You Need A Smart Strategy To Solve This Math Olympiad Game
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- เผยแพร่เมื่อ 10 ก.พ. 2025
- Thanks to James M. for this suggestion! You roll a dice up to 3 times, stopping after any roll you wish. You get paid the value of the roll you stop. If you play optimally, what is your expected payout? (more precise wording in video)
My blog post for this video:
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Source: this was a problem in a book of Olympiad style questions.
math.stackexch...
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Presh calling out the haters before starting the problem hahah
I loved that clapback lol
lol😂😂😂😂
For the general solution:
1 roll: E=3,5
2 rolls: E=17/4=4,25 as presented in the video.
3 rolls: E=14/3 as presented in the video.
4 rolls: E=89/18 = 4,944.
5 rolls: E=277/54=5,13.
For more rolls than 5, you only keep the 6 at the first rolls, and the general solution is therefore:
n rolls (n>5): E=6-(47/54)*(5/6)^(n-5) which goes towards 6, when n goes towards infinite.
super proud of myself for figuring it out and it’s thanks to watching this channel and trying to do the puzzles with pen and paper and sometimes in my head. I am confident it has helped me on some academic tests as well
Regarding dice vs die as the singular of "dice", there is a difference between British and American usage. "Dice" as the singular is more common in British English than American, so it's not surprising that the Oxford English Dictionary lists "dice" as an acceptable singular form while the American Heritage Dictionary does not. (Much like British speakers says "maths" while American speakers say "math".) (Note that when the Oxford English Dictionary refers to "Modern Standard English" it is referring specifically to British English.)
Personally I prefer die as the singular over dice, it removes any possible ambiguity over whether you mean one or multiple (eg does "I dropped the dice" mean you dropped one or two things? "I dropped the die" definitely means you dropped one thing.) In fact as far as I can tell there's no particularly good reason to use the word "dice" as the singular form, any sentence that uses "dice" as a singular could have instead used "die" and been clearer.
Anyway that's just my pedantic preference.
Doug Rosengard didn't know that. Thanks for the information. Especially since I didn't even know that their was a single lexicon for American English.
If we are talking language ..... don't forget the multiple uses of the words Die and Dice (from to stop living to cutting up into small pieces)! Context can be everything .... "alea iacta est"
James Juld Not sure why you’re getting snarky with me, when did I say there is only one lexicon for American English? I just said that the word usage tends to differ between British and American dialects.
Dave W Interestingly the plural of die long ago was “dies” but over time the pronunciation evolved into “dice” for some reason even though “dies” would be the more naturally consistent form.
Somewhere I read that the spelling was changed to 'dice' to preserve the pronunciation, because people started pronouncing the s at the end like a z.
Always reroll a one, two, or three.
Reroll a four if it's on the first roll, keep on the second roll.
Keep a five or six on either roll.
I understand it mathematically, but realistically, would you actually reroll a 4 on the first roll? Just curious
If I knew this, for sure! If not, probably not.
yuo usui It depends on how many games I'm playing, but for a single game I'd probably keep the 4. A bird in the hand is worth 2 in the bush.
The point is you need to figure out the expected value so you know what is optimal value to stop rolling. It may be somewhat intuitive as to when to stop rolling, but still it should be mathematically determined.
As a gambler, yes. You know the last round equals 3,5. That means during the second round you roll 1,2,3 again and you keep 4,5,6. In the first round you obviously keep 6 and roll 1,2,3 again. That only leaves 4 and five to decide. I couldn't do the math in my head but I knew the chances of getting a payout was higher if I kept rolling so I decided to split the difference.
Luckily the EV wasn't 3,9, otherwise I would have been fucked.
Instead making the decision to stop or proceed by comparing against the expected value and using the linearity of the expectation operator, I calculated the probabilities you could do better by continuing at each stage. If this probability was greater or equal to 0.5, then I continued to the next roll.
I basically framed the problem as a big tree of potential rolls with 6*6*6 = 216 branches. Stopping, is equivalent to passing down the roll value to the final level (3rd roll) in the tree. I calculated the expected value by counting the number of level 3 branches with a value, multiplying it by that value, then dividing everything by the total number of branches (216).
E = ((36+24+12)*6 + (36+24+12)*5 + (24*12)*4 + (12)*3 + (12)*2 + (12)*1 )/216 = 4.6667
The variance can be calculated from VAR[X] = E[X^2] - E[X]^2
E[X^2] = ((72)*6^2 + (72)*5^2 + (36)*4^2 + (12)*3^2 + (12)*2^2 + (12)*1^2 )/216 = 5136/216 = 23.78
E[X]^2 = 4.6667^2 = 21,78
VAR[X] = 2.00 -> STD = 1.41
The dice controversy will never die!
Many people spend their life arguing via the net. My response nowadays is to ignore them - these videos are great.
