show Alan the answer is 1/11 and not 1/6. He's already made up his mind, he thinks the answer is 1/6. Anything other than 1/6 he'll claim is a different question.

[mickeycrimm]

Mickey, you're wasting your time rolling the dice, if you think it's going to show Alan the answer is 1/11 and not 1/6. He's already made up his mind, he thinks the answer is 1/6. Anything other than 1/6 he'll claim is a different question.

Yes, I know I'm wasting my time with Alan. But a lot of other people read this forum. All we can do is throw the evidence out there and let people make up their own minds. Not everyone will be convinced the answer is 1 in 11. As a matter of fact, I think if you proposed this question to the entire public most of them will say it's 1 in 6. But it doesn't make them right. I have the results from the first 100 decisions which I will be posting shortly.

[mickeycrimm]

While shopping in Walmart today I bought a yahtzee set just to pirate the cup and dice. So I'm ready to run the first sample space of 100 decisions. I'm conducting this experiment just as the question below dictates. This was the original question and it is the only one that matters for this empirical study. I'll get back to you with the results sometime tonight.

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?

It only took about 40 minutes to get out a 100 decisions. I'm posting a screenshot of the results. Every time I looked under the cup and one of the dice was a 2 I made a mark on the post-it on the left. Every time the other die was also a two I made a mark on the post-it on the right. In 100 decisions the other die was a two just 9 times. That's an average of once every 11.1 decisions. The sample space is still small but the pattern has already developed. I'm doing it in 100 decision blocks so we can see what the variance looks like per every 100 decisions. I'll be doing another 100 decision sample space after this post.

[Alan Mendelson]

There is no conditional math. The only condition is that one of two dice is known to be a 2. Therefore only six possible results remain on the second die.

Because it is only a two-dice problem the possible results will always be 1/6. Regnis got it. Only you "super smart math people" trip over your own two feet because you failed to read and understand the question and you jumped ahead with convoluted math that isn't called for.

[Count Room]

There is no conditional math. The only condition is that one of two dice is known to be a 2. Therefore only six possible results remain on the second die.

Because it is only a two-dice problem the possible results will always be 1/6. Regnis got it. Only you "super smart math people" trip over your own two feet because you failed to read and understand the question and you jumped ahead with convoluted math that isn't called for.

Alan: It's only if you know exactly WHICH of the two dice are a "2" that the probability improves to 1/6 for the other die. When the problem states "at least one of the dice is a 2", you don't know WHICH of the two dice would be a "2", lowering the probability to 1/11.

Here are ALL of the combinations in which at least one of the dice is a "2" and not knowing WHICH one (read top to bottom for each set):

Die A: 1 2 3 4 5 6 2 2 2 2 2
Die B: 2 2 2 2 2 2 1 3 4 5 6

That's 11 different combinations, right? Only one is a 2-2 (above in bold). Odds are 1/11.

[mickeycrimm]

Some slight variance in the 2nd sample space. The hard 4 came in ten times this time around. So we have the other die being a two for 19 times out of 200 decisions. That's an average of once in 10.53 decisions. Hey, I just couldn't get hot with the hard four's tonight. What can I say.

[regnis]

I should add that while Kentucky derby day has historically not been a good day for me, this year I hit 12 of 14 races, a load of pick 3's, two pick 4's and a pick 5. I made more in that one day than you can possibly make as a full time gambler. You can look up what those exotic bets paid if you want to be sick. That being said, I still prefer the safety of my part time job with no risk of loss. Do I hate dealing with clients--often yes. But I also hated gambling at times and dealing with the people involved in gambling.

Regnis, congratulations on your big Kentucky Derby score. But dare I ask, would your pile of losing tickets from over the years be big enough to fill a barn?

Thanks Mick. And I have large chests full of losing tickets that are organized and free of footprints that I keep for at least 7 years. Probably several barns. Luckily, they amount to far less than the wins, notwithstanding my worst year ever last year. I have over 50 years in this game, about 15 of which were my sole source of income when I soured on law.

