# Winning the lottery

several times before I understood, though Cecil’s second and third explanations finally made it through my skull. I now accept the counter-intuitive results regarding doors, playing cards, sex of children, etc.

But I’m still unclear as to why the lottery example is different. As I understand it (which is to say, shakily), in all the other examples the odds of your original guess being right do not change, but the odds of unchosen options appear to increase as incorrect options are deleted. Why does this not apply in the lottery example?

Please speak slowly and use small words. Thanks.

I’ll give it a go. I promise the words will be simple (it’s just the maths that may be dodgy!).

When Monty opens a door, the odds change because he KNOWS where the prize is.

When 9,998 lottery entrants scratch their cards, they DON’T know beforehand that they’re losers. This makes the difference (honest).

Now it is incredibly unlikely that only you and one other player are left after 9,998 people try to win. This is why you feel your odds are better - something unlikely has just happened. So try this:

Assume there are 10,000 lotteries (I know, I know, just humour me). You have a ticket in each game. In 9,998 of them, one of the early scratchers claims a win. In the other two games, you and the other player are left looking at each other. Your chances are 50% in each remaining game.
(It may help to imagine that if each player was given one winning number in each lottery, then you would now win one game, and so would the other player).

Now I hope that made sense, because here is the sting in the tail. If the organiser KNOWS which is the winning ticket, and he removes 9,998 losers because he KNOWS they are, then you should definitely swap for the last ticket!

Why doesn’t the sun come out at night when the light would be more useful? (Pratchett)

The lottery example was different because it did not include somebody with prior knowledge. Chance was not involved in turning cards or doors in the other cases- in the lottery example every time a card was scratched it involved another operation of the laws of chance. It also was subtely different because unlike the cards or doors, the lottery case was not looking at all possible alternatives, but only one other, equally slim outcome.

Consider how the 3 examples operate-

Doors: 1 out of 3 times you will pick the right door and the emcee will open 1 of the wrong doors; 2 out of 3 times you will pick a wrong door and the emcee will open the 1 remaining wrong door, leaving the correct one shut

Cards: 1 out of 52 times you will pick the right card and the dealer will turn over 50 of the 51 wrong ones; 51 out of 52 times you will pick the wrong card and the dealer will turn over the 50 remaining wrong cards, leaving the correct one concealed

Lottery: 9998 times out of 10,000 the winning ticket will be one of the cards already scratched off; 1 out of 10,000 times the winning ticket will be the unscratched one you don’t have;1 out of 10,000 times you will have the winner

In the first two examples you are comparing your results to all possible outcomes; in the lottery example you are only comparing your 1 in 10,000 result to one other 1 in 10,000 result, not to the whole set of possible results.

Thanks for the response, but I’m still having trouble wrapping my brain around this. Consider the following lengthy clip from Cecil’s article:

======================================

Cecil is well aware the answer to the lottery question is “no”–if there are only two tickets left, they have equal odds of being the winner. The difference between this and the Monty Hall question is that we’re assuming Monty knows where the prize is, and uses that information to select a non-prize door to open; whereas in the lottery example the fact that the first 9,998 tickets are losers is a matter of chance. I put the question at the end of a line of dissimilar questions as a goof–not very sporting, but old habits die hard.

The answers given to Jordan Drachman’s questions–2 in 3 in both cases–were correct. The odds of the original card being an ace were 1 in 2 before it was placed in the hat. We are now trying to determine what card was actually chosen based on subsequent events. Here are the possibilities:

(1) The original card in the hat was an ace. You threw in an ace and then picked the original ace.

(2) The original card in the hat was an ace. You threw in an ace and then picked the second ace.

(3) The original card was a king; you threw in an ace. You then picked the ace.

In 2 of 3 cases, the original card was an ace. QED.

The second question is much the same. The possible gender combinations for two children are:

(1) Child A is female and Child B is male.

(2) Child A is female and Child B is female.

(3) Child A is male and Child B is female.

(4) Child A is male and Child B is male.

We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child’s sibling is male. QED.

=======================================

Now, in the Monty Hall scenario, Monty knows which doors give false answers, and can open them selectively. But in the above two scenarios, there is no omniscient “Monty” character to finagle the outcome, yet the odds are the same as with a Monty.

