# How to explain that winning the lottery is not a 50/50 distribution?

When casually discussing with my 13 yo child about probabilities, he told me

there is a 50% chance to win at the lottery

To what I said

no, there is a 1 chance over 90 million

(I roughly estimated $$_{7}^{49}\text{C}$$ which I think is more or less the lottery here)

To what he replied

no: either you win, or you don't. That's the probability of the fact to win.

He is obviously wrong, me being the educated father and him the child with silly ideas. I am now sitting and thinking about a counter-argument.

On the serious side, I am trying to quantify his answer from a mathematical perspective but I believe that the whole premise of his reasoning is wrong (but I am not sure where).

Note: I am asking the question here and not on Math SE because it is in my opinion more a matter of how to explain math to children (and accessorily, to their parents), more than a question about probabilities.

Note 2: I should have made it clear that we are there after many discussions on probability so he understands the "number of positive outcomes" / "number of all possible outcomes". What he said was more like an invitation for discussion about the concept of "winning or not, as a single event".

• Have you asked him to explain what he thinks “probability” means? Oct 19, 2020 at 19:03
• Get 10 or more coins, and tell your child that winning means all heads face up when you toss the coins. Point out that there are two outcomes, win and not win. Then ask which is more likely to occur, win or not win, if you toss them. Explain that 50% chance to win means equal chance of winning and not winning, not that there are two outcomes. Oct 19, 2020 at 19:50
• @RustyCore Your "definition"of probability applies only when all outcomes are equally likely. And that's exactly the 13-year-old's problem. He sees two outcomes, win and lose, of which one is favorable, and he concludes that the probability is ½. Oct 19, 2020 at 21:32
• Note that "no, the chance for X is 50%: either you X or you don't" is a quite common meme / joke (at least on twitch.tv). He might be joking. (Which doesn't make this question invalid, of course!) Oct 20, 2020 at 9:23
• @Noiralef Indeed, half of the fun of the joke is that while everyone knows it's a ridiculous way to calculate the probability, very few people can actually tell you why :P Oct 20, 2020 at 12:34

Your child is using the Principle of Insufficient Reason, which states that if we have no information about something other than the set of possible outcomes, then we should assume that all outcomes are equally likely.

This principle is behind basically all of statistics, probability theory, and statistical mechanics, although it is often disguised in some way.

The Principle is not applicable in this situation, since we do know how many winning tickets there are in the lottery, and how many tickets are possible. But we can still apply the Principle to the lottery, in the sense that we assume that each ticket is equally likely to win.

Your child would be right if we knew nothing about the lottery, other than that there are two possible outcomes "win" or "lose". However, we do know quite a lot about lotteries in real life, so even knowing that it is a lottery already tells us that the two outcomes (win and lose) are probably not equally probable.

So, your child's reasoning would be right (or at least justifiable) if you said, for example: "Let's play a game. Either you win or you lose. What is your chance of winning?". Or if, for example, he was an alien who had just arrived on earth and had no idea what "lottery" meant, and someone said "Let's play the lottery. Either you win or you lose. What is your chance of winning?"

• This is a very good answer.
– JRN
Oct 20, 2020 at 7:46
• It answers the "why?" but not the "how to explain?" Oct 20, 2020 at 10:54
• Based on my in-class experiences, I disagree. A person in this situation literally doesn't understand the "count equally likely outcomes" clause. Oct 20, 2020 at 14:07
• Note that only (some forms of) frequentist statistics are based on this. Bayesian statistics allows you to choose another prior. Oct 20, 2020 at 16:05
• This concept is called classical probability. It is neither frequentist nor Bayesian; it underpins both. Without it, there is no concept of simulating a random draw from a probability distribution, so it is impossible for statistical calculations to have any meaning. Oct 20, 2020 at 23:36

I don't think that talking about probabilities formally would be to any benefit for your son. However, you could simulate a lottery at home, using a die. Say that a player wins if they guess right the next outcome. So, at each roll, there is a $$1/6$$ probability that the player wins. Roll the dice many times - say 50 or 60 - and write down the number of wins. If your son is not Donald Duck's super-lucky cousin - whose name I cannot recall - then you will give him some food for thought.

After your game, you could ask him something like: "Why didn't you win about half of the times but only $$X$$?" - substitute $$X$$ with the number your son won in your game.