Wouldn't one dice be a douse? :-)
Mark Wheeler good point, nousely made.
And a singular sheep is a shoop
Mark Wheeler are you supposed to be funny? Cause you're not.
What happened? Did your dog run away? Again?
Rolled in a little douse coup?
Really fun one! Would love to see more EV related problems
I understand the answer presented here, and it seems from all of the comments that everyone agrees:
- Roll a 1, 2, or 3: roll again.
- Roll a 5 or 6: take the money.
- Roll a 4 seems to be the point of contention. Some say take the money; others say roll again if it is the first roll.
There is another way to explain this, and probably how most everyone would think of the problem.
With every roll, you have a 1 in 6 chance of rolling any one of the numbers. This means, for example, on the first roll you have a 1 in 6 chance of rolling a 6; and on the 2nd and 3rd rolls you have the same 1 in 6 chance of rolling a 6. So, the logic for each roll would be dictated by the following:
- If you roll a 1, you have a 5 in 6 (~83%) chance of rolling a higher number on the next roll. Also, you have a 0 in 6 chance ( 0% ) of rolling a lower number on the next roll.
- If you roll a 2, you have a 4 in 6 (~67%) chance of rolling a higher number on the next roll. Also, you have a 1 in 6 chance (~17%) of rolling a lower number on the next roll.
- If you roll a 3, you have a 3 in 6 ( 50%) chance of rolling a higher number on the next roll. Also, you have a 2 in 6 chance (~33%) of rolling a lower number on the next roll.
- If you roll a 4, you have a 2 in 6 (~33%) chance of rolling a higher number on the next roll. Also, you have a 3 in 6 chance ( 50%) of rolling a lower number on the next roll.
- If you roll a 5, you have a 1 in 6 (~17%) chance of rolling a higher number on the next roll. Also, you have a 4 in 6 chance (~67%) of rolling a lower number on the next roll.
- If you roll a 6, you have a 0 in 6 ( 0% ) chance of rolling a higher number on the next roll. Also, you have a 5 in 6 chance (~83%) of rolling a lower number on the next roll.
How you should play the game? Play the odds.
Since the goal is to make the most amount of money, the higher the odds of rolling a higher number, the more you should roll again. The higher the odds of rolling a lower number, the more you should not roll again. So, for every roll:
- Roll a 1: always roll again. You can not do any worse if you do.
- Roll a 2: always roll again. You could do worse, but the odds are very low.
- Roll a 3: always roll again. You could do worse, but the odds are low.
- Roll a 4: take the money*. The odds of doing worse are even, but the odds of doing better are low.
- Roll a 5: take the money. The odds of doing worse are high.
- Roll a 6: take the money. The odds of doing worse are very high.
* - If you are a gambling person by nature; you will most likely will try to roll again (at least on the first roll) because...well...you're a gambler and are willing to take the risk. ;)
7 months later: Just want to nitpick that rolling a 6 will automatically end the game, the odds don't matter.
I wonder if the expected value can get arbitrarily close to 6 if we allow high enough number of rolls. I think it should because we are simply making it very likely that a 6 will be rolled eventually.
Of course we can. As you said, there is a 100% chance of a six being rolled within infinite rolls.
For example if you have 10rolls you are getting on average $5.71 so for 1 milion rolls you will have an expected value of ~6 (5.9991 let's say). So because of that there are only 3 rolls
clearly it should approach 6, 1 out of 6 chance of getting a 6. More than 6 rolls, you have to aim for 6 for optimal value.
So the Lim as the number of roals tends to Infinity the optimal value would tend to to the largist number on the dice.
Fairly easy to prove. Clearly no strategy can give an expected value higher than 6 for any number of rolls since the highest roll is 6. So EV with infinite rolls is
Finally a problem I was able to do without the explanation. Feels good.
We'll try to usual technique of working backwards.
Since we nail down all assumptions here, I'm going to assume that the roller's utility function is extremely linear in dollars, not appreciably inflected in this dollar range. Like, he's not saying "If I win four dollars I can pay the rent otherwise I'm scrwed. Five or six dollars is just gravy."
The expected dollar value of roll 3 is easy: It's the average number of pips showing, 3.5
On roll 2, he will roll again if he rolled {1,2,3} but not if he rolled {4,5,6}. So the expected dollar value of roll 2 is 1/2 * 3.5 + 1/2 * average({4,5,6}), or 4.25.
On roll 1, he will roll again if he rolled {1,2,3,4} but not on {5,6}, so the expected value is 2/3*4.25 + 1/3 * average({5,6}) or 4 2/3.
What about the general solution of rolling m number of k-sided dice n times?
geryon i repeated the process many times for 6 sided dice and as m gets higher and higher the expected value for 6 sided dice approaches 6. it will reach 6 as m gets infinity. now working on to finding the formula and then will try to find solution to your question. but it will be difficult.
