Two Dice Puzzle: Part Deux
They have three table games:
Table 1: The dealer places two six sided dice in a cup, shakes them, then slams the cup down on the table hiding the results. He then peeks under the cup, and states, "At least one of the dice is a two". Please place your bets. The only wagers available are "Total 4" and "Total not 4"
Table 2: You approach the table, and see that there is a die with the number two up, next to an upside down cup. The dealer says to you, "I shook 'em. Under the cup is the other die. As you can see, at least one of the dice is a two". Please place your bets. The only wagers available are "Total 4" and Total not 4".
Table 3: You approach what appears to be a standard Craps table, and see that it is not Craps, but a game where the only wagers available are, "Total 4 (at least one die a two)" and "Total not 4 (at least one die a two)". The felt also states, "All rolls without at least one die a two 'en prison'". The dealer explains that this means that any roll without at least one die a two will be "locked up" until a roll with at least one die a two. She asks you to "Please place your bets". You make a wager. The stick then dumps a bowl with five dice and pushes them to you so that you can pick two and shoot em.
Is the probability of "Total 4" at all three tables, 1-11?
Table 2: 1/6, 5/6.
Table 3: 1/11, 10/11.
Quote: DJTeddyBearTable 1: Need more info about the rules. This reeks of the same wording problem as the other thread. Did he tell you about deuces because he was obligated to tell about deuces, or for some other reason?
Table 2: 1-6, 5-6.
Table 3: 1-11, 10-11.
Table 1 is supposed to be the same as the other thread. My assumption is that the peeker only makes a statement when there is at least one die with a two.
Quote: WizardPr(tot 4)=1/11Pr(tot 4)=1/6Pr(tot 4)=1/11
Regarding
Quote: AyecarumbaRegarding
The dealer did say that he "shook 'em", indicating both dice. Is the fact that he has one die already on display, and doesn't shake them both in front of you the difference maker?.
My interpretation of table 2 was that the dealer just placed one die on a 2 face up and shook the other one under a cup. Correct me if I'm wrong.
Quote: Wizard
My interpretation of table 2 was that the dealer just placed one die on a 2 face up and shook the other one under a cup. Correct me if I'm wrong.
Table one and three the answer is obvious (Wizard's answer.) Table two we don't have enough info on what happened between the shaking and what we are seeing now. The Wizard's interpretation is one. Another is that the dealer randomly reached under the cup and picked out one of the dice. Another is that the dealer did the same thing as table 1 but pulled out a die with a two on it to prove it.
Quote: WizardQuote: AyecarumbaRegarding
The dealer did say that he "shook 'em", indicating both dice. Is the fact that he has one die already on display, and doesn't shake them both in front of you the difference maker?.
My interpretation of table 2 was that the dealer just placed one die on a 2 face up and shook the other one under a cup. Correct me if I'm wrong.
The dealer tells you, the player, that he "shook 'em". Which in "not the regular casino" parlance means both dice. However, a die with a two is already set when you walk up to the table, while the other die remains hidden under the cup. You are then asked to place your bet.
Quote: AyecarumbaQuote: WizardQuote: AyecarumbaRegarding
The dealer did say that he "shook 'em", indicating both dice. Is the fact that he has one die already on display, and doesn't shake them both in front of you the difference maker?.
My interpretation of table 2 was that the dealer just placed one die on a 2 face up and shook the other one under a cup. Correct me if I'm wrong.
The dealer tells you, the player, that he "shook 'em". Which in "not the regular casino" parlance means both dice. However, a die with a two is already set when you walk up to the table, while the other die remains hidden under the cup. You are then asked to place your bet.
Doesn't matter IMO what happened before you got there; you have a fixed result on the die you can see at the point you're offered the bet, so that changes the proportions because you have a result on that particular die instead of on at least one of two.
Quote: AyecarumbaThe dealer tells you, the player, that he "shook 'em". Which in "not the regular casino" parlance means both dice. However, a die with a two is already set when you walk up to the table, while the other die remains hidden under the cup. You are then asked to place your bet.
It is important to know the dealer's behavior, much like the Monty Hall problem. Can you tell us if one of these is correct:
Interpretation A: The dealer peeked at one die, which was a 2, and plucked it out.
Interpretation B: The dealer peeked at both dice, of which at least one was a 2, and plucked out a 2.