Another approach could be to get out of house to some basketball court and take shots blind-folded. According to your son, about half of your shots should be on target, however, given your son is not Steph Curry, you will have again a chance to explain him how the relative size of the basket narrows down the chances that you hit the target about half of the times.

Also, note that the two above situations are nice demonstrations of discrete (dice) and continuous probabilities (shots), in case it comes up in your discussions!

P.S.: Congratulations for engaging in discussions about maths with your children! :)

• Gladstone Gander? Oct 20, 2020 at 1:18
• The number of basketball shots is discrete, not continuous.
– JRN
Oct 20, 2020 at 7:45
• @JoelReyesNoche Maybe I wasn't clear, but the probability of shooting on target is an example of continuous probability - area etc. The process of taking many shots is used as a way to statistically approximate that probability. Oct 20, 2020 at 8:40
• Okay, so you meant the position of the ball. I agree that that would describe a continuous random variable.
– JRN
Oct 20, 2020 at 11:38
• @MichaelBächtold Indeed I assume that about the son, however, even if the case is different, this misconception gives a nice chance to discuss about relative frequencies - informally, of course - and how could to equiprobable events occur with different frequencies. As for basketball, success in a single shot is determined by a continuous quantity (area). Oct 20, 2020 at 20:36

I think so far best reaction is the top-voted comment:

Have you asked him to explain what he thinks “probability” means?

I'd address the topic from here. And as this is not a school environment when teacher has the truth and shares it with the student, I'd take the approach to try to understand him through questions:

• What probability means for you?
• Are there situations where chance is not 50-50?
• if yes, what are those?
• How are those different from lottery?
• Can we apply something similar also for lottery?
• if no, then maybe that's a time to start explaining that "probability" has a different meaning for him and for (at least majority of) rest of the world

## For the updated question:

What he said was more like an invitation for discussion about the concept of "winning or not, as a single event"

Again, without teaching, you might start asking about what's the practical use of probability if you exclude "PROBABILITY" from it and just list outcomes.

Or, just go back to definition: "probability for me is chance of winning. When I say 1 to 90M I mean I need to buy 90M tickets and I expect one of them will win. What 50-50 means for you?"

(if he still GENUINELY insists to 50% it would even worth price of 10 tickets for me to buy together than discuss why 5 of them was not winning. All other answers approach this from doing a different experiment, but that case it's all about explaining how those other events are similar instead of focusing on the lottery)

• Βασίλης Μάρκος's answer is great and would be a wonderful experience for parent and child. However, I love this answer because it addresses the way we, as humans, assimilate information and, more importantly, knowledge. Instead of providing any sort of answer, walking your son through reasoning it out himself, providing counterexamples to keep him thinking, will result in a far deeper understanding while likely giving him pride in finding the answer himself (encouraging future curiosity and discovery). Oct 20, 2020 at 19:23
• I'd hope if you're buying 90M tickets you're being careful and making sure one of them will definitely win. Oct 21, 2020 at 8:28
• For further "model-breaking" (let's call it Socratian questioning of the child's mental model): if 1 lottery ticket has 50-50 chance of winning, what are my chances if I have 2 tickets? Oct 21, 2020 at 18:58

Without going into the mathematics too deeply, I would say it boils down to this: There is only one way of winning the lottery: guessing all the numbers correctly. But there are a lot more possible ways to lose. So the "whole premise of his reasoning is wrong", as you say, because he is ignoring millions of possible outcomes!

A slightly different approach:

Let's say there are 100 lottery tickets in total and there is only one ticket that will win you the prize. If you don't buy any tickets at all, what's your chance of winning? No chance. Nothing out of a hundred is zero %. What if you could buy all the tickets? You are certain to win. You've got 100 out of 100 tickets. Your chance to win is 100 %. How many tickets do you have to buy to have a 50 % chance? How big is your chance when you buy just one ticket?

Edit: yet another attempt to get into a non-mathematical mind:

It's a banana or it's not a banana. If it's not a banana, it can be an apple, a pear, an orange or a chayote. If it's a banana it can only be a banana. So "banana" or "not banana" are not the same kind of thing. They are not equal. The same applies to "you win" or "you don't win".

