Quote:unJon

1) A challenge. I’ll be the person that peeks. I will sometimes when I look at the dice make the truthful statement “there is at least one 2.” Every time I make that statement, you can make a bet that there are 2 twos showing on the dice. I will offer you 20 to 1 odds. Should you take the bet?

2) The rule the peeker follows is:

a) if there is at least one 1, say “there is at least one 1”.

b) if a) doesn’t happen and there is at least one 2, say “there is at least one 2.”

The dice are rolled and the speaker says “there is at least one 2.” Now what is the probability that there are two 2s?

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Quote:unJonI offer my challenge posted above to the Wiz, Gordon or anyone else. Willing to escrow funds. 20 to 1 odds on the bet that there are 2 twos (I take side there are not 2 twos).

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Analysis... I think one of us dropped the ball, Unjon.

Here are all the 36 possibilities for dice landing, equally weighted.

1,1 “there is at least one 1” No Bet

1,2 “there is at least one 1” No Bet

1,3 “there is at least one 1” No Bet

1,4 “there is at least one 1” No Bet

1,5 “there is at least one 1” No Bet

1,6 “there is at least one 1” No Bet

2,1 “there is at least one 1” No Bet

2,2 “there is at least one 2” Game Open Pays Out

2,3 “there is at least one 2” Game Open

2,4 “there is at least one 2” Game Open

2,5 “there is at least one 2” Game Open

2,6 “there is at least one 2” Game Open

3,1 “there is at least one 1” No Bet

3,2 “there is at least one 2.” Game Open

3,3 Silence=No bet

3,4 Silence=No bet

3,5 Silence=No Bet

3,6 Silence=No Bet

4,1 “there is at least one 1.” No Bet

4,2 “there is at least one 2.” Game Open

4,3 Silence=No Bet

4,4 Silence=No Bet

4,5 Silence=No Bet

4,6 Silence=No Bet

5,1 “there is at least one 1” No Bet

5,2 “there is at least one 2” Game open

5,3 Silence=No Bet

5,4 Silence=No Bet

5,5 Silence=No Bet

5,6 Silence=No Bet

6,1 “there is at least one 1” No Bet

6,2 “there is at least one 2” Game open

6,3 Silence=No Bet

6,4 Silence=No Bet

6,5 Silence=No Bet

6,6 Silence=No Bet

Of the 9 times that he calls "There's at least one 2", they are both twos

Thus if he calls "At least one of the dice is a two" then the probability of both dice being a two is ONE in NINE

The answer 1.9

QED

Quote:OnceDearQuote:unJon

1) A challenge. I’ll be the person that peeks. I will sometimes when I look at the dice make the truthful statement “there is at least one 2.” Every time I make that statement, you can make a bet that there are 2 twos showing on the dice. I will offer you 20 to 1 odds. Should you take the bet?

2) The rule the peeker follows is:

a) if there is at least one 1, say “there is at least one 1”.

b) if a) doesn’t happen and there is at least one 2, say “there is at least one 2.”

The dice are rolled and the speaker says “there is at least one 2.” Now what is the probability that there are two 2s?

link to original postQuote:unJonI offer my challenge posted above to the Wiz, Gordon or anyone else. Willing to escrow funds. 20 to 1 odds on the bet that there are 2 twos (I take side there are not 2 twos).

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Analysis... I think one of us dropped the ball, Unjon.

Here are all the 36 possibilities for dice landing, equally weighted.

1,1 “there is at least one 1” No Bet

1,2 “there is at least one 1” No Bet

1,3 “there is at least one 1” No Bet

1,4 “there is at least one 1” No Bet

1,5 “there is at least one 1” No Bet

1,6 “there is at least one 1” No Bet

2,1 “there is at least one 1” No Bet

2,2 “there is at least one 2” Game Open Pays Out

2,3 “there is at least one 2” Game Open

2,4 “there is at least one 2” Game Open

2,5 “there is at least one 2” Game Open

2,6 “there is at least one 2” Game Open

3,1 “there is at least one 1” No Bet

3,2 “there is at least one 2.” Game Open

3,3 Silence=No bet

3,4 Silence=No bet

3,5 Silence=No Bet

3,6 Silence=No Bet

4,1 “there is at least one 1.” No Bet

4,2 “there is at least one 2.” Game Open

4,3 Silence=No Bet

4,4 Silence=No Bet

4,5 Silence=No Bet

4,6 Silence=No Bet

5,1 “there is at least one 1” No Bet

5,2 “there is at least one 2” Game open

5,3 Silence=No Bet

5,4 Silence=No Bet

5,5 Silence=No Bet

5,6 Silence=No Bet

6,1 “there is at least one 1” No Bet

6,2 “there is at least one 2” Game open

6,3 Silence=No Bet

6,4 Silence=No Bet

6,5 Silence=No Bet

6,6 Silence=No Bet

Of the 9 times that he calls "There's at least one 2", they are both twos

Thus if he calls "At least one of the dice is a two" then the probability of both dice being a two is ONE in NINE