A: prob = 1/6
B: prob = 1/11
Quote: beachbumbabsQuote: AyecarumbaQuote: WizardQuote: AyecarumbaRegarding
The dealer did say that he "shook 'em", indicating both dice. Is the fact that he has one die already on display, and doesn't shake them both in front of you the difference maker?.
My interpretation of table 2 was that the dealer just placed one die on a 2 face up and shook the other one under a cup. Correct me if I'm wrong.
The dealer tells you, the player, that he "shook 'em". Which in "not the regular casino" parlance means both dice. However, a die with a two is already set when you walk up to the table, while the other die remains hidden under the cup. You are then asked to place your bet.
Doesn't matter IMO what happened before you got there; you have a fixed result on the die you can see at the point you're offered the bet, so that changes the proportions because you have a result on that particular die instead of on at least one of two.
I agree with bbb. You are basically making a bet on 1 die. I think they are all 1/6 and my math prowess speaks for itself.
Quote: GWAEI think they are all 1/6 and my math prowess speaks for itself.
GWAE, BBB go to jail, go directly to jail. Do not pass go, do not collect $200. You are wrong. Did you not read that earlier thread? Is not every pixel of it forever burned into your retina, burned into you brain till you cannot eat or sleep?
Final contribution from me. As clarified so far. all 1/11. It DOES matter what happened before you got there. It is absolutely not a bet on one die unless only one die was thrown.
These were all variations on that earlier thread where both dice were shook and 'at least one of the dice was a two'. That the two was pulled from under the cup is not relevant at all. Shaking two dice with outcome 'at least one of the dice is a two' is a 1/11 shot on the pair every 6l00dy time. Get over it. It only becomes 1/6 if the call is not always about 2's, or if the static die was not thrown, or if only one die was peeked at, or if there is subterfuge such as dice tampering.
Ayacarumba is possibly one of the only two guys that came away from that earlier thread still not getting it. I actually believe that at least one of them did eventually get it, which is why the subject went so very, very quiet.
Quote: OnceDearGWAE, BBB go to jail, go directly to jail. Do not pass go, do not collect $200. You are wrong.
Try to roll a double two to get out early.
When you get out, the Wizard casino will be open and ready for you. As I stated in the other thread, I will offer the following all day long based on the roll of two dice:
2 twos: You win 9 to 1.
1 two: You lose.
0 twos: Push.
We can fuss with a peeker if you demand it, but I will insist that you either declare the conditional bet in advance or agree to a flat betting arrangement.
Quote: OnceDearGWAE, BBB go to jail, go directly to jail. Do not pass go, do not collect $200. You are wrong. Did you not read that earlier thread? Is not every pixel of it forever burned into your retina, burned into you brain till you cannot eat or sleep?
Final contribution from me. As clarified so far. all 1/11. It DOES matter what happened before you got there. It is absolutely not a bet on one die unless only one die was thrown.
These were all variations on that earlier thread where both dice were shook and 'at least one of the dice was a two'. That the two was pulled from under the cup is not relevant at all. Shaking two dice with outcome 'at least one of the dice is a two' is a 1/11 shot on the pair every 6l00dy time. Get over it. It only becomes 1/6 if the call is not always about 2's, or if the static die was not thrown, or if only one die was peeked at, or if there is subterfuge such as dice tampering.
Ayacarumba is possibly one of the only two guys that came away from that earlier thread still not getting it. I actually believe that at least one of them did eventually get it, which is why the subject went so very, very quiet.
Unless I'm reading scenario 2 incorrectly, one die is displayed. For all we know, the dealer SET that die on a 2 to start. A picture of a die on the felt, showing a 2, would be the same thing. AT THAT POINT, (which is where you walk up) it's now 1/6 what result is under the cup. If the dealer had peeked and said, I will be pulling 1 die out from under the cup that's a 2, what are the chances that they're both 2's, I would say 1/11. But one distinct die has been removed from the question by Aye's conditions, in this scenario only, and eliminated the question of whether THAT die is a 1,2,3,4,5, or 6. Those possibilities have disappeared from the bet, replaced by the certainty that it is a 2.
It's the same principle of getting an ace in single deck BJ. The probability (without considering any other cards, including the dealer's) of getting another ace are no longer 4/52; they're 3/51. The odds change as a distinct possibility has been eliminated.