Assume a lottery with only three tickets (absurd, but then, what isn’t in higher mathematics?). The odds of my ticket winning are 1/3. The odds that the winner is one of the other tickets is 2/3. One ticket is scratched. Maybe it’s the winner, which makes further calculation moot. But, if it happens to be the loser, doesn’t that now make the other ticket a 2/3 favorite?

I’ll stop now – may brain is beginning to hurt…

nebuli –

You posted while I was still writing. OK, that makes a little more sense. Thanks.

– Beruang

Off topic but addressed in the Article:

The thing about the sexes of the children is flat out wrong. If you have two children, there are only three combinations: both girls, both boys or one girl and one boy. Cecil thinks there is a fourth combination: one boy and one girl. How you can distinguish this from the third is beyond my comprehension. In any event, the sex of one child has no effect on the sex of another, and knowing the sex of one child gives you no statistical advantage for knowing the sex of the other.

Wow! This is really going to change how I play craps, now that I know that the dice can come up 4-3 but not 3-4. Off to Atlantic City, I am!

Livin’ on Tums, Vitamin E and Rogaine

The odds are computed for the set of both dice, not for each individual die becoming a 3 or 4 (since the result you want is that one die is a 3 and one is a 4). If you were playing craps any other way, you’d probably need to use different colored dice and pretend that the color of the die somehow affected what number it would roll. Their identity is not relevant.

Simple… #1- both girls; #2- both boys; #3- older one a girl, younger one a boy; and #4- older one one a boy, younger one a girl. You can think of it as only the 3 combinations you mention, but then you can’t ignore the fact that one combination-the third- is twice as large as either of the first two. It just makes the math simpler if you break it down into the 4 equal sized groups I cited.

It doesn’t matter how old the children are, or what their names are, or what color their hair is. Identity is not relevant. The total set consists of only 3 possibilities for the relevant characteristic - which is sex. In most everything, we use identity as a characteristic to build a probability matrix. In the case of unknown children however, identity is not a factor. If there is sufficient interest I will prove this.

I will write a computer program that will compute a set of 100 pairs of children, with random sex assignments. To ensure that it is random, the program will count the number of each sex it assigns and throw the set out and recompute if the total exceeds 55% either way. Thus we will have a sampling that the last census indicates is about accurate for humans.

The user will be shown the sex of one child for each pair, and will have to guess the second. Use Cecil’s rules for this if you wish. Anyone who gets better than 55% correct wins a rubber chicken. Anyone seriously interested?

Anyone who has thought through this still think it needs to be done?

As I said in my previous post, you can do the problem by considering only three combinations- if you don’t ignore the fact that the 3rd option, one of each sex, is twice as large as either of the other two combinations. So, if your program produces a sample close to 50 families with one girl and one boy, 25 families with two girls and 25 families with two boys it should work fine. The only reason for breaking the mixed group into two was that it tends to be easier to appreciate odds when one is dealing with options of equal likelihood.

I’ve always wanted a rubber chicken, Cooper. Post the source code of your program so folks can make sure you’re taking out all the families with two sons (the original problem told us that at least one of the kids is a girl, but not which one), and I’ll post the patch to make the program play by Cecil’s rules.

I might have something by tomorrow, but don’t count on it. Probably be next week things are mostly hectic. Incidentally, the only IDE I have is visual basic so you will be stuck with that. If you can’t compile perhaps a third party could so we know I didn’t put something different in the EXE than what the source says.

Nebuli:
How is the set twice as large? As I said, identity is irrelevant.

Yoop, I didn’t parse your whole message.

You want 100 examples where you know the first child is a girl? That doesn’t make any sense. There will be a 50% chance of the second child being male since the program will be totally random (unless you believe uncle cecil, in which case it somehow becomes 2/3s that the child is male). How about you just flip a coin 100 times and admit Cecil made a mistake?

What I proposed is that you get 100 sets of children, and the program tells you what one of them is in each case. The one it tells you could be a boy or girl (50% chance) and the second one will be a boy or girl (50% chance). This way, you basically have 100 chances of doing the gist of the original question, which is guess an unknown child’s sex knowing only the sex of their sibling.