Finding the probability that roll n+1 > value_n is the hardest part I think because you will get some roundings in your formulas
What does m and k means?
integers
I'm curious, too...
I love how responding to the die/dice debate only increased the amount of comments about die/dice
Before last roll: leave a 4,5 or 6, but when it is 1,2 or 3 roll again, because rolling again gives expected value 3.5.
So, the expected value with 1 optional roll to go is (3×3.5 +4+5+6)/6 = 51/12
With 2 optional rolls to go: leave a 5 or 6, but when rolled 1,2,3 or 4: roll again.
So, the expected value with 2 optional rolls to go is (4×51/12 +5+6)/6 = 14/3
Really well explained :), I intuitively thought of your solution, but you explained the maths great
Finally I get one of these probability problems right!
I love your (die)agrams
That's so banter mate
A Great example of Dynamic programming. I will use this problem in my class. Thanks.
OMG my mind is actually blown, that was so cool.
I solved it differently, but got to the same strategy.
So I take the result of a first roll and calculate what's the probability of rolling a higher number in the 2 rolls that's left.
6 is the highest so obviously keep it, don't need to calculate.
For a 5, you would have to roll 6 to get better value. The probability of not rolling a 6 in two rolls is 5/6*5/6 = 25/36. So probability of rolling a 6 is 11/36. Less than a half, so you're more probable to not roll a 6 than to roll it. So you keep the 5.
For a 4, the probability of not rolling a higher number is 4/6*4/6 = 4/9. So there's 5/9 probability of rolling a higher number. More than a half, so you're more probable to roll a higher number than not.
So for 4 or less you continue and for 5 or 6 you keep the value.
And for the cases you continue, the same calculation - what's the probability of rolling a higher number in the last roll.
Once again, if you roll 6 you obviously keep it.
For 5, there's 1/6 probability of rolling a higher number in the last roll. Less than a half, so you keep it.
For 4 it's 2/6 (1/3). Again, you keep it.
For 3 it's 3/6. A half. But the probability of rolling a smaller number is 2/6 (1/3). So it's more profitable to roll the last roll - you have only 1/3 probability of getting worse value but you have 1/2 probability of getting better value.
So for 4-6 you keep the value and for 1-3 you roll the last roll.
thank you for posting all these amazing puzzles. me and my lady solve these together all the time. we both came up with different ways of solving this.
easy way to do this is, you divide the number of sides of dice by number of rolls left. if your roll is higher than quotient then you keep. if the number is lower then reroll.
Wow these are tough. This is the first I got right(Rounding up!) only because it was one I had a lot of experience in. Grew up in a family that loved games. I was surprised and in denial in a few of the other vids I watched in this channel.
The expected value of a dice roll is 3.5. This means that on the second roll, if you roll a 1, 2, or 3, you would want to roll a third time.
The expected value of the greater of two dice rolls is a little bit trickier. We'll use the definition of expected value. Let G be the greater of two dice rolls. Then:
E(G) = 1*P(G=1) + 2*P(G=2) + ... + 6*P(G=6)
= 1*1/36 + 2*(3/36) + 3*(5/36) + 4*(7/36) + 5*(9/36) + 6*(11/36)
= 4.47.
This means that on the first roll, if we roll a 1, 2, 3, or 4, we would want to roll a second time.
What's our total expected value? We can once again use the expected value formula and do a case-by-case examination. Let N be the number we stick with. Then:
E(N) = 1*P(N=1) + 2*P(N=2) + 3*P(N=3) + 4*P(N=4) + 5*P(N=5) + 6*P(N=6).
P(N=1): we get this when the first two rolls are sufficiently low and the third roll is a 1. This will be (2/3)*(1/2)*(1/6) = 1/18.
P(N=2): we get this when the first two rolls are sufficiently low and the third roll is a 2. This will also be 1/18.
P(N=3): Also 1/18.
P(N=4): we can get this when the second roll is a 4 or when the third roll is a 4. This will be 1/18 + (2/3)*(1/6) = 3/18.
P(N=5): this can happen at any of the three rolls. We get 1/18 from the third roll, 2/18 from the second roll, and 3/18 from the third roll. This nets us a probability of 6/18 = 1/3.
P(N=6): Also 6/18 = 1/3.
Note that these probabilities add up to 1, so I'm pretty sure my math is right. n_n
We finally get the expected value:
E(N) = 1*1/18 + 2*1/18 + 3*1/18 + 4*3/18 + 5*1/3 + 6*1/3 = 14/3 = 4.66... .