Quote: WizardTry to roll a double two to get out early.
Nice analogy Wizard. If any fool ever wants to take this 9 to 1 wager, please help me find a way of being in on your action.
Quote: beachbumbabsUnless I'm reading scenario 2 incorrectly, one die is displayed. For all we know, the dealer SET that die on a 2 to start.
Hi BBB,
ONLY if the rules are not being followed, else you are still wrong. You need not consider the number of ways that you can arrive at the second die being a two from where you are now: You MUST consider the count of ways that you could have arrived at this situation, and that is 11 and in one of those 11 situations, the die under the cup is a 2.
It IS monstrously counter-intuitive. but it IS 1/11
Ayacarumba clarified that. Both dice were thrown together, presumably UNTIL the banker was able to truthfully say that one of the dice was a two and so pull one deuce from under. That you come to the table after the event is NOT relevant to the 1/11 UNLESS the dealer is a liar and did indeed SET the two and only throw one die. EVEN THOUGH you now know that a specific die was a two, it is TOTALLY irrelevant, because it could have been either.
Quote: OnceDearNice analogy Wizard. If any fool ever wants to take this 9 to 1 wager, please help me find a way of being in on your action.
Same here!
Quote: OnceDearHi BBB
Ayacarumba clarified that. Both dice were thrown together, presumably UNTIL the banker was able to truthfully say that one of the dice was a two and so pull one deuce from under. That you come to the table after the event is NOT relevant to the 1/11 UNLESS the dealer is a liar and did indeed SET the two and only throw one die. EVEN THOUGH you now know that a specific die was a two. TOTALLY irrellevant, because it could have been either.
Hi OnceDear!
I saw Aye's clarification, but I'm not seeing your rationale in saying the remaining die is still 1/11. 5 of those possible throws have NOW been eliminated. The order now matters, if you prefer. 2/1 2/2 2/3 2/4 2/5 2/6 are now the only possible results. Just like throwing the dice until at least one 2 appeared changed the odds from 1/36 for 2,2 to 1/11. Possibilities were eliminated in both cases.
Count the number of equally probable ways that we could have arrived at the situation that we find ourselves in.
That's your denominator.
5 possibilities to get here were NOT in fact ever eliminated. You think they were, but they were not. No five actual possibilities can be dismissed, because BOTH dice were peeked at.
First and second has no meaning here. If anything, both were equally likely to be first and both were equally likely to be second, doubling up that bunch of 5 chances until both came to rest. But that makes for un-needed complexity.
To anticipate your next question, I say again:-
Count the number of equally probable ways that we could have arrived at the situation that we find ourselves in.
Quote: OnceDear
Final contribution from me.
I was gonna take that bet, but before I could, you posted 3 more times XD
Love it. Don't ever change =)
Quote: FaceI was gonna take that bet, but before I could, you posted 3 more times XD
Love it. Don't ever change =)
Got me good. If you see me respond to Alan M in this thread, please suspend me for a week.
$:o)
Quote: OnceDearGot me good. If you see me respond to Alan M in this thread, please suspend me for a week.
$:o)
Hi OnceDear.
I still would like to bet lunch with the Wizard. I like the idea of being paid 9 for 1 for what I think is a 1 out of 6 chance of showing any of six numbers on a single die.
Unfortunately, I arrived about midnight on Friday night, and had an early meeting with a client and could not hook up with you guys at Monkeyfest.
I will be back in Vegas around the 25th of May.
Quote: AlanMendelsonI still would like to bet lunch with the Wizard. I like the idea of being paid 9 for 1 for what I think is a 1 out of 6 chance of showing any of six numbers on a single die.
I'd like to give you that chance. Do you have Skype? Maybe we could do it over that.
But if we did it over Skype when would we have lunch?
I look forward to having lunch with you -- win or lose.
Quote: WizardI'd like to give you that chance. Do you have Skype? Maybe we could do it over that.
(response to Wizard)
Beware camera angle and the variance of too few rolls in play. That and the fact that Alan could easily have tried this on his own if he had ANY inclination to wonder if he might be wrong.
Quote: AlanMendelsonHi Wizard. I don't have Skype.
But if we did it over Skype when would we have lunch?
I look forward to having lunch with you -- win or lose.
You should get it, both not just for this reason.