I actually should have more losing tickets, but I am stubborn and refuse to play a lot of boxes or wheels whereby you automatically have more losing tickets. While the frustration of having an exacta the wrong way is always there, and large hits can be made had you boxed, I just hate making a higher number of losing bets. It may have actually hurt me over the years, but that is my style. Ans of course, when I am right, all my money is on the right way rather than half or less.

[jbjb]

There is no conditional math. The only condition is that one of two dice is known to be a 2. Therefore only six possible results remain on the second die.

Because it is only a two-dice problem the possible results will always be 1/6. Regnis got it. Only you "super smart math people" trip over your own two feet because you failed to read and understand the question and you jumped ahead with convoluted math that isn't called for.

No. You guys are just stupid. I showed you how it's 1 in 11. Stick with infomercials.

[Alan Mendelson]

Alan: It's only if you know exactly WHICH of the two dice are a "2" that the probability improves to 1/6 for the other die. When the problem states "at least one of the dice is a 2", you don't know WHICH of the two dice would be a "2", lowering the probability to 1/11.

Here are ALL of the combinations in which at least one of the dice is a "2" and not knowing WHICH one (read top to bottom for each set):

Die A: 1 2 3 4 5 6 2 2 2 2 2
Die B: 2 2 2 2 2 2 1 3 4 5 6

That's 11 different combinations, right? Only one is a 2-2 (above in bold). Odds are 1/11.

You guys are having a real problem with reality, aren't you?

There are only two dice. You can't use the same die twice to figure the answer to the question.

If I tell you that at least one of the two dice is showing a 2, then the answer must be with the remaining die.

It doesn't matter which of the two dice we say is showing a two. Try to get this to sink in: It doesn't matter which of the two dice we say is showing a two.

Why (please tell me) do you think it matters with a two dice problem?

If we have two dice and one of them is known to show a 2, then only one die remains. (Hint: 2-1=1) And on that remaining die (it doesn't matter which one it is) there are only 6 faces. Of those six faces, only one face is a 2.

What the fuck is the difference which die it is in a two fucking dice problem?

[mickeycrimm]

This is not a question about rolling dice and seeing results.

But, Alan, here is your quote from post number 39 "To everyone who thinks I'm wrong: please get yourself two dice, a cup, shake rattle and roll, and then figure what the odds are for the second die to also be a two."

Hey, I took your advice, went out and spent $9 to get a cup and dice, spent a couple of hours rolling the dice, and the results so far are very close to 1 in 11. Why did you advise people to do that if you were just going to say it's not a matter of rolling dice and seeing results? T

[Count Room]

This is not a question about rolling dice and seeing results.

But, Alan, here is your quote from post number 39 "To everyone who thinks I'm wrong: please get yourself two dice, a cup, shake rattle and roll, and then figure what the odds are for the second die to also be a two."

Hey, I took your advice, went out and spent $9 to get a cup and dice, spent a couple of hours rolling the dice, and the results so far are very close to 1 in 11. Why did you advise people to do that if you were just going to say it's not a matter of rolling dice and seeing results? T

Even Rob has told us to "Always do what you say you're going to do." Roll them bones!

[mickeycrimm]

[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?

Alan, the above question is the original question asked. Two years ago I found some college math professors online and submitted this question to them. They all agreed that the answer to the above question, repeat, the above question, and not any other question, is 1 in 11. I talked one of them into coming onto this site and explaining the answer, which he did. You quickly discounted him and stuck to your answer of 1 in 6.

Alan, you have some pretty heavy journalist credentials. I'm sure you know the right channels to go through. Would you submit the above question, written exactly as it is above, to several math departments of universities in southern California? Then publish their responses here?

If even one college math professor tells you that the answer is 1 in 6 I will kiss your ass on main street at high noon with Rob Singer and everyone else watching.

