I think that me (NO party) and @ShitakiIntaki (YES party) found a "reduction" of a snake eyes problem (link at the end). Reduction - A most simple task, which already causes divergence in opinions. And it feels to me that people are likely to take the same sides here as they took on the original market.

The problem is:

I will throw a coin. If Heads, one person is chosen and he loses. If Tails, 9 people are chosen and all of them win.

Given you are chosen to play, what is the probability you lose?

It is not an infinite series of events where some of throws cause continuation of the game, it is just a round with one throw.

The market resolves NO, if at the market closure I think that the answer is 50%.

Resolves YES if I am convinced in any other number.

My current position:

Some people intuitively say that there are 9+1 places, so 1/10 is the probability to lose given you are chosen. But those are not spacial places, not eggs and boxes where you solve permutation problem. Those are different results of different branches on a probability tree. Only people within the second group are equal to each other, but none of them is equal to the person that is on the other branch of the tree.

The phrase "given you are chosen to play" only limits the amount of cases we look into, but it does not equalize those cases. And those cases are not equal.

At start we know that a person is chosen, that is given, so "The person A is chosen" is 100% base of our tree.

The tree splits in two halves depending on a toss.

"The person A is chosen AND Heads" is 100%*50%=50%. There is one more step in thos branch. There is only one possible person to end up being in this case, so let's call this "place" an index 1.

"The person A is chosen AND Heads AND he is assigned index 1" has 100%*50%*100%=50% probability

The other branch "The person A is chosen AND Tails" is 100%*50%.

But it branches further, there are nine equal indices (2..10) the person A could end up with. Here they are equal between each other, because they have the same parent branch.

"The person A is chosen AND Tails AND he ends up with index 4" has 100%*50%*1/9. There are 9 such branches. The person A has lower probability to be any specific index 2..10 than index 1.

The final calculation would be: sum up over all cases [result of the case * probability of that case]. Probability of the case is its the result's weight.

0.5[prob of case 1] * 1[ratio of losses in case 1]

+ 0.5*1/9*0 *9

The probability is 0.5

I see WHY people vote for 1/10. They feel like there is a preprepared pool of 10 people, and one of those is Guaranteed to be in lose case and the rest would end up in winning case. Like a permutation problem. But there is no pool of potential players. Our calculation does not depend on the size of a pool, it can be infinite or undefined size (if I for example just pick people from the street). And there is no equality of places between index 1 and index 2, which is needed for permutation problems. And there is not pool existing, because the draft of those 9 people might not even happen. There is 50% those people-places do not even exist, so they cannot have a weight of more than 50%.

Same logic is applied to the Snake eyes problem.

The probability of game ending in round 1:

1/36

The ratio of dead people in that case:

1

Game ending in round 2:

35/36*1/36

The ratio of dead in that case:

2/3

And so on to infinity.

This is an infinite sum which CONVERGES.

1/1 * 1/36

+2/3*35/36*1/36

+4/7*(35/36)^2*1/36

...

The ratio of dead people we do not weight by the amount of people in that case, we weight by the probability this case was reached. Weighting by the people count would be absurd, because it is already figuring inside the ratio. And it would give an infinite weight to the very last "never ending case".

To be pedantic we can account for that infinite case too (although the sum converging already hints something).

The probability of that case is an infinitely small number. Let's mark it ~0.

The ratio of dead people in that case is strictly zero. 0.

When we calculate the last term it would be 0*~0=0. When we add that to 0.52 we will not change the answer.

I am using ratios, because we are asked about probability. Probability is the ratio of suitable cases to all reachable cases. I have seen people trying to work with absolute values instead and trying to fit in expected counts of dead people. But I've noticed a common mistake I. The original market.

My infinite sum looks like this:

dead/players*prob + dead/players*prob ...

They tried to divide the expected number of dead people by the expected number of players. That is incorrect. Look why, you cannot convert a/b+c/d+e/f into average(a,c,e)/average(b,d,f).

1/2+2/3+3/4 != (1+2+3)/(2+3+4)

I simplified "expected number" into "average" for the sole purpose of the mistake being visible. That approach breaks the infinite series of fractions and make a new one, which is meaningless.

To sum up. THE WEIGHT of a case is the probability of the case being reached by the game process, not the amount of people in that case (because as I said in the first part, those people-places are not equal).

Only in simple school cases, where all boxes are equal and can be reached from a single probability parent branch in one, we can say that ending up in any of those boxes has equal probability.

• If you reached this part of the text, don't forget that 50% here corresponds to NO. That is done to have consistent parties names.

• Snake eyes