• Once understanding comes from this exercise, you can then add a zero and repeat the exercise until you have a number that approximates the odds of winning the actual lottery. Then you compare that number to the odds of winning if you don't buy a ticket. Hopefully they will come to the conclusion that the only way to win, is to not play the game. Oct 20, 2020 at 19:27
• "Hopefully they will come to the conclusion that the only way to win, is to not play the game." In order to do that, you'd need to delve into expected values taking into account the probability of winning, the price of the tickets, and the prizes you can potentially earn. Oct 21, 2020 at 3:01
• If the odds of winning are essentially zero, the other items are irrelevant. Oct 21, 2020 at 14:14

Taking the ed part of the question:

1. Don't feel like you have to convince the kid of everything immediately. Give him time.

2. In particular, watch out for him just trolling you.

3. If you do decide to engage him, you could do so by suggesting that if he really thinks it's 50-50, that you'll take the opposite side of the bet for 2:1 odds. Ask him to put some of his lawn-mowing money down...and see if he really believes in his thesis.

4. (Lapsing into math), of course the probability of each outcome is not 50% just because there are two possibilities. (back into ed) Perhaps you can come up with similar bets (e.g. bet that the next house door you see will be open or closed...it's a binary result, but with unequal probabilities). Pepper him with a couple examples like that. But mind (1) and (2). ;-)

P.s. Still feels more like a math question than a how to communicate question, given you didn't answer it yourself and then discuss the comms. Also given the responses here.

# Make it Personal

Take a marshmallow (or some small candy that you know he likes), show it to him, then put it into one hand behind your back and say: "If you pick the hand with the marshmallow, you can have it. Now, what is the probability you will choose correctly?" Then let him guess.

Then, get a jar of marbles (like, 100+), pick one distinct marble, and say this: "If you pick this magic marble blindfolded in one try, I'll give you this giant candy bar. If you don't, you have to mow the lawn, take out the garbage, and wash the dishes for a month. What is the probability that you will win this bet?"

If he insists it is not 50/50, then say: "But picking this marble is just like the lottery. In fact, it's much easier than the lottery. So why isn't it a simple 50/50 gamble? We agreed that the marshmallow test was 50/50, so if the marshmallow test is 50/50, and the lottery is 50/50, why isn't the marble test 50/50?"

Then you can reduce the marble test to 2 marbles, and add one marble at a time for him to learn concretely and intuitively how probability works. The stakes make the cost of losing very high, so that he cannot be flippant about his answer. At that point he knows he can't just guess. He needs to be right. Guessing the odds on the lottery has zero cost to him, because as a minor, he isn't legally allowed to play.

Ask him whether the probability of winning is the same if you bought 1000 tickets rather than one ticket.

Or, imagine a lottery with 100 tickets, of which only one was a winner. If 100 different people bought a ticket, how many would win? What does that suggest for the probability of any one ticket being a winner?

I want to offer a game to play with your son that he would almost definitely understand and would impart the principles of probability (and the futility of gambling at the same time).

First, get some tokens. These can be marbles, beads, painted rocks, Legos, bits of colored paper, or whatever else you have handy. Of the tokens, choose a defining characteristic that you can use to separate the tokens into two easily-discernable groups. For the sake of this explanation, I will choose glass beads that are either red or blue.

In a small bag, put in 1 red token and some crazy number of blue tokens. Then offer the bag to your son, disallowing him to look inside and telling him only that the bag contains blue and red tokens and if he draws a red token, he wins. You could even sweeten the deal by having some candy as a reward, which will serve to further drive the point home later.

Let him reach into the bag (again, without letting him look) and pull out a token. Regardless of whether he "wins", tell him to put his token back in the bag, then mix up the contents of the bag and have him try again. Do this as many times as you think is necessary, or until he gets bored and/or frustrated of losing over and over again.

When the game ends, sit him down and dump out the bag in front of him. Have him see that there are tons of blue tokens and only 1 red token. Demonstrate that, in a sea of blue tokens, the chances of him picking a red token at random is incredibly slim, so even if the only possible outcomes are either winning or losing, it's much more likely to lose than to win.