[mickeycrimm]

I should add that while Kentucky derby day has historically not been a good day for me, this year I hit 12 of 14 races, a load of pick 3's, two pick 4's and a pick 5. I made more in that one day than you can possibly make as a full time gambler. You can look up what those exotic bets paid if you want to be sick. That being said, I still prefer the safety of my part time job with no risk of loss. Do I hate dealing with clients--often yes. But I also hated gambling at times and dealing with the people involved in gambling.

Regnis, congratulations on your big Kentucky Derby score. But dare I ask, would your pile of losing tickets from over the years be big enough to fill a barn?

Thanks Mick. And I have large chests full of losing tickets that are organized and free of footprints that I keep for at least 7 years. Probably several barns. Luckily, they amount to far less than the wins, notwithstanding my worst year ever last year. I have over 50 years in this game, about 15 of which were my sole source of income when I soured on law.

I actually should have more losing tickets, but I am stubborn and refuse to play a lot of boxes or wheels whereby you automatically have more losing tickets. While the frustration of having an exacta the wrong way is always there, and large hits can be made had you boxed, I just hate making a higher number of losing bets. It may have actually hurt me over the years, but that is my style. Ans of course, when I am right, all my money is on the right way rather than half or less.

Regnis, you're a pretty rare breed. I think successful horse bettors are the rarest of all AP's. I knew an old horse bettor years ago that had supported himself for over 50 years at the trade. He once told me that out of the thousands of horse bettors he knew in his lifetime there was only about a dozen, that if he owned the horsebook, he would tell he didn't want their business.

[Alan Mendelson]

[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?

Alan, the above question is the original question asked. Two years ago I found some college math professors online and submitted this question to them. They all agreed that the answer to the above question, repeat, the above question, and not any other question, is 1 in 11. I talked one of them into coming onto this site and explaining the answer, which he did. You quickly discounted him and stuck to your answer of 1 in 6.

Alan, you have some pretty heavy journalist credentials. I'm sure you know the right channels to go through. Would you submit the above question, written exactly as it is above, to several math departments of universities in southern California? Then publish their responses here?

If even one college math professor tells you that the answer is 1 in 6 I will kiss your ass on main street at high noon with Rob Singer and everyone else watching.

What's interesting Mickey is that two years ago, when this was being discussed on the WOV forum, and then here, I put the question WORD FOR WORD to 8 different dice experts. They happened to be the dealers at two tables at Caesars Palace. Six dealers and two floormen. They all had the same answer: 1/6

Do you know what they told me when I suggested the answer might be 1/11 ? They all said the same thing: in this casino you can only count each die once. So, if one die is showing a 2, you can't rotate that die to show any other number -- it's a 2 and it can't be considered anymore.

So why don't you and all the others who say the answer is 1/11 go to a casino and ask the dealers this very same question. Let me know what they say. If you happen to get a dealer at a casino who says the answer is 1/11 get me the casino's name and the badge name (full name would be better) so I can visit with that dealer at that casino.

[mickeycrimm]

What's interesting Mickey is that two years ago, when this was being discussed on the WOV forum, and then here, I put the question WORD FOR WORD to 8 different dice experts. They happened to be the dealers at two tables at Caesars Palace. Six dealers and two floormen. They all had the same answer: 1/6 Do you know what they told me when I suggested the answer might be 1/11 ? They all said the same thing: in this casino you can only count each die once. So, if one die is showing a 2, you can't rotate that die to show any other number -- it's a 2 and it can't be considered anymore. So why don't you and all the others who say the answer is 1/11 go to a casino and ask the dealers this very same question. Let me know what they say. If you happen to get a dealer at a casino who says the answer is 1/11 get me the casino's name and the badge name (full name would be better) so I can visit with that dealer at that club.

