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Topic: Dice throw (Read 2164 times) |
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almost dead
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we have a dice that we can throw up to three times , after each time , we have the option to go ahead and try again or stick to the number which showed up. What is the best strategy to maximize our expected gain?
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towr
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Re: Dice throw
« Reply #1 on: Jan 14th, 2010, 4:51am » |
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I would guess, throw once, then if you expect to get higher with two dice, throw again, then if you expect higher with one die, throw again.
So if you get a 5 or higher keep it else throw again, if you get a 4 or higher keep it, else throw again.
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pscoe2
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Re: Dice throw
« Reply #2 on: Jan 14th, 2010, 5:04am » |
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on Jan 14th, 2010, 3:39am, almost dead wrote:| we have a dice that we can throw up to three times , after each time , we have the option to go ahead and try again or stick to the number which showed up. What is the best strategy to maximize our expected gain? |
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I think we can use probability here...
Like.. assume the person got 6 before...the probability of getting better is null so he wont for sure...
If he gets 5...the probability to get a 6 in the next 2 turns is 1/6+5/36 = 11/36
something on these lines...wht do u ppl think
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| « Last Edit: Jan 14th, 2010, 5:08am by pscoe2 » |
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MathsForFun
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Re: Dice throw
« Reply #3 on: Jan 14th, 2010, 5:19am » |
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If the first throw is either 5 or 6, then hold. If the second throw is 4, 5 or 6 then hold.
The expected value of the last die is (1/3 * 5.5) + (2/3 * 1/2 * 5) + (2/3 * 1/2 * 3.5) = 14/3 (approx 4.66)
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| « Last Edit: Jan 14th, 2010, 7:49am by MathsForFun » |
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ThudnBlunder
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Re: Dice throw
« Reply #4 on: Jan 14th, 2010, 5:57am » |
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How can you 'maximise gain' if you don't know the payoffs, eg. what is the cost of throwing?
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towr
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Re: Dice throw
« Reply #5 on: Jan 14th, 2010, 6:21am » |
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I think it is fairly safe to assume he means to maximize the expected value of his final dice roll.
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mmgc
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Re: Dice throw
« Reply #6 on: Jan 17th, 2010, 7:11pm » |
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on Jan 14th, 2010, 5:19am, MathsForFun wrote:If the first throw is either 5 or 6, then hold. If the second throw is 4, 5 or 6 then hold.
The expected value of the last die is (1/3 * 5.5) + (2/3 * 1/2 * 5) + (2/3 * 1/2 * 3.5) = 14/3 (approx 4.66) |
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Can you please explain your answer, I have following doubts regarding this :
1. Based on what logic 5, 6 is decided for first throw and 4,5,6 decided for second throw ?
2. Since, all three throws are independent events --how come the expected value of last throw is dependent on earlier throws.
I think the expected value of any throw is 3.5, so any time he gets more than this ( 4, 5,6 ), he should stick to it.
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pex
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Re: Dice throw
« Reply #7 on: Jan 17th, 2010, 8:42pm » |
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on Jan 17th, 2010, 7:11pm, mmgc wrote:| I think the expected value of any throw is 3.5, so any time he gets more than this ( 4, 5,6 ), he should stick to it. |
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You've got the right idea here, you just need to take it one step further.
- The expected value of one throw is 3.5, nothing we can do about that.
- Assume we have two throws. If we discard the first throw, the expected value will be 3.5; thus, we discard only lower values: 1, 2 or 3. The expected value is now 1/2 * 5 + 1/2 * 3.5 = 4.25.
- Finally, if we have three throws, we know that the expected value will be 4.25 if we discard the first throw. Therefore, we will discard any lower value (1, 2, 3 or 4) on the first throw; the expected value is 1/3 * 5.5 + 2/3 * 4.25 = 14/3.
