My doctor asked me if I would want to be saved if I got into a serious accident. Not knowing how to answer, I told her I would want someone to role a 21 sided die (like in D&D). If there was a 1, I would want to die. If there was a 2, I would want to live, if there were a 3 to re-roll, and so on. Now if this happened, what are the chances that they would try to save me? It seems like it would be 1 in 2, but someone who is a lot smarter that me says that’s not the case. What are the odds?
The ‘and so on’ part is kind of important. I mean, if 18 is ‘strap my ass to a rocket-sled and see what happens’, someone might invest in a weighted die.
I don’t even know where you get one in two from, but the odds of them trying to revive you given only what you’ve told me so far tops out around 5%, even with the re-rolls, unless I’m grossly misunderstanding conditional probabilities. So multiply 5% by whatever the odds of the revival attempt being successful are (CPR is successful 20% of the time, for example) and you have the odds of survival assuming they put your plan into action.
You can’t make a 21 sided dice where every side has an equal chance of coming up. You can only have 4, 6, 8, 12, or 20 sides.
First of all, there are no 21-sided dice. Second, I don’t understand the game. What are the outcomes for numbers 4-21?
I did not know this. This possibly invalidates my answer.
Firstly, 21-sided dice are uncommon: you may be thinking of 20-sided dice, in the shape of an Icosahedron.
However, if you had a 21-sided die, with each face equally likely to come up, and numbered 1 to 21, and 7 faces meant “die”, 7 faces meant “live” and 7 faces meant “re-roll”, then the odds would be 1 in 2 for each of “die” and “live”. In each case you would sum the infinite sequence:
1/3 (decision on first roll)
1/9 (one round of re-rolling)
1/27 (two rounds of re-rolling)
1/81 (three rounds of re-rolling)
etc.
The limit of the sum of each sequence is 1/2.
I am not sure, but I think he means a repeating pattern of face values: die, live, re-roll, die live, re-roll, die, live, re-roll,…etc. That, presumably,is why it needs a dice whose number of faces is divisible by three. It will work with a regular cubic dice, or a dodecahedral one.
I am not good at this sort of probability puzzle, but it looks to me as though the odds indeed are 1 in 2.
You could have a 21-sided die, though it’s not all that practical. Make a prism with a regular 21-sided polygon on each end, then make each end concave so that the prism can’t land on the end. You can make dice with any number of sides this way (at least in theory).
If you were using an icosohedral die, then 7 sides could be “die”, 7 sides could be “live”, and 6 sides could be “re-roll”. The sums of the infinite series would be harder to calculate, but they are obviously equal and must add together to make 1.
I infer that the “and so on” means that the pattern repeats in a simple fashion. On any whole-number multiples of 3, the die is re-rolled. On a result of 1, 4, 7, 10, 13, 16 or 19, the patient wants to die. On a 2, 5, 8, 11, 14, 17 or 20, the patient wants to live.
Assuming the fact that there is a fair 21 sided die (note that 20 sided dice are commonly used to play D&D), the likelihood of life and death is 7:7, or 50% death. If it’s instead a fair 20 sided die, the odds are still 7:7. On an unfair die, who knows? The only nagging question is does the emergency responder have the patience to wait out a potentially infinite series of multiples of 3? You would bleed out waiting for assistance if it took too long.
The OP does mention, albeit in parenthesis as if assumed, that the die is “like in D&D”. If this is the case, we’d be talking about a 20-sided die.
So, assuming from there, you could start poly-face 1 with die, face 2 with live, face 3 with re-roll, then repeat until you fill up all 20 faces.
If started in this succession, you’d get
7 faces with “Kill me”
7 faces with “Resuscitate me”
and 6 with “Re-roll”
What then?
Given a fair die, the re-rolls are probabilistically equivalent to having that particular face of the die not exist at all. So, in your scenario, you have 7 living possibilities, and 7 dying possibilities, for a total of 50-50. This is why, as someone mentioned above, there is no difference to having a 21 one sided die with 7 live, 7 die, and 7 do-overs, and 7 live, 7 die, and 6 do-overs.
Right. If the limit of the series is S, then
S = (7/20) + (7/20)(3/10) + … (7/20)((3/10)^n)
Multiply both sides by 10/3:
(10/3)S = (7/20)(10/3) + S
Subtract S from each side:
((10/3)-1)S = 7/6
i.e.
(7/3)S = 7/6
i.e.
S = 1/2
writing 20-sided die rolling out of my DNR form
I think you mean ‘convex’ here (i.e. pointed outwards, so if the die lands on the point, it falls over onto a flat face. A die can sit on a concave face, resting on the edges.), but otherwise it’s a good point. You can make a (theoretically) fair die without using a Platonic Solid.
Of course :smack:
You don’t have to sum an infinite series, just observe that the chances of living or dying are equal and the chances of rerolling forever are 0 to see it is 1 in 2. Just so long as the number of faces for live or die are the same and there is at least one of each. Of course, if only face said live and 20 said die, then obviously the odds against living are 20:1.
Was browsing thinkgeek today and ran across a set of dice with d3, d5, d7, d14, d30, d100.
I’m surprised no one’s hit on this particular nitpick yet: I would put the chances that they would try to save you at near 100%, since what you say you want, based on a die roll, probably wouldn’t be seen as sufficient reason for them not to try to save you. 
That said, I agree with njtt interpretation of the problem. This being hypothetical anyway, I don’t think it’s relevant whether or not it’s physically possible for a 21-sided die to exist. (I suspect one could be made, but of course it wouldn’t be one of the regular (Platonic) solids.)
Or, your children file a wrongful death lawsuit and subpoena the actual die the EMT’s used, which, of course, has been lost or given to the EMT’s son to play D&D with and he added it to his impressive die collection and can’t remember which one it was that daddy gave him last month. Since the evidence has been spoilated, you claim that the court should presume that the die was weighted in favor of death… hijinks ensue.
In this case you only need to reroll long enough for copperwindow to die. So it’s not really going to come out as 1 in 2.