vpFREE2 Forums

The Dice Challenge

#1

You have two dice in a cup. You shake them up then slam the cup down on the table. Your partner peeks under the cup and says "at least one of the dice is a two." What is the probability that the other dice is a two? This question was proposed in a thread at WoV. My first thought was it's 1 in 6. But a heavy debate broke out and the strong math guys said it was 1 in 11. How could that be I asked myself. So I sat down with pencil and paper and figured it out. Sure enough it was 1 in 11.

But there was a large faction in the thread that insisted it was 1 in 6. Then the Wizard of Odds jumped in the thread and said he would take all comers for any stakes and played them as long as they liked. Anytime at least one of the dice is a two, if the other die is a two then the Wz would pay them 9 bets. If the other die wasn't a two then they would just have to pay him one bet.

Alan Mendelsen, a strong proponent of 1 in 6, insisted the Wiz would get his clock cleaned in such a game. The Wiz told him to put his money where his mouth is. Alan objected to gambling with other individuals on moral grounds. So the Wiz made a new proposal. When at least 1 of the dice is a two, if the other die is a two Alan would be awarded 8 points. If the other die wasn't a 2 the Wiz would be awarded 1 point. The first person to get to 25 points wins and the loser has to buy lunch.

Alan accepted the challenge and believes he will win the bet. I thought this was some brilliant hustling by the Wiz. I've never really given him any credit for street smarts until now. If Alan rolls the hard 4 three times he will only hae 24 points. He will have to roll it one more time to win.

My advice to the Wiz. Never give a sucker an even break. Order the most expensive thing on the menu.

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#2

Mickey ... you left out a key detail: How many Moose Drools had you consumed that day?

The standard setup of any probability is the you determine how many elements meet the desired condition, and how many are in the stated population.

For 2 dice, there are 11 configurations that have at least one two; only one of these are both twos.

< Of course Michael's prop is the more interesting aspect of your story! >

- H.

---In vpFREE@yahoogroups.com, <mickeycrimm@...> wrote :

You have two dice in a cup. You shake them up then slam the cup down on the table. Your partner peeks under the cup and says "at least one of the dice is a two." What is the probability that the other dice is a two? This question was proposed in a thread at WoV. My first thought was it's 1 in 6. But a heavy debate broke out and the strong math guys said it was 1 in 11. How could that be I asked myself. So I sat down with pencil and paper and figured it out. Sure enough it was 1 in 11.

\

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#3

I'm not a big beer or ale fan. I just use them for chasers. In the course of an evening at the bar I might consume just 2 beers but will down about 20 shots of Black Velvet. And contrary to popular opinion I'm not a full time drunk. I'm in the middle of my bi-monthly road trip so haven't touched the stuff for some days. I never touch the stuff on roadtrips. I don't have ant time for leisure. I just want to get into the towns, get the money and go back home. A typical roadtrip is 2 to three weeks.

#4

No aspersion cast ... but when "snap" judgement is off, I'm inclined to look for a cause :wink:

---In vpFREE@yahoogroups.com, <mickeycrimm@...> wrote :

I'm not a big beer or ale fan. I just use them for chasers. In the course of an evening at the bar I might consume just 2 beers but will down about 20 shots of Black Velvet. And contrary to popular opinion I'm not a full time drunk. I'm in the middle of my bi-monthly road trip so haven't touched the stuff for some days. I never touch the stuff on roadtrips. I don't have ant time for leisure. I just want to get into the towns, get the money and go back home. A typical roadtrip is 2 to three weeks.

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#5

This is similar to the famous Monte Hall problem where contestants are given the chance to select a different door after additional information is revealed (door 2 is opened to reveal a goat). A famous mathematician would not accept that it is correct to switch doors based on the new information on door 2.

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#6

I have my own one die game I like to play. As you know, the odds of rolling a 2 with one die is 5 to 1 against. However, I think I can roll a 2 in four rolls. I'm taking all comers for any stakes that fit my bankroll. Yes, I know I will be taking way the worst of it but I guess maybe I'm getting bored and just want some action. This will be a prime opportunity for someone to lift a ton of dough off me.

#7

MC wrote: I have my own one die game I like to play. As you know, the odds of
rolling a 2 with one die is 5 to 1 against. However, I think I can roll a
2 in four rolls. I'm taking all comers for any stakes that fit my
bankroll. Yes, I know I will be taking way the worst of it but I guess
maybe I'm getting bored and just want some action. This will be a prime
opportunity for someone to lift a ton of dough off me.

Yeah, I'm sure you're gonna go broke. You figure to win 51.78% of the time --- with a fair die.

Back when I played backgammon, we wondered if "drugstore" dice were fair. These were the ones that had little holes drilled out to indicate the number of pips on each dice. Since a 2 only had two holes drilled out and it was opposite a 5, which, of course, had three additional holes drilled out, we figured the 2 was less likely to show up because it was "heavier."