Tell you what -- we can do it over the phone. I'll even let you do the shaking. I suggest "lunch money" stakes of $5 a bet. You may end it at any time.
I would rather concede defeat then make a money-bet against any individual for any bet. That way I would sleep better. This is a deep personal conviction instilled in me since I was a child.
Quote: WizardIt is important to know the dealer's behavior, much like the Monty Hall problem. Can you tell us if one of these is correct:
Interpretation A: The dealer peeked at one die, which was a 2, and plucked it out.
Interpretation B: The dealer peeked at both dice, of which at least one was a 2, and plucked out a 2.
A: prob = 1/6
B: prob = 1/11
Perhaps this is the crux of the issue for me. I understand that there are 11 combinations of two dice that include at least one two. However, the situation is that now; one of the die values is fixed, and I know it. It is then that I am asked to place a wager. On the face of it, there seems to be only six combinations possible, one of which is another two.
Wizard, can you please explain why the two methods have different probabilities? The unseen die in Interpretation A is not shaken again. It is not apparent to me why the dealer peeking at one or both makes a difference.
Say you walk into a 0 HE casino. You walk up to a caps table and you bet the hard 4 at 11-1. Now the casino offers a new bet, the shooter throws 1 dice at a time. Whatever the number is no matter what it is you can now place a new bet on the 2nd dice. For this example let's just say they throw a 2. Is the casino still going to give you 11-1 on that 2nd dice or are they going to pay you 6-1?
I think there is no way they are giving you 11-1 on 1 dice. I think this is basically the same senerio as 1 because you are placing the bet after you know the outcome to the first die.
Eta I did not see that other thread. By the sounds of it this was all beat to death over there.
Quote: AlanMendelsonI'm sorry, I can't do even $5 a bet. It has to be something finite such as "lunch." As I explained before I cannot bet money against another person. I wasn't raised that way, and it's how I've lived my life.
I would rather concede defeat then make a money-bet against any individual for any bet. That way I would sleep better. This is a deep personal conviction instilled in me since I was a child.
How about this. We do it for $2 a bet, which must be spent on a lunch, until one of us is down $50 or more. At such low stakes I am lowering my odds to 8 to 1. Since you feel a fair bet is 5 to 1 (or 6 for 1), then I hope you won't mind.
Quote: GWAEI just don't get why it would still be 1-11.
Say you walk into a 0 HE casino. You walk up to a caps table and you bet the hard 4 at 11-1. Now the casino offers a new bet, the shooter throws 1 dice at a time. Whatever the number is no matter what it is you can now place a new bet on the 2nd dice. For this example let's just say they throw a 2. Is the casino still going to give you 11-1 on that 2nd dice or are they going to pay you 6-1?
I think there is no way they are giving you 11-1 on 1 dice. I think this is basically the same senerio as 1 because you are placing the bet after you know the outcome to the first die.
Eta I did not see that other thread. By the sounds of it this was all beat to death over there.
Think about it as:
There are 11 different cards with pictures of the 11 different combinations that include at least one 2.
1 - Total 4 (2/2)
2 each - The other 5 combinations (2/1; 2/3; etc.)
The dealer peeks under one of the cards and declares, "At least one of the dice is a two"
What is the probability that he peeked at the total four?
Quote: GWAEI just don't get why it would still be 1-11.
Say you walk into a 0 HE casino. You walk up to a caps table and you bet the hard 4 at 11-1. Now the casino offers a new bet, the shooter throws 1 dice at a time. Whatever the number is no matter what it is you can now place a new bet on the 2nd dice. For this example let's just say they throw a 2. Is the casino still going to give you 11-1 on that 2nd dice or are they going to pay you 6-1?
I think there is no way they are giving you 11-1 on 1 dice. I think this is basically the same senerio as 1 because you are placing the bet after you know the outcome to the first die.
Eta I did not see that other thread. By the sounds of it this was all beat to death over there.
When you throw the dice one at a time, the odds are 1 in 6. When thrown together there these are the possible outcomes that involve at least one two, including order:
2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2.
Note how one of the 11 is 2-2.
Quote: WizardWhen you throw the dice one at a time, the odds are 1 in 6. When thrown together there these are the possible outcomes that involve at least one two, including order:
2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2.
Note how one of the 11 is 2-2.
See I get that but I read the example as they throw the die and then the person comes up with the bet. It is not the person makes the bet then throws the die. Since the bet is made after the throw then I feel the results of 1 of the die don't matter. It is purely the 2nd one that is relevant.