Fun problem!
you should not roll again if in the first roll you have a 4
and i will explain why
there is 50% possibility to get 1,2,3 and thus make your position worse since you will roll the 3rd time for a 3,5 win average expected
33.3% possibility to get a 5,6 and get more
and 16.6% to get another 4
and since you say to maximize profit
had a 4 and go again to 4 does not give me any more profit
so if you add up you have 66.6 possibilities against you so you should always keep the 4 in the first roll
the correct solution
okey
lets make the tree and end up the story once and for all we will expand the tree and see that the calculations of perch are wrong
1st roll we have 6 outcomes that are 1 x (1,2,3,4,5,6,)
we keep only the 5,6 and we roll for the 4 other outcomes so so far we have 2 endings
2nd roll for every 4 outcome we have 24 new outcomes so that makes our tree to go to 24 more possibilities, that are 4 x (1,2,3,4,5,6,)
now we want to keep the 4(4,5,6) and roll for the 4x(1,2,3) so we have 12 endings in the second roll and we roll for 12 positions
3rd roll 12 position we expand so we go to 72 outcomes that are split into 12x (1,2,3,4,5,6)
so all the possible outcomes of the came are 2+12+72=84 outcomes
that are all have the same possibility to happen
and what we have as payouts
12 x (1,2,3) 16x(4,) 18x*5,6) so if you add up all the possible payots you go
334
so if we take the average 334/86= 3,88
which is the correct answer and not the imaginary 4.6666
did you figure out where perch made the mistake
lets also see the 2nd 3rd roll
2nd roll we have 6 outcomes 1x(1,2,3,4,5,6)
we keep the 3 (4,5,6( and we expand the other 3 for a new 18 outcomes which will have 3 x (1,2,3,4,5,6)
so we end up with 21 possible outcomes all with the same possibilities of happening
so lets see the payout its 78/21 =3,71 and not 4.2
the problem with perch is and the ERROR
is that if a branch is still expanding and you want to talk about it with the final values to collapse it you can not take the averages and put it there but not also bring back the number of outcomes
while it is true that 2nd roll has 1/6 possibility to have the numbers 1,2,3,4,5,6
it does not have 3/6 possibility to have the average payout of 3,5
thats 18/21 possibility to get that payout
1st roll you keep 4,5,6
so lets make the tree
6 outcomes we keep the 3 and we roll for the other 3
2nd roll 18 outcomes we keep the 4,5,6 the 9 outcomes and we roll for the other 9
3rd roll 54 outcomes
so we have a total 63 outcomes and a total payout of 294
for an average 4.66666
ANIKHTOS
The stopping points don't have equal probability. The earlier roll has greater probability for its roll values because you expand the tree.
Say you have 1 to 6 for roll 1.
You leave 4-6 and expand 1-3, and you end at the second roll.
Then the probability for any of 4-6 in the first roll is 1/6. But when you expand, the probability of 1/6 in the first roll for any of 1-3 is divided to the expanded six, so the probability to stop in any specific place on the second roll is 1/36.
i was going to construct the tree and count outcomes
but having branches die and not continue does not allow you to just count
do you know any theory
when you construct a tree like this
how you count the brachess??
ANIKHTOS
Let b_i be the total number of branch ends when taking into account throws i and up.
Then
b_k = n and
b_{i - 1} = b_i*floor(e_i) + (n - floor(e_i))
where k is the number of rolls, e_i is the expected value for throw i and onwards (like in the video) and n is the number of sides the die has.
I got the same result with a different method.
If you roll a 4 on the first roll, there is only a 1 in 4 chance you will roll 3 or less on the next two rolls, so you should roll again.
If you roll a 5 on the first roll, there is a 44% chance you will roll 4 or less on the next two rolls, since you are already in the top 33% you should keep the 5.
So, first roll average is 5.5, occuring 33.3% of the time.
The second roll logic is the same as in the video, 3 or less roll again, because 50% of the time the third roll will be above 3.
So, for the 2nd roll you have 4, 5 or 6, which occurs 50% of 66.6% of the time. So, average 5 occuring 33.3%
On the 3rd roll, 3.5 average, which occurs the remaining 33.3% of the time, so your odds are
5.5/3 + 5/3 + 3.5/3 = 14/3 = 4.666
The best explanation so far for me, i guess because of the visuals
I actually remember this question as one of the first questions in my math lessons for the section about chance
Mesdames et Messieures -
Cast, roll, throw or toss your dices! Rien - ne vas plus!
Some folks call them 'dices'.
I reckon that that's nice
Interesting point: If the game were to last 4 or more rolls, until you get to the last 3 rolls you should only ever stop on a 6, as while you still have n rolls to go, the probability of still rolling a 6 is 1 - 5^n/6^n. This is just over half when n = 3, and only gets bigger as n does.
The question and solution is a nice use of the logic of expected value. It could be a little more interesting if you had to pay to roll - e.g., pay 50 cents to roll again.
i wrote a bot to play games in an online casino for me and had to save a very similar problem for this.
time is at 0:54, here comes my math
last step:
3.5$ because that's the average of a single row
previous step:
50% change to get 4,5,6 -> 5$. you end the game.