Then explain that the lottery is the same way. In order to win the lottery, you need to buy a winning ticket (which is a red token) and not a losing ticket (which is a blue token). Explain how in the lottery there are millions of losing tickets to the point where he would have to find the red token not in a small bag, but in a huge garbage bag that was packed full of blue tokens but still only had one red piece. Ask him if he wants to play that game, and you shouldn't be surprised if he says no. :)

One thing to make the "game" more like the lottery is to give him 10 pieces of candy to start with, and every time he loses, he has to give up a piece of candy, and every time he wins, he gets 5 pieces of candy. Even though he will lose far more often than he wins, encourage him to keep playing when he does win to drive home the feeling of gambling. What seems like a sweet deal at first will quickly turn into a no-win situation. Of course, if you go this route, be a good parent afterward and give him some candy anyway. ;)

This notion of "probability distribution" can be a good way to put words to the notion that just because you have X possibilities, doesn't mean they all have to be equally likely... and that saying they are is in fact a very specific thing to say, something you'll only do under specific circumstances (either they actually are equally likely, or you have no information to say they aren't and thus follow the Principle of Insufficient Reason).

And at that point, at future times this comes up (and I expect it would; I find probabilities very hard to keep a mental grasp on) you could answer something like "yes, the odds of winning are 50/50 if there are 2 possibilities if they follow a uniform probability distribution. Do they?".

Funny kid.

He would be right if there was only winning and losing, but, given your numbers, there are actually winning and over 90 million varieties of losing. He's only calculating the probability of losing in one way.

Wildly edited:

This seems like a confusion in semantics. I say just tell him as above. But if he insists that the idea of winning or losing as a single event is useful one, there's a mathematical perspective.

Consider a smaller number, a 1/3 chance lottery with 3 choices. He can enumerate the total number of pairs he has to win against: every choice (3) multiplied by the number of pairs to compare for a choice (2). Then, consider the winning pairs, which is the total number of choices minus one. Don't spoil that it's 2 winning comparisons in 6 possible comparisons.

(n-1)/((n-1)*n) if he wants.

where n is the number of lottery number combinations there are.

From here you can suggest to him exactly what it does mean for there to really be the two possibilities of winning or losing, where winning is a single event and so is losing; and the lengths he'd have to go to in order to find something like the probability of winning the lottery, while defining what that is if he doesn't know.

It seems that in the context of the right answer, from an outside point of view, he's using the principle of insufficient reason; he's simply made a logical error solving the right question. But also within the realm of possibility is that he's found a sort of glitch in the way you could understand the question that leads to some bizarre statements about the real thing.

• Memes aside, a lot depends upon how the lottery is described too. Oct 21, 2020 at 2:28
• (n-1)/((n-1)*n) reduces pretty cleanly to 1/n, might be clearer if it's also stated that way Oct 25, 2020 at 18:36

Counterexamples:

• Tomorrow rains or not. But it doesn't rains half of the days (at least, in my city).
• A treat (candy) may or may not like, but mostly likes.
• You can evaluate correctly or incorrectly a sum, but do not fail in 50% of them.
• ... (choice the ones more near to your children).

You could explain the fact to him like that: Each ticket is either a winning one or a losing one, and there is only one winning ticket. How many tickets there are that you can buy? If the answer is more than two and there is only one winning ticket, then the probability is not 50/50, because if the total number of tickets is 100, then the chance that buying one you get the winning one is 1/100.

A simple proof of that is that only a handful of people had ever won the lottery, not half of the players.

If it was 50/50 the probability of a win, then half of the ticket sold must be a winning one, which is obviously not true, since the hypothesis is there is only one winning ticket.

Also, if he was right, that must apply to each lottery player, and if he has a 50/50 chance of a win, also the other N players would have the same winning probability, and for that to be true either half of the tickets are winning tickets or the total probability is more than 100%, which cannot be.

There are already a lot of good answers here, but here's one way to put it in simple terms that even a younger child should be able to comprehend.

Your son said either you win, or you don't which means he's seeing two possible outcomes and then dividing 100% by 2. You could explain that there are actually many more possible outcomes because there are more ways to lose than ways to win.

So if we take a simple lottery where you pick a number from 1 to 10 and one number is selected as the winning number, instead of you either win, or you don't the situation is either you pick the winning number, or you pick the first losing number, or you pick the second losing number, or you pick the third losing number, and so on.... This illustrates that there are more than two outcomes, even though nine of the ten outcomes are all losing. So instead of dividing 100% by 2, you should divide it by 10 to get a 10% chance of winning.

• I like this answer, and it's probably how I would have explained it to my kids when they were younger. I would say that the outcomes aren't really just "win" or "lose", but more specifically "you lost and you're the proud owner of losing ticket #2556437". Jan 14, 2021 at 18:50

There's a way in which your kid has hit on an important attribute of probability. When he says,

no: either you win, or you don't. That's the probability of the fact to win.

