Dice Games and Children Solved

In my previous post, we considered the three-dice game where you bet a dollar on a number coming up. You get your money back plus a dollar for each time your number comes up. If your number does not come up, then you lose. Since the probability of any number coming up once is 1/6, then the probability of winning seems to be simply three times that of one throw so the probability of winning seems to be 0.5

You place your bet and then notice the bright lights and fancy decorations in the casino and wonder how the manage to pay for them when giving such good odds. You also suddenly realize that if 6 dice were thrown, then by your analysis, the probability of winning at least once is unity! This is obviously not right.

The problem is that like most people you have focused on the wrong analysis, plus you have not done the sums correctly. Going on intuition, most people can estimate outcomes reasonably, but no accurately. Our innate ability to predict outcomes has evolved to help us determine if a lion is likely to pounce on us if we take a certain path to the water hole. We do reasonably well at that sort of analysis, which requires a quick response and need not be overly accurate.

However, in estimating probabilities such as the three dice problem, unless one has studied some introductory probability theory, intuition will likely give the wrong answer, and certainly it will not allow you to compute the exact answer.

Paradoxically, the easiest way to compute the exact expected payoff is to forget about yourself and look at the game from the point of view of the operator. Throwing three dice (or spinning the corresponding wheel of fortune) created to possibility of 6*6*6 = 216 possible outcomes. Assuming you have bet on a single number, how many of those combinations will result in the operator paying? You claimed one number. That leaves 5 numbers that do not get a payoff. The same is true for all three dice. Therefore the operator does not have to pay you on the average 125/216 or about 58% of the time. The payoffs you do get work out the same.

If you are ambitious, you can work it out from your side by starting with the three ways to win once, three ways to win twice, and one way to win three times your bet (there is only one way to have all three numbers be the one you choose.). Either way, the house has a bigger edge than advertised for many slot machines.

In the previous post, I said this puzzle is related to the question about children. If I have two children, at least one of whom is a boy, what is the chance the other is a boy? Since I have not told you whether the younger or older one is a boy, the possibilities are: BB, BG, GB, GG; where the one on the left is older. However, the option of GG is eliminated. Of the three choices left, the second child is a girl twice as often as a boy. You can do the same thing with two coins of unequal value. Toss them and report that at least one is heads. What is the probability the other is heads?

Note that if I had said I have two children, and the older one is a boy, the probability changes. Similarly if I toss a quarter and dime and say the dime is heads, the quarter is equally likely to be heads or tails.

If you understand these examples, you are ready for the Monty Hall paradox. If not, then just remember to avoid betting on the three-dice game.

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  • Noud-Roemer Hesselink


  • http://twitter.com/Gallifrey103 Alexander Sigsworth

    Oh, my head…

  • virtual machin

    I play with you as soon as you want and you will lost all you money without any doubt :-)
    ! x 1/216 time I win 4$ => 4×1/216 $
    3 x 5/216 time I win 3$ => 3×15/216$
    3 x 25/216 time I win 2$ => 2×75/216$
    1 x 125/216 time I lost 1$ => -1×125/216$

    i have a expected gain equal to 4+45+150-125)/216 =74/216 = 0,34 $
    ie if I play with you 10000 times I gains approximatively 10000*0.34=3400$.

    if you are not convince just run the following little perl script:

    sub gainjeux {
    my $choix=shift;
    for ($i=0,$occurence=0; $i$d[1]) {
    $temp=$d[0]; $d[0]=$d[1]; $d[1]=$temp;
    if ($d[1]>$d[2]) {
    $temp=$d[1]; $d[1]=$d[2]; $d[2]=$temp;
    if ($d[0]>$d[1]) {
    $temp=$d[0]; $d[0]=$d[1]; $d[1]=$temp;
    if ($occurence==0) {
    } else {
    printf “@d “.”=> %4i :: total=%4in”, $gain, $total;
    return $gain;

    print “I bet in number 1nn”;
    for ($n=0;$n<$nbcoups;++$n) {