50% change to get 1,2,3 -> you move to the last step and win 3.5$
average: (5+3.5)/2 = 4.25$
first step:
50% change to get 5,6 -> 5.5$. you end the game
50% change to get 1,2,3,4 -> you move to the second step and win
average: (5.5*(1/3)+4.25*2/3) = 4.666
This was a really interesting video I actually really enjoyed it, however I what would make the argument that this game is flawed because the player never has a chance to lose money. I propose that you redo the video with the following rules: proceed as you normally would with the rules as stated in the video except, if you take the 2nd or the third roll and roll less then the previous rule you must give that amount of money back to the house and can take no further roll.
my guts told me how i should roll the dice, and that the earn should be close to five, but i couldn't think of a way to prove or disprove it
Nice video, keep the good work :)
Yep. That's exactly how I did it. Expected return is 4 2/3 dollars per game.
Yes, people complain about dice/die; it's inherently unclear. Clarity of language is more important that maintaining your pride and your refusal to accept that you're wrong.
If we expand this game to infinitely many rolls, then the expected value will be 6, so we earn maximal possible value. After 5 rolls the formula of expected value will be E_{n+1} = 5/6 * E_n + 1, and limit of E_n is 6. Nice :)
On the third throw the average is (1+2+3+4+5+6)/6 = 3½
Therefor on the second throw stop if you roll a 4, 5 or 6. The average for the second throw onwards is then (3½+3½+3½+4+5+6)/6 = 4¼
Therefor on the first throw stop if you roll a 5 or 6.
The average for the game is then (4¼+4¼+4¼+4¼+5+6)/6 = 4⅔
1. Rolling 5 or 6 initially is less likely to be improved than worsened, so the best strategy would be taking the roll. The probability is 1/3 and the expected value is 11/6.
2. If you initially roll 4, then the probability of never improving the roll is only 16/36, so you should reroll.
2a. If the reroll is no worse, take it. The EV is 1/6 * 1/6 * (4+5+6) = 15/36.
2b. If the reroll is worse, roll again. The EV is 1/6 * 1/2 * 1/6 * (1+2+3+4+5+6) = 21/72.
3. If you initially roll 3 or worse, reroll.
3a. If the reroll is above 3, take it. The EV is 1/2 * 1/6 * (4+5+6) = 15/12.
3b. If the reroll is below 4, roll again. The EV is 1/2 * 1/2 * 1/6 * (1+2+3+4+5+6) = 21/24.
The total EV = 11/6 + 15/36 + 21/72 + 15/12 + 21/24 = (132 + 30 + 21 + 90 + 63) / 72 = 336/72 = 14/3.
You got a like for the definition of dice alone. xD Well done!
The man who talks with proofs instead of just arguing is Presh Talwaker
My mind worked a little differently. I assumed that the "real game" didn't start until the 2nd roll, since any player who didn't roll a 6 would choose to roll again. So when i did the math, i assumed that there were only two rounds. Knowing that the average roll of a 6 sided die is 3.5, i assumed that players who rolled 4 or 5 would realize that their possible payout would more than likely be lower than what they already had and stop. Players who rolled 3 or below choose to roll again as their likely payout is going to be higher than what they had. So when i did the math for the average payout, i used the values 6, 3.5, 3.5 for a total of 4.333 repeating.
I already worked this out when playing yahtzee, because of the 'chance' move :p
I am not good at probability. More puzzles like this please.
Or we can simply do -
1/3*(5+6/2) + 2/3*1/2(4+5+6/3) + 2/3*1/2(1+2+3+4+5+6/6) because - expected value to stop in -
3rd roll - 3.5
2nd roll - 4+5+6/3= 5
1st roll - 5+6/2= 5.5
I would like to know what does math symbols stand for, for example at 3:19 v stands for "value" and i stand for "iterations" or something like that?
He defined it in that mathematical sentence. You don't need to know other definitions, just what he said. "Vi" is value before roll i. So the value before roll 3 is the same as saying value when i = 3 which is the same as V3.
You are right, it is not necessary, but I feel it help me understand easier. Also I want to know if that means something or it's jut random letters he pick
Thats cool and is what I like about maths, great backward concept
English is not my native language, but what I learned back in school is that "dice" is plural, but still refers to one single cube; similarly to how "trousers", "pants", "scissors", etc. refer to one single item.
In that sense, "a dice" and "a die" would be both incorrect (in the same way that nobody who knows correct English says "a trousers" nor "a trouser").
I don't know the etymology of "dice", but doesn't "dice" actually originally refer to the 21 dots that are painted ("dyed") on a singular cube?
Strange, I didn't bother working this out but my gut instinct was this strategy. Basically when you have 1 roll left you "stay" on an upper half" if you have 2 rolls left you "stay on the upper third" makes sense I guess.