I think we can agree to disagree on this and we can probably agree that you are much smarter than I and are probably the one that is correct. I just don't get, possibly I am just interpreting it wrong.
Quote: AlanMendelsonHi OnceDear.
I still would like to bet lunch with the Wizard. I like the idea of being paid 9 for 1 for what I think is a 1 out of 6 chance of showing any of six numbers on a single die.
Unfortunately, I arrived about midnight on Friday night, and had an early meeting with a client and could not hook up with you guys at Monkeyfest.
I will be back in Vegas around the 25th of May.
I will be in Vegas on the 25th and would love to get my lunch bought Alan. PM me if you want to buy me lunch :-)
Quote: WizardHow about this. We do it for $2 a bet, which must be spent on a lunch, until one of us is down $50 or more. At such low stakes I am lowering my odds to 8 to 1. Since you feel a fair bet is 5 to 1 (or 6 for 1), then I hope you won't mind.
This sounds okay but I need to understand how the bet and play will be structured.
How about this (and please fill in the blanks):
We will cap the bet at $50 for lunch.
We will shake or roll two dice. Either in the open or in a cup with a witness. I don't mind if it's in the open, and I wouldn't mind if we were at a real craps table to use the casino's dice.
If at least one of the two dice is a 2 the bet is on.
If the other die shows a 2 I am awarded _____ points.
If the other die does not show a 2 you are awarded ________ points.
Whoever reaches ________ points first wins.
Please advise.
Quote: AlanMendelsonThis sounds okay but I need to understand how the bet and play will be structured.
How about this (and please fill in the blanks):
We will cap the bet at $50 for lunch.
We will shake or roll two dice. Either in the open or in a cup with a witness. I don't mind if it's in the open, and I wouldn't mind if we were at a real craps table to use the casino's dice.
I'm already conceding you can shake your own dice at home. Just announce the roll to me over the phone.
Following is my suggestion on the contest. I think it is better to go by margin of victory. With you accumulating big chunks of points at a time, there is no different between you reaching 17 and 24 points, for example, as a goal, under your way.
Quote:If at least one of the two dice is a 2 the bet is on.
If the other die shows a 2 I am awarded ___8__ points.
If the other die does not show a 2 you are awarded ___1_____ points.
Whoever reaches is ahead by ____25____ points first wins.
If at least one of two dice shows a 2 the bet is on.
If the other die shows a 2, Alan is awarded 8 points.
If the other die does not show a 2, the Wizard is awarded 1 point.
Whoever is ahead by 25 points first wins.
If that is correct, then yes, I am happy to participate. But I want to do it in person and then have lunch.
This could actually be accomplished with the two of us observing a craps game at any casino. We don't even have to bet in the casino game or throw the dice. Personally, I think that would be best.
Quote: AlanMendelsonTo be sure I understand:
If at least one of two dice shows a 2 the bet is on.
If the other die shows a 2, Alan is awarded 8 points.
If the other die does not show a 2, the Wizard is awarded 1 point.
Whoever is ahead by 25 points first wins.
If that is correct, then yes, I am happy to participate. But I want to do it in person and then have lunch.
This could actually be accomplished with the two of us observing a craps game at any casino. We don't even have to bet in the casino game or throw the dice. Personally, I think that would be best.
Okay. Except I prefer to use our own dice as it will take too long at a craps table.
Quote: AyecarumbaThrowing one die, then making a bet on the unknown outcome of the second is exactly where I am at with this. Unlike the 11 card scenario, my position is that by peeking and sharing, the dealer has, in effect, removed five of the possible outcomes. The die he refers to cannot be any other value except 2. Can someone explain why the five other faces of the die that is now a two are still considered when computing the probability of Total 4?
He could either be identifying the die on the left as a two, leaving 6 possibilities for the die on the right, or the die on the right is a two, leaving 6 possibilities for the die on the right.
You don't know which, so of the 11 possibilities you can't eliminate either the left-two or the right-two possibilities.
The only thing the dealer revealed by announcing a two is of the 36 possible two die combinations, he is looking at one of the 11 that has at least one two.
If the dice were green and red, and he identified which color die was a two, then you coild eliminate the 5 possibilities where that color die was not a two.
