"Don't Cast Your Pearls Before Swine" (Jesus)
Maybe the confusion arises from how the problem is presented. I believe the ideal way is to view it in its "complete" form,
for which we could assume the following distribution of probabilities: 40% that it lands in the first spot, 15% in the second spot,
10% in the third spot, and 35% in the remaining spots (e.g., 9%, 9%, 9%, 8%). This totals to 100%.
Then, if it did not land in the first spot, that 40% should be redistributed proportionally to the remaining spots to maintain the total of 100%.
That is why the 15% becomes 25%, the 10% becomes 16.7%, and the 35% becomes 58.3%.
It can be observed that if it does not land in the first, second, or third spots, there will be a 100% probability that it lands in one of the other spots.
That's what I think, but I might be wrong in how I'm approaching it.
Sincerely,
Cac
Luck is what happens when preparation meets opportunity.
That's exactly how I thought of the problem as well. The prior probabilities and the sample space are well-defined, as is the function that maps events "1st", "2nd", "3rd", and "more than 3rd" to prior probabilities. I'm not sure where the confusion comes in yet.Originally Posted by Cacarulo
Three boxes add up to 65%. A sequencer would know what that 65% represents.
For example, if he performs a single riffle shuffle and tries to catch an Ace, he might bet on three boxes.
To learn more about the expected value (EV) of his bets, he would want to find the percentage chance of where the Ace lands.
So, he focuses on the three boxes with his money, with 40%, 15%, and 10% as his answer, and he doesn't care much about where the remaining 35%.
In that case, the expected value of the three bets in boxes 1, 2, and 3 given the prior probabilities of the target ace appearing in each one is
(0.40)(0.5208)(x) + (0.15)(0.5208)(y) + (0.10)(0.5208)(z),
where x, y, and z is the fraction of one's bet B one should place on boxes 1, 2, and 3 respectively. 0.5208 comes from the fact that the expected value of the hand to the player when he knows he's starting with an ace is 52.08% in a single-deck game. We have x+y+z=1 by definition, as well as B=0.4208% of one's bankroll from the article "42.08%" by James Grosjean and Previn Mankodi, representing an optimal fraction of one's bankroll of 42.08%. By inspection, the expected value above is maximized when x=1, y=0, and z=0, so that we obtain (0.40)(0.5208)(1) = 0.20832, so that (0.20832)(0.4208) = 0.08766 is the proportion of your bankroll you should bet on the 1st box when you know an ace is coming on one of the 1st, 2nd, and 3rd boxes with probabilities 0.40, 0.15, and 0.10.
Since the sequencer in this scenario presumably wants to bet something on all three boxes, he may want to keep things simple and bet 1/3 of 0.4208% of his bankroll in each box. This would yield (0.40)(0.5208)(1/3)+(0.15)(0.5208)(1/3)+(0.10)(0.5208)(1/3) = 0.11284 as his expected value.
Alternatively, he may want to bet his half of normal large bet of 42.08% of his bankroll in box 1, a quarter of 42.08% of his bankroll in box 2, and another quarter of 42.08% of his bankroll in box 3 (x=1/2, y=1/4, z=1/4). Rounding this down to the nearest ten percent, this translates to 20% of one's bankroll in box 1, 10% of one's bankroll in box 2, and 10% of one's bankroll in box 3.
It is entirely possible that I'm wrong; these are just my initial thoughts about the scenario.
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