I almost got your answer, except I said you should roll again if you get 5 on the first roll. There is a 2/6 chance, or 1/3 chance, that you will get 5 or 6 on the next roll, which would pay either the same or better. However, the chances of getting a 5 or 6 on the second OR third roll would be 1 - (2/3)(2/3) = 9/9 - 4/9 = 5/9. That means there is a greater than 50% chance of getting either the same payout or better if you roll again after getting 5 on the first roll, which I would say is worth it considering the stakes are so low. Maybe if you were to run a simulation and repeat this game hundreds of times, your strategy would have the best average payout, but as a one-off strategy, in reality, I'd roll again if I got a 5 the first time.
Finally, one I can actually solve
How did you solve it
there is another question that i dont know how to solve AGAIN .
there is a blue cube. the area of the cube is 8.4.9 = 288. there is a rope. the rope should contact with the a and b corners which is they are both in the opposite positions. i mean they dont have any contact . if 1 of the corner is in the bottom section other corner must be in the top. if one of the corner in the left side other must be in the right side. if one of the corner is on the front other must be at the back. what will be the least lenght of the rope that can contact these corners .?
A ) 21 B) 17 C) 15 D) 13
The 2nd roll is easy. The expected value for the 3rd roll is 3.5, so you should stop if you roll 4, 5, 6. Your expected value for the second roll if you play this way becomes average of 6,5,4,3.5,3.5,3.5 which is 4.25, so for the first roll you should stop on 5 or 6. So the expected value is the average of 6,5,4.25,4.25,4.25,4.25 = 4.67
It took me about a minute to write this explanation, so I'm now wondering why the video lasts for 8 minutes after the question is given.
Oh, 1 whole minute just to discuss vocabulary. :p
Yeah, that Pr*Er is a really complicated way of explaining for dice. I'm sure it has the benefit of being more generalizable, but it was way easier for me to explain it by just replacing the unsatisfactory roll values with the expected value for continuing, and then just averaging.
Great video! How would the answer differ if you were able to roll 4 times? 5 times? n times?
Keep only 6 if 6 or more rolls left.
Keep 5 and 6 if 3 to 5 rolls left.
The rest you know from video.
Now explain dice rolling strategy for King of Tokyo.
Die is German; dice is Spanish.
Dr. Gerbils Die is also a place in France with a famous wine "Clairette de Die", which is an off-putting name to me!
Die IS Definitely not german
@@surfermaresme2 dice is 'say' in spanish
dice means Würfel in german. ó.ò
@@musik350 Yes, it does.
1d6 is the correct singular for a six sided die.
I actually got this one ! (But using a different method)
There are several strategies to use depending on your goal. If your goal is to maximize your average pay-out, you should go Presh’s way. But...
If your goal is to maximize the chance that you’ll end up with a higher roll than an opponent who’s playing the same game, you’ll use a different strategy.
If your goal is to end up with the highest roll among three competitors, you use an even different strategy.
Why?
Congrats 1M subscriber
This is the same as playing Yahtzee when you only have Chance left... First roll stick with 5 or 6. Second roll stick with 4, 5 or 6. Third roll, whatever you get. Expected values at 5.5, 5 and 3.5. So the average of those? OK - Now I will watch the video!!
I got it right. I'd thought about this before. In Yahtzee when you are just left with chance. Roll the 5 dice, hold those that come up 5 or 6. Roll the rest, hold (as well) those that come up 4, 5 or 6 and finally roll the remaining.
If in Yahtzee "chance", your first roll produces (for example) three 2s, then I'd rather try and collect two more 2s, because every extra Yahtzee! (in this case: five 2s) gives you also 50 bonus points (if I remember correctly).
yuri renner Good point, I was assuming you had already put zero for Yahtzee..
The way the problem is wriiten, It seems like if you choose roll again, you give up any money you would have won if you stopped rolling. So it seems like the most you can win is $6. You have a 50 / 50 chance of rolling any number on at least one of the rolls.
Dear Presh. I see your numbers but I'm struggling with this idea. Can you help me where I'm getting it wrong with this logic? If you rolled 5 in the first draw would it not be more beneficial to decide to roll again? Because there is 5/9 (i.e. more than half the) chance of rolling another 5 or 6 in the next 2 rolls. So there's a more chance that you will not lose out by chosing to try again when 5 came out on the first draw. If you have better chance of getting even or win than lose, wouldn't you try?
