First, my short survey into sleep is still active and it would be great if you could take part. The link is here. Thank you.

On Friday I posted this puzzle…I want to play a game with you. We have three unbiased dice, and each of the two of us have a target number. My target number is 9 and your number is 10. On each round of the game we throw all three dice and add up the numbers. If I get my target number I get 1 point, and if you get your number you get a point. The winner is the person with the most number of points after 20 rounds. There are 6 ways of getting 9 (1,2,6: 1,3, 5: 1,4,4: 2,2,5: 2,3, 4: 3,3,3) and 6 ways of getting 10 (1,3,6: 1,4,5: 2,2,6: 2,3,5: 2,4,4,: 3,3,4) and so it seems to me that it is a fair game. Want to play?

If you have not tried to solve it, have a go now. For everyone else, the answer is after the break.

No, it is not at all fair because there are many more ways for the number 10 to come up. For example, the combination 3, 3, 3 can only come up one way (with all three of the dice showing a 3), whereas 2,4, 4 can come up three ways. As a result, a total of 10 is far more likely than a total of 9. So yes, you should have taken the bet.

Any other answers or ways of explaining it?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for theKindle (UK here and USA here) and on the iBookstore (UK here in the USAhere). You can try 101 of the puzzles for free here.

1. My way of explaining it:
“Sure, I’ll play. Since we both agree it’s fair, I’m sure you won’t mind swapping numbers, will you? 9 is my lucky number this month.”

1. Anders says:

You wanted to swap even though the initial bet would have been in your favour?

2. Oops.

2. Anders says:

typo I think. “There are many more ways for the number 9 to come up” should read “…for the number 10 to come up”

3. The answer should read: “… there are more ways for the number 10 to come up.” Not “many” more ways — just two more ways: 27 versus 25.

4. I agree with Anders. The number 10 has more chances to come up, and Richard is therefore giving his competitor an advantage. Talk about altruism. Maybe I should send him my bank account number. It could use some altruism right about now!

5. It’s slightly overstating it to say that “As a result, a total of 10 is far more likely than a total of 9”. Out of the 216 possible permutations (6x6x6), 25 of them sum to 9 whilst there are 27 that produce 10.

1. How long is a piece of string? How much is many?

2. “How long is a piece of string? How much is many?” – Find me a situation where 2 is many and I’ll eat my hat (Side note – if I have put a cake on my head does that count as a hat?)

3. LOL. I seem to remember there is a tribe in South America whose members declare anything greater than one as “many”. So there you are. Let’s celebrate the eating of the hat! Will it be with pepper sauce or with shiitake mushrooms?

I’d have to agree with the hat. After all, there was a time women had bird cages as part of their hats (I think ^_^).

4. It’s only 8% more, and that doesn’t count as “many” in any normal sense of the word. In any event,why use imprecise terms like “many” when the answer can be easily expressed as a number?

5. Make no mistake, I agree with you. Even more so because of the use of “far more likely”. Nevertheless, in real life, a lot is contextual, and depending on the context, two can indeed be a lot.

6. Dave says:

If you’re being slapped about the head with a garden spade, two is plenty of whacks. It won’t feel like “not many” at all.

7. But you aren’t being slapped round the head by a spade. I suggest it’s good for your power of reasoning not to introduce foolish, inappropriate examples. However, if you do want to continue with this analogy, I would suggest that the difference between 27 and 25 slaps round the head with a garden spade is going to make a minimal difference to your entry into the mortuary.

6. wraakh says:

This is gonna turn into one of those things we don’t agree on isn’t it?

1. No, that’s WRONG. We ARE going to agree.

7. dharmaruci says:

obviously 10 will come up more often. the higher the target number the more time the dice have to reach it. it is like betting on two runners one of whom has 9 secs to complete a race while the other has 10.

1. So do you think 19 will come up more often than 18?

2. The distribution curve is symmetrical. It peaks at 10 & 11 (27 permutations each) and declines on each side to just 1 permutation for 3 & 18 respectively. So 9 & 12 are equally likely.

3. dharmaruci says:

obviously we’d need more than three dice throws for larger numbers. but the principle as stated is sound.

4. No the principle is not right at all. I suggest you get three dice and see just how many times the sum of 18 comes up compared with a sum of 10.

8. Kristian says:

(1+2+3+4+5+6)/6 = 3.5 per dice
3*3.5 = 10.5, which is closer to 10 than 9.

I also counted 27 ways to get 10 and only 25 ways to get 9, but I think sums and averages is a funnier approach.

1. Jerry says:

Very elegant!

2. martinlong1978 says:

It breaks down when you have 1 die. 3.5 is the average throw, but not the most likely. As it happens, when you have more than one dice, the average coincides with the peak distribution, but that doesn’t necessarily have to be the case.