Nebuli: Nevermind I see your point. I agree the result of the program should be 50% mixed, 25% all boys, 25% all girls. This is what will happen as long as it is truly random. It will then reveal the sex of the first child from each family, and you’ll guess the second. Everyone in agreement?

Oh and sorry to be a spazz and keep posting like this, but its just occurred to me the best way to do this would be write it in qbasic since all windows people at least have a qbasic. This way you can just copy, paste and run it and you’ll know it works like everyone sees it does.

Last post, I swear!

Ok I know I promised I wouldn’t post again, but I finally figured out how I was wrong.

Thanks nebuli! Really, I understood I was wrong after I got the 50% mixed 25% boy/boy girl/girl families. My problem was figuring out how eveyrone else was wrong to. So off to the gym I went!

On the rotary calf machine it hit me! I almost skipped the cross-trainer and ran home I was so excited to post it. Ok here, goes:

We have two children, A and B (yes identity is relevant, everything has identity). As posted above, the possible result set is:

``````A   B
``````
1. F F
2. M M
3. F M
4. M F

Now, we are told that one of the children is a Female. So we know that 2. is out. That leaves:

``````A   B
``````
1. F F
2. F M
3. M F

Odds are 2 out of 3 the child is male, right! Wrong (duh). You know that either A or B is female. If you pretend the female one is child A, then you eliminate option 4. Otherwise you eliminate option 3. No matter what, you have to eliminate one of them.

It needn’t be said but I will anyway: its irrelevant that you don’t know which child you are told the sex for - you are told the sex for one of the children and so you must always eliminate two of the options.

We still need the program?

Cheers all.

Cooper, you said

I think I’d have to disagree -but I’m not as positive this time need to think about it some more. But if I’m understanding the original problem correctly we can’t eliminate any of the 3 options from consideration based on the info given. As I see it, any one of the 3 options is equally likely to occur, so

• 1/3 of the time it will be option 1, and the other child will also be a girl

-1/3 of the time it will be option 3 and the other child will be a boy

-1/3 of the time it will be option 4 and the other child will be a boy.

Looks like 2 out of 3 to me, but I’ll work on it. Thanks for the offer of the program, but I think simple drawing slips of paper, cards, etc. will do fine for something as simple as this (that’s what I used to reach the, to me, astounding conclusion that Marilyn vos Savant hadn’t totally flipped out with her 3 door problem).

BTW, a slight tangent for whatever it’s worth- I was doodling around on some scrap paper, and if my math is right, your original assumption of a 33% gg, 33% bb, 33% mixed breakdown would occur if one third of the families had identical twins.

Cooper,

you posted 'We have two children, A and B (yes identity is relevant, everything has identity). As posted above, the possible result set is:

A B

1. F F
2. M M
3. F M
4. M F

Now, we are told that one of the children is a Female. So we know that 2. is out. That leaves:

A B

1. F F
2. F M
3. M F

Odds are 2 out of 3 the child is male, right! Wrong (duh). You know that either A or B is female. If you pretend the female one is child A, then you eliminate option 4. Otherwise you eliminate option 3. No matter what, you have to eliminate one of them.

It needn’t be said but I will anyway: its irrelevant that you don’t know which child you are told the sex for - you are told the sex for one of the children and so you must always eliminate two of the options.’

As Cecil said, the odds depend upon the question. But assuming you’re just told that one child is female, it is indeed 2/3 that the other one is male.

I’m afraid your error above occurs when you say (unscientifically) ‘If you pretend the female one is child A,…’.
It’s difficult to break thru your misunderstanding but try this:

Probability changes when facts are KNOWN. If a child (that’s EITHER child) is female, then it’s more likely that the other is male.
If you KNOW (not pretend) the FIRST child is female, then indeed one possibility drops out. But you don’t know anything about the female child…

Why doesn’t the sun come out at night when the light would be more useful? (Pratchett)

No!

You do not have to know the children’s names in order to have fair odds at guessing their sex. You know that EITHER A or B is the female, so you cannot save BOTH options 3 and 4.