He's describing a component that goes into computing probability: the number of possible outcome states. If instead he had said,

no: either you win, or you don't. That's all the things that can happen.

He would have been completely correct.

The number of possible outcome states are important when he wants to describe a system with more than two outcomes; say, winning, losing, and tieing.

Consider a game where two players have to guess heads or tails when a coin is flipped. There are now 3 game states to model.

1. $$P(\text{"a tie"}) = 1\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$$
2. $$P(\text{"a win"}) = (1\cdot\frac{1}{2}) \vee (1\cdot\frac{1}{2}) - P(\text{"a tie"}) = (1 - (1- \frac{1}{2})(1- \frac{1}{2})) - \frac{1}{4} = \frac{1}{2}$$
3. $$P(\text{"both lose"}) = 1\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$$

What's great about this game is you can easily enumerate all possible outcomes to check your math:

H/T P1 P2 TIE WIN LOSE
H   H  H  1
H   H  T      1
H   T  H      1
H   T  T          1
T   H  H          1
T   H  T      1
T   T  H      1
T   T  T  1


(H/T is the result of the flip, P1 and P2 are the guesses of player 1 and 2, TIE, WIN, and LOSE are marked with a 1 if that game state occured)

Now, play the game with him and ask him to predict how many games he will win by guessing correctly while you guessed wrong (i.e. he wins but doesn't tie).

What's nice about this formulation is that it offers a way for his intuition to be incorrectly bolstered by the $$P(\text{"a win"}) = \frac{1}{2}$$ result, aligning with his preconception that "you either win or you don't".

Once you've played the game, reevaluate the game states. What's missing? Why did he win half as frequently as he thought he would? (Hint: while we have described all possible game states—winning, losing, and tieing—we haven't described all possible player states. What are all the possible states he may find himself in as a player? What are the probabilities of those?)

Through exploring this with him he can come to understand that not all game states are equally likely. The heart of probability is about knowing what can occur (which he already understands) as well as how frequently it does occur (which he's yet to grok).

This game is a great tool for exploring probability because with one coin and two players it's quite simple, but if you add players and coins it can get much more complicated to model how frequently someone will win by guessing the number of heads flipped (eventually leading to the need for game theory to describe dominant strategies and counterstrategies).

Try it with two coins to show him a game where not only all game states have different probabilities, but all player states do as well.

I don't play the lottery. When someone asks why, I resort to the old joke: "I figure my chances of winning are about the same whether I play or not". The point being that there isn't much difference between 0 and 0.000000001

But my brother (who does play the lottery) objected: He said that if you don't play then you can't win, but if you do play then you can win. So he looked at it from a binary point of view: can't win or can win. Like your child it seems.

FWIW, he's a smart guy and understands probability, but figures a few dollars per year are a small price to pay for just the possibility of winning.

Present a deck of playing cards. Fan out to show all the cards. Red you win. Black you lose. fifty fifty right?

But what happens if I do this:

Lay one red card down.

Lay nine black cards on top of it.

Turn over and shuffle.

Still think your odds are fifty fifty?

Standard joke aside. Explain with an info-graphic, and also describe the operation of the lottery properly. YouTube has plenty of these.

It depends how many prizes are offered too, and how lots are sold. Oh, you mean if I only have one lot, what are the chances of winning the major prize?

Does winning have to satisfy a requirement to profit from the win? Buying exactly half the lots should give you a 50% chance of winning the major prize.

If you spend as little as \$100 on Lotto like games, you should have a fair chance of getting something back, finding the 0.50 depends upon the game.

Of course if the lottery is conscription it may genuinely be 50%.

This is why the Monty Hall problem is generally hated, because it is frequently not properly stated.

The fallacy lies in the outcomes being mapped, and not their likelihoods.

Try that 10 coin example with loaded coins, let's say depleted uranium on one face and polystyrene on the the other, maybe don't use the Russian Roulette example.

I know that if I had to write code for two outcomes I might have to spend the same amount of effort upon each.

50/50 is a common default whenever there is a sense of uncertainty. I like the generalisation by data science guy Rob Arthur that humans tend to have only three settings -

• Impossible
• 50/50
• Certain

“I think human beings are just sort of hardwired to do view in only those three categories"

The other aspect to consider is that the advertising and the intensity of the discussion around the lottery may also have contributed to an availability bias. When we know and hear about events, they become familiar and intensity of the discussion cements the event into our memory more intensely. Events that are familiar are assumed to be more common. Humans are biased to the familiar and exciting, so we assume that it is common too. Winning the lottery is far more exciting than the concept of wasting money on the ticket.