Well let's look more into it
probability to get each result
1-4: 2/3*1/6 = 2/18
5-6: 1/6+2/3*1/6 = 5/18
Indeed 10/18 chance to end up with even or better and only 8/18 chance to end up worse, but half of that chance to end up even or better is ending up even. Chance to end up better is only 5/18 < 8/18 and better can only go up by 1, but to downwards we have much more to go 6/18 chance to go down by at least 2, there both probability and distance are greater than for going up. Even without calculating EV it is clear this is not good idea but let's calculate EV for rerolling anyway in relation to our current 5. 2/18 * (-4-3-2-1) + 5/18 * (0 + 1) = -10/18 + 5/18 = -5/18
First throw I'll only accept 5 or 6. If lower, throw again. Second throw I'll accept 3-6.
if we had more dice wouldnt the 1/e=37% rule apply, ie ignore 37% of rolls and then you have to take the one thats higher than the highest pre 37 roll. as described in the port-a-potty numberphile video? think it was called "the mathematical way to choose a toilet"
Quality stuff
That means that so many people used to use 'dice' for singular, that Oxford made it official because they got tired of explaining, lol
If you were starving and there was only a shop that sold donuts for $6. Then you should always roll again unless you get a 6. Because you have nothing to lose.
That was an easy one.
Even if I used Excell because I hate keeping track of numbers in my head.
1:10 savage
But how do you prove that it is the optimal way??
Is using an "average payout" for a threshold to roll/not always the best?
I try some of my own algorithm and test it for 300 data and always losing. Can you explain?
Hi Presh, Thank you for the challenges you put forward to us. could you demonstrate this using spreadsheet. I tried spreadsheet and i think i am making mistakes ,because i consistently get better results if i continue for anything less than a six on first roll. to randomize i used ROUNDUP((RAND()*10) /1.666)
col A for first roll, b for ..... and i used 3 senarios :
IF(A6>4;A6; IF(B6>3;B6;C6 ) )
IF(A6>5;A6; IF(B6>3;B6;C6 ) )
IF(A6>5;A6; IF(B6>4;B6;C6 ) )
best results is from 2nd scenario then 3rd and at last the 1st scenario.
I cannot figure it out. Thanks
Sorry , i figured it out.
An average of about 5 dollars a game, 42% of that being 6, 28% of five, 17% chance of a 4, 9% chance of 3, 3% chance of 2 and 1% chance of one, roughly.
A roll of 1 has an 83.3% chance of gaining value
A roll of 2 has a 66.7% chance of gaining value and only a 16.7% chance of losing
A roll of 3 has a 50% chance of gaining value and 33% of losing
A roll of 4 has a 33% chance of gaining and 50% of losing
A roll of 5 has a 16.7% chance of gaining and 66% of losing.
A roll of 6 has a 0% of gaining and 83.3% chance of losing.
Of the 36 rolls with a leading 4, a third of them will result in gains after the second, a sixth of them will have no change after the second and half of them will lose after the second. Deducting the gains, we have 24 remaining. Of these, if the third die is rolled, six are losses, four match and fourteen gain. Half of the losses correspond to a second roll of four. If we instead hold at four, of the 18 possible results, three lose, three match and twelve gain. Therefore, the best odds are attained if the second roll is held for values of 4, 5 and 6.
If r1= 5 or 6, hold
If r1= 4, 3, 2 or 1, reroll.
If r2= 4, 5 or 6, hold, else reroll.
Take a 5 or a 6 and the first roll and 4-6 on the second. The expected value is 14/3.
What if you cast a 1 each and every time?
For a general case for N rolls and a dice of D faces I made a simple python script you can try ;)
Go there and paste the lines below:
www.programiz.com/python-programming/online-compiler/
Code:
# Number of rolls at max
N = 20
# number of face on the dice
len_dice = 6
dice = range(1, 1 + len_dice) # all the dice values
mean_result = 0 # mean result at the "0th" roll
# compute the mean result for each roll
for n in range(1, N + 1):
results = [face if face > mean_result else mean_result for face in dice]
mean_result = sum(results) / len_dice
print(f'For a {len_dice}-dice, the optimal with {n:3d} rolls is {mean_result:.10f}')
I did a java code and the result is 4.6666 after 99999999 rolls; the code is as follow if you have a Java 1.7 or later compiler:
import java.util.Scanner;
import java.util.concurrent.ThreadLocalRandom;
public class Practice {
private static double dice;
private static double n;
private static double total;
private static double percent;
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
n = reader.nextInt();
for(int i = 0; i
Ruben Izembrandt I think your script is broken. I crunched the numbers by hand (well, calculator) and just 5 rolls was enough to get the expected value up above 5.
andymcl92, a small sample (like 5) will not give anywhere near as accurate an answer as a large sample (like 99,999,999) when it comes to probability. For instance, if I took a fair coin and flipped it a certain number of times, the expected value (if heads is 1 and tails is zero) should be .5. But, I could flip it 5 times and get all heads, so my expected value would look like it is turning out to be 1. However, if you were to flip it 1,000 times, it would definitely work out closer to .5.