Good grief, Alan. You're calling them experts? If you really want to know how much those dice table flunkies know about dice combinatorics then the next time you are at the dice table ask them this question "When rolling three dice, how many combinations add to ten?" I'll even give you the answer so you can compare it to their answers. It's 27. I can show you the math if need be. So go ahead and ask them this question then watch their blank stares.

[Alan Mendelson]

Hey Mickey instead of asking bullshit questions about three dice, how about arguing with this:

"in this casino you can only count each die once."

[RS__]

Alan.....dealers, floor people, and pit bosses are generally not experts on probability. To say they are experts on dice is like saying the cooks at McDonalds are expert chefs.

Hell, one time my boss backed off our biggest bettor because he was winning.....he always played the same way: Hop hard ways, regular hard ways, horn bets, etc. for the max he could, and play until he was stuck $20k.....he could be up $40k but he'd still lose it all back, if it took 1 hour or 10 hours.

I asked the other dealers I was dealing with the same question, word for word. Three dealers total. One almost instantly said 1/6 because the other die had 6 sides. Another waited a few moments, his brain working, and said 1/36 because that's the chance of a 2-2 being rolled. The other was still thinking, said 1/12. I asked why 1/12 and he said because you could roll a 1-2, 2-2, 3-2, 4-2, 5-2, 6-2, 2-1, 2-2, 2-3, 2-4, 2-5, 2-6. Then thought about it for a second more and said no, it's 1/11 because he double counted 2-2.

I asked a boss and he just said "it's random, dice are random". I said yes they're random but what do you think the probability is or chance? Like flipping a coin and it landing on heads is 50% or 1/2. Again, he said its random.

I asked them if you are dealt a blackjack, what's the probability the first card is a ten or face card. The guy who said the dice prob was 1/6 said it was 4/13 for the BJ problem, because 4 of 13 cards are tens. The other dealer had some convoluted math and came up with an answer worse than Rob could come up with. The one who said the dice are 1/11, that there's a 1/2 chance your first card is a ten if you have a blackjack, because there are 2 combinations to get a BJ, either Ace+Ten or Ten+Ace. He was 2 for 2.

I asked a few other probability questions, riddles, etc. as did they. It became apparent the guy who said the dice probability is 1/11 actually knew how to do math, while the other two were a little slow, to say the least.

The fact you think dice dealers and floor people are considered "experts" is a bit telling.



I'd be careful about asking a group of math professors any question and accepting (sort of) the answer from just one. I had a computer science professor who gave us a homework assignment that was literally impossible in the language we were using. I don't remember the specifics now, but he wanted us to write a function which would use pass-by-reference which, done the way he wanted, was impossible (per the book and any expert or beginner). I asked for an example of code which would be sufficient, and the code he gave did not meet the criteria in the assignment (pass by reference). I ended up getting a 100% on the assignment, but only after many emails back and forth as well as getting the head of the department involved. The head of the department said something like, "This assignment was impossible. The language simply doesn't work that way."

[Alan Mendelson]

In this case the dealers have a better idea of the actual problem. And they are not using one die to have multiple values. In other words, they have a better grip on the reality of the question.

You "1/11 guys" still don't understand how you are misinterpreting simple English.

I guess when you take probability classes they forget to teach how to understand the English language.

I gave you the question that would have an answer of 1/11.

[a2a3dseddie]

How about this:

In a standard game of craps considering only the rolls where "At least one of the dice is a 2." What are the odds of being paid on a Hard 4 bet?

Let's see, you CAN'T have an "easy 4" since you know "At least one of the dice is a 2." so 1,3 or 3,1 have been ruled out.

So, your only loss comes if a 7 appears... 2,5 or 5,2

The odds of a 5 appearing on a single die is 1/6 (the same as another 2 appearing).

Do you think it's an even money bet by disregarding the possibility of a 5,2? And what wording in the original question would make you disregard the possibility of 5,2?

Or are the odds actually 2-1 against you? (2,5 or 5,2 versus 2,2)