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Ant_Man
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Re: Dice throw
« Reply #8 on: Jan 18th, 2010, 2:00pm » |
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I don't think pex's expected values are right. You should look at the probabilities for getting each value if you throw 2 dice and keep the highest value. For example, you can get a six if both dice give sixes (1/6*1/6=1/36) or if one gives a six and the other gives anything lower (1/6*5/6*2=10/36). So the total probability of getting a 6 is 11/36. You can find probabilities for the other values and then get an expected value. For rolling two dice the expected value is ~4.47. For three, it is ~5.25. For one die it is obviously 3.5. Thus towr's strategy is right.
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Obob
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Re: Dice throw
« Reply #9 on: Jan 18th, 2010, 6:25pm » |
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Except you don't throw the 2 dice at the same time. Pex's analysis seems right.
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mistaken_id
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Re: Dice throw
« Reply #10 on: Jan 18th, 2010, 11:26pm » |
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How are they different(pex and towrs solutions)
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Obob
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Re: Dice throw
« Reply #11 on: Jan 19th, 2010, 12:50am » |
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Pex and towr's solutions are the same (although towr's analysis isn't complete); antman's is different.
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towr
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Re: Dice throw
« Reply #12 on: Jan 19th, 2010, 2:07am » |
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I don't quite remember if I used the right analysis. But I did double-check with a brute-force search 
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mmgc
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Re: Dice throw
« Reply #13 on: Jan 19th, 2010, 3:18am » |
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on Jan 17th, 2010, 8:42pm, pex wrote:
You've got the right idea here, you just need to take it one step further.
- The expected value of one throw is 3.5, nothing we can do about that.
- Assume we have two throws. If we discard the first throw, the expected value will be 3.5; thus, we discard only lower values: 1, 2 or 3. The expected value is now 1/2 * 5 + 1/2 * 3.5 = 4.25.
- Finally, if we have three throws, we know that the expected value will be 4.25 if we discard the first throw. Therefore, we will discard any lower value (1, 2, 3 or 4) on the first throw; the expected value is 1/3 * 5.5 + 2/3 * 4.25 = 14/3. |
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Got your point...Thanks...But how do we say that 4.66 is the maximum expected value we can get ...
May be with some other combination the expected value is more than this
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pex
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Re: Dice throw
« Reply #14 on: Jan 19th, 2010, 3:23am » |
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on Jan 19th, 2010, 3:18am, mmgc wrote:Got your point...Thanks...But how do we say that 4.66 is the maximum expected value we can get ...
May be with some other combination the expected value is more than this |
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Given that we take the optimal decision at every point, I don't see how it could be improved.
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towr
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Re: Dice throw
« Reply #15 on: Jan 19th, 2010, 3:41am » |
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on Jan 19th, 2010, 3:18am, mmgc wrote:Got your point...Thanks...But how do we say that 4.66 is the maximum expected value we can get ...
May be with some other combination the expected value is more than this
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You can look at it this way, if you reroll the die for certain numbers, it's equivalent to replacing those numbers on the die with the expected value of the new roll.
So for 2 dice rolls, we start with (1,2,3,4,5,6) and replace some of those numbers with 3.5 (the expected value of the next roll). Which numbers do we replace to get the maximum expected value?
If we replace a number greater than 3.5 the average goes down, so we shouldn't do that. If we replace numbers smaller than 3.5 the average goes up, so we should do that.
So our new "die" is (3.5, 3.5 ,3.5, 4, 5, 6), with expected value (average) of 4.25. We can't get any higher, because there are no more numbers on the die that we can replace without decreasing the value of the die.
For three dice rolls, you similarly get (4.25, 4.25, 4.25, 4.25, 5, 6) with the average of 4.66...
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| « Last Edit: Jan 19th, 2010, 3:42am by towr » |
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Ant_Man
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Re: Dice throw
« Reply #16 on: Jan 19th, 2010, 5:15am » |
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I agree with towr and pex now. I forgot that if you roll again, you have to take that roll. Guess I am a noob.
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