Logic Riddles for Chess Players (6)

by Arne Kaehler
7/27/2020 – The main theme for this logic riddle is the world of Combinatorics. Although it requires some calculations, it is a lot of fun to try to correctly compare probabilities. Of course, we also have the solution for Logic Riddle #5 available!

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The fair fair?


Once a year, a big chess fair takes place on the chessboard, organized by the black king. It consists of some fun rides, a great variety of food stands (mostly fried liver for some reason) and the highlight: four games of chance! If you are older than 18 years, you can join the games and try your luck!

The name of the four games are:

  • The ten white pawns
  • Black or White
  • Numbers of fortune
  • The dice

The Ten White Pawns

There is one box, containing ten white pawns. Each pawn has a different number written underneath it. The numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. Four random pawns are picked from the box to build a four-digit number.

You can bid one ducat to guess the correct number in the correct order. If you guess the number correctly, you will win 4000 ducats.

Black Or White

A spinning wheel which shows the colours black and white equally — a roulette wheel without the zero — has to be turned ten times. If you are able to get the white colour ten times in a row, you will receive 1000 ducats. One attempt costs 1 ducat.

Numbers Of Fortune

Here you can see four boxes, where each box holds ten white pawns with a different number from 0-9, just like in the game "The Ten White Pawns". Now one pawn is chosen from each of the four boxes. If a pawn has a number that had already been picked, it will be exchanged for a different pawn. This continues until four pawns are picked with four different numbers.

Whoever can guess the four numbers correctly, without needing to guess the particular order of the numbers, wins 500 ducats, by paying 2 ducats.

The Dice

Four dice are rolled (numbers 1-6) and if you can guess the correct numbers being rolled, without any particular order, you can win 500 ducats for an entry fee of five ducats.

Which of the games is by far the most profitable, comparing the entry fee and the amount you can win?

The games in table form:

Game Name Entry per game Winning
The Ten White Pawns 1 Ducat 4000 ducats
Black Or White 1 Ducat 1000 ducats
Numbers Of Fortune 2 Ducats 500 ducats
The Dice 5 Ducats 500 ducats

The riddles are pretty interesting to solve and require mathematics. I worked for an Online Casino for a couple of years, and all the games you could play online had a certain "RTP" percentage, which means “return to player”. 

If you do the math correctly, you can find out the RTP % of each game.

Solution to the last riddle: How many colours?

Thank you once again for your participation in the comment section! It shows that the logic riddle section is enjoyable.

Certainly you were all correct. The answer is simply “four colours”. How this might look like is shown in the picture below:


Arne Kaehler, a creative thinker who is passionate about board games in general was born in Hamburg and learned how to play chess at a very young age. Through teaching chess to youth teams and creating chess content on YouTube, Arne was able to extend this passion onto others and has even made an online chess course for anyone who wants to learn how to play this game. Currently, Arne blogs for the English news page of ChessBase and focuses on creating promotional and entertaining articles.
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JoshuaVGreen JoshuaVGreen 8/3/2020 01:17
Since this has been up for about a week, I figure it's safe to post my calculations now.

The Ten White Pawns: Assuming that the high digit is allowed to be 0, there are 10 possibilities for the first digit, then 9 for the next, then 8 for the next, then 7 for the final. That makes 10 * 9 * 8 * 7 = 5040 four-digit numbers that can be constructed, and we should win 1/5040 games. We pay 5040 ducats to play 5040 times, expecting to win 5000 ducats back once, hence we get 5000/5040 = 50/63 of our money back.

Black or White: We have to get 10 white spins in a row. Each spin has a probability 1/2 of being white, hence we have (1/2)^10 = 1/1024 chance of winning. We pay 1024 ducats to play 1024 times, expecting to win 1000 ducats back once, hence we get 1000/1024 = 125/128 of our money back.

Numbers of Fortune: Here we are essentially choosing four distinct numbers from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and their order doesn't matter. The number of such subsets is given by (10 choose 4), where (n choose k) = [n * (n - 1) * (n - 2) * ... * (n - k + 1)]/(k!). In particular, (10 choose 4) = (10 * 9 * 8 * 7)/(4!) = 210 and our probability of winning is 1/210. We pay (2 * 210) ducats to play 210 times, expecting to win 500 ducats back once, hence we get 500/(2 * 210) = 25/21 of our money back.

The Dice: Here we have to consider strategy. If we guess all numbers the same, then there's only way to win -- all dice must show that number. We increase our chances by guessing four distinct numbers. In that case, there are 4! = 24 ways that the dice could agree with our guess as our numbers can appear in any order. There are (6^4) possible dice rolls, but 4! of them win for us, hence our probability of winning is 4!/(6^4) = 1/54. We pay (5 * 54) ducats to play 54 times, expecting to win 500 ducats back once, hence get 500/(5 * 54) = 50/27 of our money back.
charlesthegreat charlesthegreat 7/30/2020 08:40
4 dice are rolled once for 5 ducats. Some numbers can be repeated, so this reduces the RTP (in fact , same as Ten white pawns!)
JoshuaVGreen JoshuaVGreen 7/30/2020 01:15
My comment giving my computed probabilities doesn't seem to have been accepted (yet?). Here's my summary:

"The Dice" provides the best return (provided that the player guesses distinct numbers) and, in fact, yields a positive expected value for the player (as does "Numbers of Fortune"). We should play this one. Of course, if we win too much too quickly, the fair operators may realize that the games have positive expected value and/or think we're cheating. Therefore, we should play some of the losing games every now and then to keep our winnings small enough to stay under the radar.
charlesthegreat charlesthegreat 7/28/2020 08:38
I'd play the numbers of fortune... 119:100 rtp
JoshuaVGreen JoshuaVGreen 7/28/2020 01:14
If we assume that the entry fee is always lost and that a leading digit of 0 is allowed for "The Ten White Pawns," then I get the following RTPs (well, fractions):

The Ten White Pawns: 50/63

Black Or White: 125/128

Numbers Of Fortune: 25/21 -> Positive expected value, though not as good as "The Dice"

The Dice: 50/27 (assuming the player guesses four distinct numbers, lower otherwise) -> Positive expected value and the best return! -- I'll play this game.

Of course, if we win too much too quickly, the fair operators may realize that the games have positive expected value and/or think we're cheating. Therefore, we should play some of the losing ones every now and then to keep our winnings small enough to stay under the radar.
Zurubang2 Zurubang2 7/27/2020 05:55
@lajosarpad the same box.
lajosarpad lajosarpad 7/27/2020 05:49
"If a pawn has a number that had already been picked, it will be exchanged for a different pawn." From which box?
guyhaw guyhaw 7/27/2020 04:08
'The dice' is correct but the return/cost figures is wrong … should be 50-v-27
asoni asoni 7/27/2020 03:25
The dice 4!/6^4 ?
JoshuaVGreen JoshuaVGreen 7/27/2020 01:36
Also, what does

"You can bid one ducat to guess the correct number in the correct order."

mean for The Ten White Pawns? For all the other games, the phrasing makes it clear that the entry fee is always lost. (For example, winning "Black or White" means a net profit of 999 ducats.) However, "bid" suggests a wager, and a wager isn't normally lost on victory.
JoshuaVGreen JoshuaVGreen 7/27/2020 01:05
In The Ten White Pawns, suppose the number is built from the top digit down. Then, if the first pawn drawn is 0, do I really get a four-digit number?