None of us thought it was nearly unbalanced enough to ruin Mickey's 51.78/48.22 advantage. Which as bar bets go, seems pretty fair! (Be careful, Mick. it wouldn't surprise me if you get posts explaining what size bets you can safely take with an x% RoR assuming a $1,000 bankroll.)

The dice we use for backgammon (and craps) had the holes filled up with
something of another color that weighed the same amount as plastic
drilled out. Backgammon dice have rounded corners to make them roll (so
they're less easy to control) while casino dice have sharp corners to
make them land more quickly so the casino can win more bets per hour.

Bob

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#8

Mick wrote: Alan Mendelsen, a strong proponent of 1 in 6, insisted the Wiz would get
his clock cleaned in such a game. The Wiz told him to put his money
where his mouth is. Alan objected to gambling with other individuals on
moral grounds. So the Wiz made a new proposal. When at least 1 of the
dice is a two, if the other die is a two Alan would be awarded 8
points. If the other die wasn't a 2 the Wiz would be awarded 1 point.
The first person to get to 25 points wins and the loser has to buy
lunch.

Alan accepted the challenge and believes he will win the bet. I thought
this was some brilliant hustling by the Wiz. I've never really given
him any credit for street smarts until now. If Alan rolls the hard 4
three times he will only have 24 points. He will have to roll it one
more time to win.

I don't know Alan Mendelsen --- but if he is a big fan of Shack's and hasn't been able to spend any time with Mike, then he might gladly buy Shack lunch just for the opportunity to chat with Shack for an hour or so.

If this is true about AM, then he is actually free-rolling Michael! AM might win the bet, in which case he gets his lunch and Shack pays. He might lose (probable). But he still gets a lunch with Shack that he was willing to pay for anyway.

This is all speculation. I don't know the relationship between the two men at all. But it's very possible that both sides correctly figure they have the advantage!)

(BTW --- Wiz, Shack, Michael, and Mike are all the same guy! I used all names interchangeably when he was my co-host on GWAE, but it's fair to assume that not everybody knows that.)

Bob
                 
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#9

This is what Alan is going to be up against in the dice challenge. The Wiz is expected to get to 25 points by the 90th roll. Excluding the one combination that makes 2-2, there are 10 combinations out of 36 that will gain the Wiz one point. Thats an average of one point every 3.6 rolls. So 3.6 times 25. Points is 90 rolls.

Meanwhile, Alan's expectation is to get to 25 points in 144 rolls. There is only one hard 4 on the dice, so the frequency is 36 times 4 = 144.

Alan will be a major dog in this contest. He will have to average a hard 4 about every 20 rolls or so to have any chance of winning.

#10

This is my own simple explanation of why the answer to the question is 1 in 11, not 1 in 6. A two dice game is perfectly symmetrical. There are 36 total outcomes (6X6):

1 combination makes a total of 2.
2 combinations make a total of 3.
3 combinations make a total of 4.
4 combinations make a total of 5
5 combinations make a total of 6.
6 combinations make a total of 7.
5 combinations make a total of 8.
4 combinations make a total of 9.
3 combinations make a total of 10.
2 combinations make a total of 11.
1 combinations make a total of 12.

That totals to 36 possible outcomes. Why is there 2 combinations that make a total of 11? You specifically have to roll a six and a five. Wouldn't that be just one combination? No. The reason is because the 11 can come in two different orders. Die A can be the six and die B can be the five. Or, Die A can be the five and die B can be the six. It doesn't matter if the roll is six-five, or six-four or 2-5 or 3-4. The same is true for all of these examples.

Here are the combinations that the Wiz will gain 1 point with:

2-6, 6-2
2-5, 5-2
2-4, 4-2
2-3, 3-2
2-1, 1-2

The 1 in 6 crowd says that "if at least one of the dice is a two then it is 1 in 6 that the other die will be a 2. Take a look at the above ten combinations again. Notice that in all of them "at least one of the dice is a 2."

This is why the Wiz will beat anyone who gambles with him at this game.

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#11

My money is on Shack, speaking of which, he is a proponent of Kelly:

The Kelly Criterion - Wizard of Odds http://wizardofodds.com/gambling/kelly-criterion/

http://wizardofodds.com/gambling/kelly-criterion/

The Kelly Criterion - Wizard of Odds http://wizardofodds.com/gambling/kelly-criterion/ The Wizard of Odds on the Kelly Criterion.

View on wizardofo... http://wizardofodds.com/gambling/kelly-criterion/
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#12

Bob, I figure you've probably heard of Alan Mendelson. He's a B celebrity and a pretty strong businessman. But when it comes to gambling and the math of it he is somewhat unusual. I once explained to him in detail how accumulator slots like the IGT Vision Series work and how one can exploit short term advantages on them. He called me a crackpot with bad science. He said I was trying to pull a fast one on him. LOL!

