Tentative title, too long to fit the titling rules: Let A be a natural number. An experiment is defined as A doubling in count, becoming zero, or becoming any natural number in between zero and 2A. Question: After n tries, what is the probability that A will be X, where X is any natural number between zero and 2A.
This is derived from a videogame, Path of Exile, where a gamble is presented to the player:
-In this game, there exists divination cards. Each individual card can be collected and stacked up to a certain maximum number, and upon reaching a full stack of cards they can be redeemed for a reward deleting the cards from the inventory of the player who gathered them.
-The player is presented with a gamble that he can repeat as many times as he wants as long as the number of cards to be gambled are equal or less than half the maximum numbers that can be collected in a stack, or half minus one for an odd maximum number of collected cards in said stack.
-For each time the player gambles, the number of cards on the stack can
*(1) double and become equal to the maximum number of cards possible for a stack or, in the case of an odd maximum numbers of cards in a stack, become equal to the maximum minus one card,
*(2) be lost (become zero, however the stack is no longer in the players inventory) or
*(3) the number of cards in the stack can become any natural number in between zero and the maximum of cards or, in the case of an odd maximum numbers of cards in a stack, become any natural number in between zero and the maximum of cards minus one.
Given that these cards are valuable for a player they may be tempted to gamble their cards in order to increase the total number of cards they ultimately own. In order to calculate the expected value of this gamble, first we must calculate the probability of each outcome given the initial number of cards offered in the gamble and how many times a gamble will happen.
COMPLETED WORK SO FAR
In order to tackle this problem I have calculated the probability for an experiment that starts with only one card and where the gamble is attempted three times. To simplify this calculation, i have ommited the upper outcome limit for every gamble.
First Gamble:
The initial number of cards to be gambled is one. The possible outcomes is either 2, 1 or 0. Each one of these outcomes has equal weighting and the probability of each ocurring is one in three.
Second Gamble:
Would the first gamble had resulted in 2, the possible outcomes of the second gamble are 4, 3, 2, 1, 0. Each one of these outcomes has equal weighting and the probability of each ocurring is one in five.
Would the first gamble had resulted in a 1, the possible outcomes is either 2, 1 or 0. Each one of these outcomes has equal weighting and the probability of each ocurring is one in three.
Would the first gamble had resulted in a 0, the only possible outcome of the second gamble is 0, thus the experiment concludes.
Third gamble:
Would the first gamble had resulted in 2 and the second gamble had resulted in a 4, the possible outcomes of the third gamble are 8, 7, 6, 5, 4, 3, 2, 1, 0. Each one of these outcomes has equal weighting and the probability of each ocurring is one in nine.
From this we can conclude a single outcome where the result of the experiment is 6. The probability that the first gamble had resulted in a 2, the second gamble had resulted in a 4 and the third gamble had resulted in a 6 is equal to the compound of the probability of each of the gambles independently. To say, the probability is 1/3 * 1/5 * 1/9.
After manually completing the probability tree, only two experiments concluded in a total number of 6 cards as the outcome:
*The first one corresponds to the experiment where the first gamble results in a two, the second results in a four, and the third results in a six. This outcome has a 1/3 * 1/5 * 1/9 chance of ocurring.
*The second one corresponds to the experiment where the first gamble results in a two, the second camble results in a three, and the third gamble results in a six. This outcome has a 1/3 * 1/5 * 1/7 chance of ocurring.
From this we conclude that the chance of the experiment resulting in a six is equal to the sum of these two experimental outcomes, aproximated to 1.69% chance of ocurring.
After calculating the probability of every outcome for the experiment, these are the probabilities of each outcome:
The same method can be applied for any number of gambles and with any number of starting cards.
QUESTION:
Can this process be generalized?
At first glance, the probability of the experiment resulting in an outcome can be generalized as a sum of multiplications.
To say, for the experiment, the chance of outcome equals 1 is:
P(1) = 1/3 * 1/5 * 1/9 + 1/3 * 1/5 * 1/7 + 1/3 * 1/5 * 1/5 + 1/3 * 1/5 * 1/3 + 1/3 * 1/3 * 1/5 + 1/3 * 1/3 * 1/3 = 11.17%
the chance of outcome equals 3 is:
P(3) = 1/3 * 1/5 * 1/9 + 1/3 * 1/5 * 1/7 + 1/3 * 1/5 * 1/5 + 1/3 * 1/3 * 1/5 = 5.25%
NOTES:
The probability of an experiment with outcome 1 is equal of that of an experiment with outcome 2. this happens because of the nature of the outcome of the first gamble having a maximum of equal to the double of the starting numberof cards making the maximum outcom always an even number.
Excel worksheet with my completed work
This is how the gamble is presented inside of the game