Also, probability is a really hard concept for humans to understand and comprehend, especially when it comes to massive numbers such as 90 million. I really like Professor's Spiegelhalter's argument that probabilities are unintuitive. Humans are guided by intuition and emotions for most of our lives, but when it comes to probablities, we really need to manage and ignore our lifelong bias to intuition and emotion, something that is very difficult to do. Especially for your 13 year old.

Learning how to manage these intense feelings is probably a useful part of the discussion and education about probabilities.

"there is a 50% chance to win at the lottery"

In the context of his answer he's absolutely right so instead of these other answers where you challenge his views ... "explain probability to me" ... you need to focus on understanding the context of his answer.

He's right because the chance of you EVALUATING if that ticket is the right ticket is 50/50 ... 1/2 ... one half.

First you guessed a set of numbers ... one out of millions ... then you evaluated if that guess was right ... they are two different events.

The scenario of the guess being right is different than the chance of guessing the right number.

There are millions of other tickets so the chance of your evaluating ALL tickets being a winner or not is the number of tickets multiplied by 1/2.

For example if I said I'm thinking of a number, a digit, between 0 and 9, guess what it is, at first you're 50/50 of being right or wrong ... 1 over 2 ... one guess where you were right or wrong ... right + wrong = 2 possibilities.

But if you're wrong and you choose again the fraction is now 1 over 3 ... right + wrong, the possible results of your first choice, PLUS the wrong guess you already made, for a total of three outcomes for the one guess ... 1/3 ... and if you keep going you get to 1/10 ... the number of possible guesses for all the numbers that can be chosen is 1/10 ... 10%.

So the chance of winning the lottery is figuring out how many possible guesses there can be ... not if the one you chose is right or wrong.

If there are ten number in the lottery it's 1/10 ... 10% ... ten "per cent" is 10/"per cent" = 10/100 (cent =100) = 0.1.

If there are a hundred numbers in the lottery it's 1/100 = 0.01

If there are a thousand numbers in the lottery it's 1/1000 = 0.001

Lets say you can choose six different numbers and each one can be between 0 and 99 - a hundred choices.

The chance of choosing one of those six right is 1/100. The chance of choosing the second of those six right is 1/99 ... because there's one less number to choose from. The chance of choosing the third of those six right is 1/98 ... because there's again one less number to choose from.

The chance to choose all six is 1/100 * 1/99 * 1/98 * 1/97 * 1/96 * 1/95 ...six SETS of choices, each SET of choices being one out of one hundred, one out of 99, etc.

• "He's absolutely right" no, he's not.
– eps
Oct 20, 2020 at 14:36
• "I'm thinking of a number, a digit, between 0 and 9, guess what it is, at first you're 50/50 of being right or wrong" - what you refer here is Principle of insufficient reason already explained in another answer. Also it's explained why it's not true. Oct 20, 2020 at 15:32
• No clue what you're getting at with the "choose a number" thought experiment - the odds of guessing 0-9 correctly in one guess is 1/10, not 1/2. Also, the scenario of a guess being correct is no different from the scenario of guessing correctly. Choosing a specific number 0-9, you have a 1/10 chance of it being correct. Before having chosen a number, you have a 1/10 chance of guessing correctly. Oct 20, 2020 at 18:11
• Not only are you wrong about the probability of getting first guess of a one digit number correct as others have pointed out, you also say that each subsequent guess after a first incorrect guess has a lower possibility of being correct (you say it becomes a 1/3 chance, then on until a 1/10 chance), which is the exact opposite of correct. If the number doesn't change between guesses then the chance of subsequent guesses being correct goes up. The first guess is 1/10, then 1/9 (because there are only 9 options if the first number was wrong), until it is a 100% chance on the tenth guess Oct 20, 2020 at 18:21
• Your thought experiment is making the exact same error the child is - incorrectly distributing the prior probabilities over all possible outcomes. The only difference is with the thought experiment we have to assume a uniform prior, with a fair lottery we know it for certain. But I believe the only logical prior is uniform in the guess a number experiement. Oct 20, 2020 at 22:49