Since he doesn't identify which one it is, you can't know which 5 possibilities to remove.
An important aspect of this problem is how it is setup—that is, how are the conditionals determined. This is not usually given; therefore assumptions must be made…which then produces 40 page threads.
Quote: Dalex64He could either be identifying the die on the left as a two, leaving 6 possibilities for the die on the right, or the die on the right is a two, leaving 6 possibilities for the die on the right.
You don't know which, so of the 11 possibilities you can't eliminate either the left-two or the right-two possibilities.
The only thing the dealer revealed by announcing a two is of the 36 possible two die combinations, he is looking at one of the 11 that has at least one two.
If the dice were green and red, and he identified which color die was a two, then you coild eliminate the 5 possibilities where that color die was not a two.
Since he doesn't identify which one it is, you can't know which 5 possibilities to remove.
Thanks Dalex64! While I can understand your explanation, I can't shake the feeling that having the information that at least one die is a two, prior to being asked to compute the probability, makes a difference. According to the Wizard, how the dealer peeks makes a difference, but I am not clear why.
Think about the Crazy Casino's Table #2 described above. You approach the table with one die already set aside with the two face up. The dealer assures you that he shook both dice, but you didn't witness it. Theoretically, the die under the cup should be a two once every eleven times, but the dealer looks pretty lazy, and you are pretty sure he doesn't shake them both all the time.
Does it make a difference to you, the bettor? You have the same known starting information in both cases: A die under a cup, and a die that is identified as the one the peeker saw in order to determine that at least one die in the pair was a two".
What is unkown, is the randomness of the shaking beforehand. In the extreme Table #2 case, it is possible that the first die wasn't shaken at all. Since we know that one of the dice is a two before we place our bets, my contention is that it doesn't matter how we got to that point. The unknown now, is what is under the cup. There are only six possible outcomes, each equally likely.
How about this: The dealer has two dice and two cups. He shakes one die in each cup, then slams both down simultaneously. He then lifts one cup to reveal a two. He then says, "Place your bet on Total 4". What is the probability of Total 4? i think the Wizard would say this is the same as the "single die peek" method, so the odds are 1 in 6; but I'm not clear why this is different than peekikng at both (resulting in 1-11 odds).
Quote: Ayecarumba...I'm not clear why...
Because with 2 dice, either one - EITHER ONE - can step forward and say "I AM THE FIRST DIE and I AM A DEUCE." Kinda doubles the chances that a die can step forward as the first deuce. (Not exactly double, because there's one case where they're both deuces.)
Wow. Now THAT'S putting your money where your mouth is!Quote: WizardI'm already conceding you can shake your own dice at home. Just announce the roll to me over the phone.
Quote: IbeatyouracesQuite simply, roll both dice SIMULTANEOUSLY and exclude any non-2 rolls, it's 1/11. Roll ONE DIE until a 2 shows and then try to roll a second 2 is 1/6.
but if you are excluding any non 2 rolls then you are guaranteeing a 2 on at least 1, therefore you are betting on the 2nd die with a 1/6 chance of being a 2.
Quote: GWAEbut if you are excluding any non 2 rolls then you are guaranteeing a 2 on at least 1, therefore you are betting on the 2nd die with a 1/6 chance of being a 2.
Again, you don't know which die has the 2. Roll two dice and record 2-2's to 2-1's, 2-3's 2-4's, 2-5's, and 2-6's.
There are 36 total combos of two dice, only 11 have a 2 in them and only 1 has both 2's in them. 25 of the 36 have no 2 at all.
Quote: IbeatyouracesAgain, you don't know which die has the 2. Roll two dice and record 2-2's to 2-1's, 2-3's 2-4's, 2-5's, and 2-6's.
There are 36 total combos of two dice, only 11 have a 2 in them and only 1 has both 2's in them. 25 of the 36 have no 2 at all.
Where I am coming from is that as soon as the peeker identifies the die with the two on it, they are no longer both "random" The peeker has established that one die is a two, by observing that at least one die is a two; it no longer matters if it is the first or second die in the combinations 1-2, 2-2, 3-2, 4-2, 5-2, 5-2, 2-1, 2-3, 2-4, 2-5, 2-6, as five of those combinations (whether they be the first or second) have been eliminated. Only six combinations remain:
2 (the die the peeker noted) + 1,2,3,4,5 or 6.
