If I am offered two bets to choose one from, and either wins $100 or losses $100, I would want to know the chances of winning in order to decide which one to pick. For example, if one bet has a 90% chance of winning, and the other has a 30% chance of winning, I will pick the 90% one

Question: why is the bet with the higher probability "better"? Assuming that I very much want to win the $100.

Attempt: this makes sense from a Bayesian point of view. 90% is my confidence or belief that I will win $100. Therefore, I will pick the bet that I am more confident or has higher belief in.

However, this makes no sense using the frequentist framework since 90% chance of winning is interpreted as the proportion of wins if I repeat the bet a "large" or infinite number of times. Since I am only doing this once, what is the reason to choose the one with the higher chance?

  • a) just shut up and calculate? b) imagine the problem is about two guns with the probability reflected by the number of chambers loaded with a bullet. now the reason is simply your wish to remain alive. c) probability is label on the sign by the entrance to a bottomless rabbit hole. – nir Apr 16 '16 at 20:27
  • d) evolution? imagine two species of lemmings — one wired in such a way that individuals of the species tend to make choices that are likely to keep them alive while the other species is wired in such a way that individuals tend to make choices that lead to certain death. – nir Apr 16 '16 at 20:41
  • I find your example to be question begging: you've characterized the games in terms of "chance of winning", then ask "why are probabilities related to my goal (of winning $100)?". They are related because that is the way you've framed the question. – Dave Apr 18 '16 at 20:26

Your question highlights an important issue with respect to different interpretations of probability. The frequentist has great difficulty attaching a probability to a single event. Strictly speaking, there cannot be a frequency of a single event, so frequentists tend to wave their arms a lot and claim that we can consider something to be an instance of a long run of trials in principle. Such a claim is not plausible in general. What is the probability that Hillary Clinton will become the next US president? (This is written in 2016 with the primaries still in progress.) Such an event, if it happens, will be unique: there cannot be a long run frequency of it. If you try to assess its probability as a frequency you cannot get any sensible value for it. What is the frequency of a female president? Zero. What is the frequency of a democrat president? About a half? What is the frequency of a democrat president given that the previous president was a democrat? Less than a half. Whatever frame of reference you choose will give you a different answer. Frequentists might try to tough it out and say that such an event does not have a probability, because it lacks a frequency, but this is implausible. You can bet on Hillary to win (or not win) and betting odds imply probabilities.

The virtue of the epistemic approach to understanding probability is that the derivation of the probability calculus can be made without reference to frequencies or possibility spaces, but by starting from simple assumptions about how decisions are made under uncertain information, and about what constitutes a bad decision. We all have to make decisions, and we almost always have to do so with imperfect information. Frequently we make bad decisions, and often this happens because we have failed to quantify the uncertainty properly. One approach to deriving probability is to take a bet as a paradigm case of a decision under uncertainty, and a Dutch book as a paradigm case of a bad decision. (A Dutch book is a combination of bets that results in you losing, no matter what happens.) A fairly remarkable result, first proved by Bruno de Finetti, is that from this consideration only, if you wish to avoid making bad decisions, your calculus of uncertainty must conform to the probability calculus. Of course, this is a fairly limited concept of 'bad decision' - it corresponds to nothing more than maximising expectation value, and as many theorists have pointed out, it is not always the best strategy to maximise expectation value. Nevertheless it is a powerful concept: it shows that we can derive probability theory from decision theory.

This provides you with the answer to your question: probability is used to make decisions because properly understood it is exactly that quantity that allows us to make decisions under uncertainty without falling into straightforward errors of judgement.


The question here really is "Does probability predict the future or assume from the past?" in a situation where you have no information to go on from past knowledge there is no statistical probability at all making the decision a matter of "odds" not probability, you can mathematically decide the assumed probability based on the number of choices. Note that this is logical not statistical. Logical meaning that you can without a doubt create a logical assessment of chance but a statistical assessment requires past knowledge. Taking this into account and removing the statistical rationale behind the decision the "probability" based on logical reasoning will without a doubt become a 50/50 strictly due to there being two choices.

Probability becomes useful in these decisions only when the information used to create the probable situation is itself hidden from the population at large. If there are too many variables for your average person to asses accurately the chances of winning then your logical probability shoots up based on the niche information you wield. This works in matters of chance as well as skill.

Lottery: If two options of lottery tickets are available to you both with a chance at winning $100 but either has an additional stipulation unknown to you then the chances of you winning any money at all is 50/50. Say Ticket A's additional piece of information is that even though the statistical odds of any given person winning $100 are 25%(I don't like to do math with 30%) every 10th winning ticket has a multiplied payoff of x4 winning you $400 meaning you actually don't win $100. That said it is completely exempt from being reveled within the statistical odds of 25%. The other has the 75% for $100(adjusted to keep the odds three times as high that you get a hundred dollars with the changed 30%->25%) But the hidden information on this ticket is that every 4th winning ticket holder is responsible for offsetting the tax on all transactions between the company and customer. This would mean that even if you win $100 there is a chance that you'll win it and then be put into a select few who are responsible for using a portion of their winnings, lets say $70 for each fourth winning ticket, to make the system cheaper for the rest of the actors involved. This means that even though you are one of the 75% of winning ticket holders you actually only won $30 and thats before the cost of the ticket itself. In this way someone who says logically that there is a 25% chance of winning up to $100 depending on which ticket you chose would be right based on only knowing there are two tickets and either could be a winning or a losing ticket. When you take into account that the tickets are not supplying you any false information by saying "One in four wins up to $400!" or "Odds are 3 to 1 for a $100 payoff!" just withholding information then the power of probability only comes into account once you have that extra information and the populous does not. In that case your logical deduction might be something like: ticket a has a 1 in 4 chance of winning every 10th win rolls giving me a likely chance of winning twice in a row and a bonus 400. I buy 20 tickets then, I win 5 $100 tickets and one $400 which is pretty much aligned with my information and I unfortunately just missed out on the second high priority win. Still $900 and if we say each ticket cost $20 with a 20% tax on lottery tickets in this world then my tickets individually cost me $22 and that 20 times is $440 meaning I still won net $460. The other ticket with a 75% chance of winning would yield 15 wins and $100 each there with 5 unlucky tickets meaning I picked up an extra toxic ticket again within reason of expectation. $100 15 times is $1500 then $70 five times is $350 which brings me down to $1150 then I pay $20 for each ticket $400 in total which means I won a grand total of $750 this means that for someone who can spend $400 on lottery tickets they could find this information to realize its way better to play through the high chance wins rather than the jackpot goal but you can see how the information hidden could easily sway the "logical probability" without your knowledge.

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