9. It seems that Richard has been changing his message, but not completely:

No, it is not at all fair because there are many more ways for the number 9 to come up. For example, the combination 3, 3, 3 can only come up one way (with all three of the dice showing a 3), whereas 2,4, 4 can come up three ways. As a result, a total of 10 is far more likely than a total of 9. So yes, you should have taken the bet.

If there are *many* more ways for the number *9* to come up, *10* should be *less* likely to come up. However, 10 is actually *more* likely to come up, not less, and that is because there are more ways for the number *10* to come up than for the number *9* to come up.

10. Ian says:

Oh, this is a clever one (well, they all are). I’ve not had chance to pop in and respond during the weekend, but it took me a while to realise anyway: had to step back and let what I was look at sort of mull in the back of my mind till I realised. Thanks for another excellent brain workout. Hope the weekend went well and the survey’s proving fascinating too 🙂

11. Alistair says:

I rapidly realised what the issue was and that the 3+3+3=9 combination was probably the one that would make 9 less likely than 10.

Tried it with my daughter over the weekend and she got 4 10’s while I only got 2 9’s.

Then I worked out all the options and it is the 3+3+3 that is the key.

1. Jerry says:

Bravo! That was what tipped me off as well.

12. I agree that the phrasing of Richard’s solution leaves clarity lacking.

Also there’s the prospect that someone of Richard’s calibre offering a bet on a game that would leave him disadvantaged. “Ah ha!”, thinks the mark (I don’t know why we call him Mark, are people called Mark particularly gullible?) “Richard would not disadvantage himself, I will take the better number and wrest the advantage from him.” but Richard, a psychomologist, has seen this coming and has _deliberately_ picked the worse number to begin with.

The sneaky deviant.

13. Lazy T says:

the three threes gave it away, the other combinations could all be cancelled out, no need for counting, simples

1. Dave says:

^I agree with this person, that’s how I did it.

14. SimonP says:

Are we missing the bigger picture here? Are 20 rounds really enough for this small bias (27:25) to become significant? For the maths experts out there, how many rounds need to be played before you get a better than 90% chance of winning?

1. Charith says:

It’s as good as any other, really; there’s no reason why 9 and 10 ever have to appear at all in any sample size short of infinite.

15. Charith says:

Oh. How dull. I knew 3 dice with 1-to-6 would favour 10, but discounted “Yes, I would” as there was no puzzle / that would be boring.

I assumed you would have a trick up your sleeve to make you win with 9, the simplest way I thought of that still keeps the dice fair is two 1-to-6 and a third with two 1s, two 2s, and two 3s; the third die is still fair; it has no bias on which of the three numbers it shows, but by looking at the numbers required to make 9 and 10 it’s clear that 1, 2, and 3 appear most often in 9 so more 1s, 2s, and 3s shifts the odds in favour of 9.

1. Alex says:

Sounds like you didn’t get it right then. Best luck next Friday!

16. I did your sleep survey, but it’s hard to remember how my sleep’s been for the past 6 months. I think the way we view our ‘normal’ sleep patterns has a lot to do with how we’ve slept in the last few nights. I’ve noticed that people can get into the way of thinking they sleep badly as a rule if they have one or two bad nights and they forget about all the nights when they’ve had good sleeps. I know this to be true because I’m one such person. It was very interesting though, are you giving a lecture at the Science Festival?

17. ralphatserendib says:

Sure, generalising, there are more ways for 10 to come up than 9 but a brute force attack on the pc shows that the advantage does not become 10’s until 50 turns are taken instead of 20.

18. I solved it quickly with this simple method that works for any number of (standard) dice:

Find the average of the Maximum Dice Total and the Minimum Dice Total. In this case 18 and 3 respectively, giving us an average of (18+3)/2=10.5

Numbers closest to this result (10.5) are more likely to be rolled; and less likely as you get further away from it.

So in this case 10 and 11 are the most likely numbers to roll.

1. dharmaruci says:

i am clearly finding other people’s idea of mathematics hard to follow (see other posts above) and also with this one. if i have one dice then 4 will come up most often (1+6/2=4)? are you sure?

2. Calgacus says:

Hey Dharmaruci, (1+6)/2=3.5 !

If your gonna pick holes in other peoples arguements, at least get you facts right. 🙂

3. This simple method is valid only when more than one die is thrown. With just one die, clearly all totals are equally likely. 🙂

4. Obviously it doesn’t work for one die. Sorry for leaving that out dharmaruci. I forgot to make my post pedant-proof… 😉

5. [I retract that last quip. I just read dharmaruci’s other posts and realised he probably wasn’t poking fun at me, but is genuinely confused about the maths. Sorry.]

Dharmaruci: As Nick says, my method doesn’t work for one die. Only 2 or more standard dice.

19. Robert Leyland says:

I you roll a 6 sided die a bunch of times the average value of the roll is 3.5, so if you roll 3 dice, the average roll is 10.5. Since 10 is closer to 0.5 than 9, 10 will be the total more often.

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