## Poll

16 votes (51.61%) | |||

12 votes (38.7%) | |||

5 votes (16.12%) | |||

2 votes (6.45%) | |||

9 votes (29.03%) | |||

3 votes (9.67%) | |||

5 votes (16.12%) | |||

5 votes (16.12%) | |||

10 votes (32.25%) | |||

7 votes (22.58%) |

**31 members have voted**

December 12th, 2020 at 3:12:01 AM
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Quote:gordonm888You are correct. I came late to this discussion and have been trying to understand what has gone wrong. I did not mean to muddle it up further. Is Charlie's solution the correct solution?

Obviously if you have a nice low one this isn't an issue, but the problem is when you have to keep a high one. One approach might have been to look at the average of each die, but this isn't correct as you need to factor in having to keep a high die. For instance consider what happens if your first roll is 33336. It's also worth working out what would happen if you didn't have to keep any dice, as this sets a lower bound for the answer.

December 12th, 2020 at 10:23:34 AM
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Quote:charliepatrickObviously if you have a nice low one this isn't an issue, but the problem is when you have to keep a high one. One approach might have been to look at the average of each die, but this isn't correct as you need to factor in having to keep a high die. For instance consider what happens if your first roll is 33336. It's also worth working out what would happen if you didn't have to keep any dice, as this sets a lower bound for the answer.

charliepatrick : Question about your choices (in spoiler)

I agree the complication is in having to hold the dice. I also agreed with your original assessment on holding the 2 break point. But my difficulty was implementing that. In your solution you have

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Work backwards to first round e.g.

5 dice held - average total = 2.5 (as they're either 0 or 1)

4 dice held = 2 + 3

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5: I agree

4: It seems by using "3" you are assuming you re-roll the last die. But what if first roll is 1,1,1,1,2. In that case don't you keep the 2? as it is less than the expected value of re-rolling and getting EV=3?

So I was trying to separate the 2 (1/4 of the time for last die) from the 4,5,6. (3/4 of time for last die)

Hold 4 dice (3=0 or 1=1) :

(2 + (1/4 * (keep 2) + 3/4 *E(re-roll 1 die)) ) =

2 + (2/4 + 9/4)

19/4???

I admit I am still trying to sort through your other numbers....maybe this effect is accounted for elsewhere, (but it seems not in the 5 dice held case as you are assuming all 0 or 1 to get EV 2.5).

Anyway, that line stood out as different than my understanding?

December 12th, 2020 at 1:42:22 PM
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I agree...Quote:chevycharliepatrick : Question about your choices (in spoiler)...

I haven't rerun the numbers but can see if all the other numbers are 2s after the first roll, then you're better off standing. The chances of this are fairly small, so it makes the answer slightly smaller than my first answer: about 7.752715.

December 12th, 2020 at 5:36:10 PM
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Quote:unJon

I have an idea on modeling this but won’t have time until the weekend. Request that you do not post answer yet.

We're just waiting for unJon to make his post.

Have you tried 22 tonight? I said 22.

December 13th, 2020 at 9:50:00 AM
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Quote:GialmereWe're just waiting for unJon to make his post.

Sorry. Weekend got away from me. Don’t hold anything up for me.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

December 13th, 2020 at 10:07:02 AM
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Quote:Ace2

I had a small error in my first calculation. If this one is confirmed as correct, I’ll show my method

6.253979...

Correct!

Very good.

If desired, I'll post the official solve after Ace2 posts his method.

Have you tried 22 tonight? I said 22.

December 13th, 2020 at 3:11:11 PM
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Quote:GialmereQuote:Ace2

I had a small error in my first calculation. If this one is confirmed as correct, I’ll show my method

6.253979...

Correct!

Very good.

If desired, I'll post the official solve after Ace2 posts his method.

Though I attempted to find a slick formulaic solution, it's quite easy to get the answer via brute force in Excel. There are only 6^2 + 6^3 + 6^4 + 6^5 = (6^6 -1)/(6-1) - 6 -1 = 9,324 permutations to review

Working backwards and using correct strategy the expected values are:

Roll 1 die : 3.0000000 = a

Roll 2 dice: 4.38888… = b

Roll 3 dice: 5.23380… = c

Roll 4 dice: 5.83386... = d

Using the same method we used to find the aforementioned values, we find the expected value for rolling 5 dice as follows:

1) List all 6^5 = 7,776 permutations. Each permutation MUST be in ascending order. This is easy to do in Excel (let me know if you need the method)

2) Make 5 calculations for each permutation:

Sum of all dice 1-5

Sum of dice 1-4 plus a

Sum of dice 1-3 plus b

Sum of dice 1-2 plus c

Die 1 plus d

3) Take the minimum of the five calculations for each permutation

4) Take the average of the 7,776 minimums to get the final answer of 6.25398…

Incidentally, if efficiency was a key consideration, you could use the same method on just a few hundred combinations instead of 9,324 permutations. For instance, rolling 5 dice there are 7,776 permutations yet only 252 combinations. However, with a computer I think it's easier to generate all permutations and run the calculation on all of them instead of figuring/grouping the combinations. If you only had a pen, paper and slide rule, then it would be much easier to use combinations

Last edited by: Ace2 on Dec 13, 2020

It’s all about making that GTA

December 14th, 2020 at 6:27:37 AM
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Yes. The interesting thing about this one is that you can't make a rule saying to reroll a die if it is x or higher.

Instead, if you roll a 3-3-3-2-2, you'd keep all the dice since the two 2s add up to less than 4.3.

If you roll a 3-3-2-2-2, however, you'd reroll the three 2s since they add up to more than 5.2.

Instead, if you roll a 3-3-3-2-2, you'd keep all the dice since the two 2s add up to less than 4.3.

If you roll a 3-3-2-2-2, however, you'd reroll the three 2s since they add up to more than 5.2.

Have you tried 22 tonight? I said 22.

December 14th, 2020 at 8:00:05 AM
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It's easy Monday. Let's have ice cream.

An ice cream cone is packed full of ice cream and a generous hemisphere of ice cream is placed on top.

If the volume of ice cream inside the cone is the same as the volume of ice cream outside the cone, what is the height of the cone?

An ice cream cone is packed full of ice cream and a generous hemisphere of ice cream is placed on top.

If the volume of ice cream inside the cone is the same as the volume of ice cream outside the cone, what is the height of the cone?

Have you tried 22 tonight? I said 22.