The answer 1.9

QED

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That’s correct for my scenario 2). Good work. That has nothing to do with my scenario 1) challenge.

Quote:gordonm888But for OnceDear to constantly denounce people on other threads who don't understand "Expected Value" and then to play at being obtuse about dice probabilities on this thread is very discouraging. No wonder Wizard lost his cool with Alan and the other flat-earthers.

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1. I'm not playing at being obtuse. I'm pointing out one of those remarkable situations where I abjectly disagree with Wizard. A worthy cause against the math and stats and logic expert. Be discouraged if you wish.

2. Did you just group Alan in with other flat earthers? I just suspended Wizard for implicitly doing that.

Quote:unJon

1) A challenge. I’ll be the person that peeks. I will sometimes when I look at the dice make the truthful statement “there is at least one 2.” Every time I make that statement, you can make a bet that there are 2 twos showing on the dice. I will offer you 20 to 1 odds. Should you take the bet?

2) The rule the peeker follows is:

a) if there is at least one 1, say “there is at least one 1”.

b) if a) doesn’t happen and there is at least one 2, say “there is at least one 2.”

The dice are rolled and the speaker says “there is at least one 2.” Now what is the probability that there are two 2s?

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Quote:unJonI offer my challenge posted above to the Wiz, Gordon or anyone else. Willing to escrow funds. 20 to 1 odds on the bet that there are 2 twos (I take side there are not 2 twos).

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Sorry Unjon. I misread para 2) as the peek rules applied to challenge at 1)Quote:unjon

That’s correct for my scenario 2). Good work. That has nothing to do with my scenario 1) challenge.

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If challenge 1 stands on its own, you would never need to pay out a bean. 1000 to one would be robbery. Only ever declare when there is exactly one 2 showing. It is within the challenge 1 rules

Quote:OnceDear

If challenge 1 stands on its own, you would never need to pay out a bean. 1000 to one would be robbery. Only ever declare when there is exactly one 2 showing. It is within the challenge 1 rules

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Bingo. I’m hoping it will help show Wiz and Gordon why ambiguity in the rules affects analysis.

Quote:billryanThe difference here might be semantics more than math.

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Much of it is. But once you get past the semantics there is a real conceptual issue in the math where people have trouble with seeing 1/11 vs 1/6.

The problem is people go back to being bogged down in the semantics.

Quote:billryanThe difference here might be semantics more than math.

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Until the semantics are ironed out, there can be no codified question and no derived explicit answer.

It's like playing a card game, not knowing the rules and losing your arse.

Quote:unJonQuote:billryanThe difference here might be semantics more than math.

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Much of it is. But once you get past the semantics there is a real conceptual issue in the math where people have trouble with seeing 1/11 vs 1/6.

The problem is people go back to being bogged down in the semantics.

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Wizard is working on a reworking of the puzzle where the semantics were ironed out. I believe he's doing that, cos like the Monty Hall problem, this is one where the answer is not as intuitive as one would expect. In his reworked version, the 1 in 11 answer is absolutely proven, but still expect AlanM and a few others to plumb for and cling on to 1/6

Quote:OnceDearQuote:unJonQuote:billryanThe difference here might be semantics more than math.

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Much of it is. But once you get past the semantics there is a real conceptual issue in the math where people have trouble with seeing 1/11 vs 1/6.

The problem is people go back to being bogged down in the semantics.

link to original post

Wizard is working on a reworking of the puzzle where the semantics were ironed out. I believe he's doing that, cos like the Monty Hall problem, this is one where the answer is not as intuitive as one would expect. In his reworked version, the 1 in 11 answer is absolutely proven, but still expect AlanM and a few others to plumb for and cling on to 1/6

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Easy to work out semantics. Just have someone else that can’t see the dice to ask the peeker “is there at least one 2” and the peeker truthfully says “yes.”