He thinks I'm a fraud. So I told him that if he ever took a Montana vacation I would be glad to let him follow me around for a few days and he can report back to everyone else whether I'm a fraud or not. He totally ignored the invitation.

#13

For the record, Alan accepted the Wizard's challenge on April 29th. He said he would be in Las Vegas on May 25th. That was last weekend which has come and gone. Alan posted that he was in Las Vegas last weekend. But nothing has been posted in any of the forums about the results of the challenge. I wonder what happened. I know the Wiz is chomping at the bit to play. Alan accepted the challenge but it appears now that he is avoiding it.

#14

MC wrote: For the record, Alan accepted the Wizard's challenge on April 29th. He
said he would be in Las Vegas on May 25th. That was last weekend which
has come and gone. Alan posted that he was in Las Vegas last weekend.
But nothing has been posted in any of the forums about the results of
the challenge. I wonder what happened. I know the Wiz is chomping at the
bit to play. Alan accepted the challenge but it appears now that he is
avoiding it.

Why would Shack be chomping at the bit?

Win or lose it's a lunch bet with somebody he doesn't know! For sure, Shack will honor the agreement and go through with the bet and the meal --- but it's hardly something to be looking forward to.

Bob

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#15

Alan has made so many snide remarks about the Wiz being wrong that I think the Wiz is motivated to prove Alan wrong. I don't know how well they know each other but Alan has been a long time member of the Wiz's forum.

#16

Last November in Bellagio for a short while I watched a man playing solo ( the table had a reserved sign on it) at a $100 craps table. He was about six one and I'd say somewhere in his late sixties/early seventies. Needless to say he had thousands on the table during most rolls.
   The dealers called him Alan.
Watched for awhile while he seemed to be winning.
Strolled past the table maybe an hour later and his fortunes had changed and so had his attitude. Sport jacket was off and he was pounding his hand on the rail with each seven out that he rolled.

At one time he said to one of the female dealers"... do you know who I am ?..."

She shook her head no...

   Was this Alan Mendelson ??

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#17

Alan Mendelsen is a known craps player. And he plays $5 video poker. I think Aces & Faces is his favorite game. It could be him. But Caesars is his favorite haunt.

#18

I don't know Alan, but the length he's gone to demonstrate "he's right" (see:
How do you interpret the "dice problem"? http://forum.alanbestbuys.com/showthread.php?3502-How-do-you-interpret-the-quot-dice-problem-quot&s=1c768edfc6b90501abd865a145dcdef5

How do you interpret the "dice problem"? http://forum.alanbestbuys.com/showthread.php?3502-How-do-you-interpret-the-quot-dice-problem-quot&s=1c768edfc6b90501abd865a145dcdef5 Let me just sum this up and return to the basic question. Just in case you missed the other threads this is the original "dice problem" as posted on the Wizard of Vegas message board: You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the ...

View on forum.alanbestbuys.com http://forum.alanbestbuys.com/showthread.php?3502-How-do-you-interpret-the-quot-dice-problem-quot&s=1c768edfc6b90501abd865a145dcdef5
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is a classic example of having "the blinders on".

I don't always get things right the first time, but my first impulse when challenged is to examine the other guys argument for validity. I doubt Alan gave Shack 2 seconds of reconsideration.

It's a 5 minute exercise to test this "at home" with a couple of dice. Keep track of rolls (200-300 probably will lead to a good indication of the correct result) counting the rolls with at least one "2" showing, and how many show 2's on both dice.

In very short order, had Alan done this, he'd be questioning his stand.

I'd suggest he's in for a good dose of embarrassment were it not for the fact that he's likely to remain blind to the truth in his determination to be right.

- H.

---In vpFREE@yahoogroups.com, <mickeycrimm@...> wrote :

Alan has made so many snide remarks about the Wiz being wrong that I think the Wiz is motivated to prove Alan wrong. I don't know how well they know each other but Alan has been a long time member of the Wiz's forum.

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#19

Great post vp wiz. Alan just doesn't get that unpaired dice come in 2 different orders. Only one combination makes 2-2 so the frequency is 36. But when one die comes a 2 and the other a 6 it comes in two different orders and the frequency is 18. Same for 2-5,2-4/,2/3,2/1. Which means an unpaired 2 comes every 3.6 rolls. Meanwhile that hard 4 only comes every 36 rolls.

#20

I've followed the threads that vp wiz put up the links to. I don't know whether to laugh or cry. Some of them have gotten bogged down in the interpretation of the original question. What they should be getting bogged down in is Wizard's challenge. Because both the dice are going to be rolled at the same time. But while they are smearing the Wizard for "failing the mensa test" and using that as fodder for why he is wrong, they have somehow forgotten about his challenge and Alan's acceptance of it. He accepted the challenge more than a month ago but seems to have totally forgotten about it. I seriously think he is now dodging the Wizard's challenge.