SgtSupaman Aaaah, sorry, gotcha now. You mean the average result after 99999999 runs of the game. I thought you meant the expectation of a game 99999999 rolls long! I was saying the expectation value of a game 5 rolls long is greater than 5. (I hadn't actually read what your script did :p )
curiousinternals did you try my code? And I been programming in Java for two weeks so I take every crtisicism gladly to improve
Here's another way of looking at this problem: make your target 5. Why 5? Because your chances of getting LESS than 5 after more than one roll is LESS than 50%. Chances of less than 5 on any single roll is 2/3 so getting less than 5 on two rolls is (2/3)squared which is 4/9, meaning you're already ahead probability-wise. Getting less than 5 after three rolls is (2/3)cubed which is 8/27. Aim for at least 5 and you will win 19/27 times or more than 70% of the time.
@Presh :- can u generalize this result for k rolls? Is it tending to 6 or something less?
If you were playing for material money, I suspect a lot of people, even knowing this, would stop at 4 the first time- you gain only 6% return by continuing, and have a very real chance of getting much less. Playing with that strategy the game is worth 4.625 vs 4.666 a reduction in return of about 1%. aka people value avoided loss more than potential gain.
Even now l don't understand, why is it incorrect to reroll 5 on 1st move. Cause l thought this way:
What is the chance of making better: it's 2/3×2/3, what is equal to 4/9, and that's deffinrtly less than 1/2, so that's why l should reroll it. Where am l wrong?
I didn't do any of the maths except the average of the 3rd roll and came to the same conclusion...
Roll 1
I was right (obviously) But intuitively !! 😁😁😁
Nice Video
Has anyone heard the riddle "if every couple never stopped having kids until they had a boy (then they stopped for good), what would the boy to girl ratio be in the world"?? This seems like a parallel to this problem but in the boy/girl riddle, people swear that it will always be 50/50. And that it doesn't matter what sex your last child was it doesn't changed the distribution) I would love if presh could make a video comparing these two riddles, and maybe assign number values to "boy and girl" and try to make it similar, but prove why it's different, or prove that it's wrong
You should look up Utility Theory
It's exactly a 33.3333...% increase. That's actually how I figured it out.
This is my logic:
There is a 1/3 probability to roll one of this combinations:
1 or 2, 3 or 4, 5 or 6.
As i said, each of these combinations has a 1/3 probability to he rolled, and as we have 3 rolls, then on average at least one number of each combination should be rolled, so as soon as you get 5 or 6, you should stop rolling. If u roll a 4 or less, u should keep playing
A more challenging problem is ......
When u play the game with "2 DICE", all other rules unchanged, what is your expected earning from that game now?
Last roll average is 7. Out of 36 chances, 2,3,4,5,&6 have 1,2,3,4&5 chances respectively. So that is a 15/36 you would roll again and 21/36 of getting 7 or more and stopping. 7-12 would have 6,5,4,3,2&1 chances respectively. From here one can figure out the weighted average for roll 2 then evaluate roll 1 accordingly
I thought the title was a guide for Dark Souls III
Before watching: weighted average of possible maximum values gives me 4.9583 so I guess take the money as soon as a 5 pops up.
Dictionaries report the use of language, they don’t prescribe it.
Where can I play this game where I get payed to roll dice?
This would be a far more interesting problem if you had to pay some amount to roll each time, and figure out what strategy earns the most money
Call them "hexahedral random number generators" and avoid the problem.
On the first roll of the douse, if you roll a 5 or 6, stop. Otherwise, roll the douse again.
On the second roll, stop on a 4, 5 or 6.
Expected value = 5/6 + 6/6 + (2/3 * (4/6 + 5/6 + 6/6 + 1/2*7/2.)) = 28/6 = $4.66.
Dice is the plural and die is the singular
Bonus question: how many rolls would you need before you can get an average return greater than 5?
You need 5 rolls.
Invenblocker Winner, winner, chicken dinner!
The follow up is a psychology problem: if a casino decided to run that game, what should be the price of an attempt for the casino to gain the most money? The two additional assumptions are that not all people will play the game optimally and that the number of people who will attempt to play the game is highly dependent on the price of the attempt.
My gut feeling is that it is a crappy casino game, because if you charge $5 I doupt people will attempt to play and if you charge $3 you need too many people playing sub-optimally to make money.
well obvious if you pay less than 3.5 you are guarantee to win if you play many time
if you have to play 3,5 then you will go to play few times to see your lack and if your lack is bad you can keep playing to get even
if you have to pay above 3.5 then simple do not play you will loose
so the correct answer is 3.5 to give incentive to gambles to play since they think they will beat the odds
but since the casinos always put the favors a bit in their favor for a guarantee win that would be 3.6 or 3.7
I think Rico is right. Few people will pay $5 for the chance to win at most $6.
$3.70 is too low. On any roll other than a 4 on the first roll, even naive analysis will lead players to optimal play.
If the player rolls a 1, 2, or 3, he is going to roll again because if he stops he has lost money.
How would rolling two dices change the outcome?