My daughter's birthday is coming up. Here's a fun party game we can play to celebrate. (Bring your wallet!)

  • I have a pair of 6-sided dice, and a standard deck of 52 cards.
  • You place your stake (say €1) on the table and get to pick either my age, or my daughter's age. Both of these numbers are known to both of us. I will play the remaining one.
  • I choose to either roll the dice, or pick two random cards from the deck.
  • We multiply the two numbers. (Aces are counted as 1.) If the product is the number you picked, you win back three times your stake (i.e. your €1 plus €2 profit). If the product is the number I have, you lose your stake. If it's neither, you get your stake back.
  • Note that both numbers are possible to make with both dice and cards.

If you want, we can also play this game after my daughter's upcoming birthday.

I cannot lose in this game in the long run. How old am I? How old will my daughter be this year?

  • 1
    $\begingroup$ @Jafe, can your daughters age still be made up from both dice and cards after her birthday? $\endgroup$
    – Gamora
    Jul 1, 2019 at 15:33
  • $\begingroup$ @Bee Yeah, all bullet points concerning the game still hold after her birthday. $\endgroup$
    – Jafe
    Jul 1, 2019 at 15:40

1 Answer 1


I think that the solution is

You are $36$ and your daughter is $2$ (she will turn $3$ at her next birthday).


If the player picks your age, roll the dice.
If the player picks your daughter's age, choose the cards.
In both cases you are, at least, twice as likely to win as your opponent which means that your earnings will be non-negative in the long run (the expectation value for your earnings is non-negative).


Using the dice, your age occurs with probability $\frac{1}{36}$
and the daughter's age occurs with probability $\frac{1}{18}$ (in both cases).
Using the deck of cards the parent's age occurs with probability $4 \times \frac{1}{13} \times \frac{4}{51} + \frac{1}{13} \times \frac{3}{51} = \frac{19}{663}$.
while the daughter's age occurs with probability $\frac{2}{13} \times \frac{4}{51} = \frac{8}{663}$ (in both cases).

Why I chose these numbers

Using just the dice, we can list all the possible products. Given that the daughter's age must be the first in two consecutive entries in this list, this leaves the possibilities for the daughter's current age as being either $1,2,3,4,5,8,9,15$ or $24$.
For all of these values, at least one of the two consecutive ages will arise as the product of the two dice in at least two of the $36$ possible outcomes. This means, to enact a simple strategy where the player's choice directly informs the parent's choice, the parent's age must occur with probability $ \leq \frac{1}{36}$ or $\geq \frac{1}{9}$. This means the parent's age must be one of $1,6,9,12,16,25,36$.
When analysing the cards, this probability reasoning must flip. Heuristically, we can get an idea for what happens by analysing the number of products when choosing two numbers from the range $1$ to $13$ and multiplying (the cards probability is subtly different but this will still give a good idea of how to proceed).
In this instance, the number of products is essentially unchanged for the values $1,2,3,4,5,6,8,9,12,15,25$ while the number of products increases for the values $16, 24$ and $36$. Hence, one of the ages must be from this group of three.
If the parent's age is $16$, this needs to be half as likely as the daughters age in the dice case but, in the case of the cards the probability of $16$ is $2 \times \frac{1}{13} \times \frac{4}{51} + \frac{1}{13} \times \frac{3}{51} = \frac{11}{663}$ while all the possibilities for the daughter's ages contain at least one value with probability $\frac{8}{663}$ so we cannot guarantee a win here.
If the daughter's age is $24$, and the parent is older than the daughter then the daughter's age is more likely than the parent's age for both the dice and the cards and the player will always pick it.

Hence the parent's age must be $36$. The daughter's age must then be, at least, twice as likely as the parent's age for the dice, in both years. This restricts the possibilities for the daughter's current age to be $2,3,4$ or $5$.
Returning to the cards, the probability for getting $36$ is at least twice as likely as getting a $2, 3$ or a $5$ but less than twice as likely as getting a $4$ or a $6$. Hence, to make it work for both years, the daughter must be currently $2$ (turning $3$ at the next birthday).

  • 2
    $\begingroup$ This was my conclusion too, but I believe it fails the requirement that the dad "cannot lose in this game in the long run." In case the player always chooses the dad's age, then the dad's gain will follow a randow walk with a 50% proba that it tends towards -Infinite in the long run... Just like if we play indefinitely head or tail with a fair coin, I have a 50% proba of losing all my fortune. en.wikipedia.org/wiki/Random_walk $\endgroup$
    – Evargalo
    Jul 1, 2019 at 15:56
  • 1
    $\begingroup$ @Evargalo My interpretation was that the expectation value was non-negative. I'm not sure that there is a solution which makes the expectation value strictly positive although it's possible that there may be a different way to approach it. $\endgroup$
    – hexomino
    Jul 1, 2019 at 15:58
  • $\begingroup$ I believe there is no solution which makes the expectation value strictly positive. I followed the same reasonning as you, and I think you got the only solution with a non-negative expectation value in the worst case. $\endgroup$
    – Evargalo
    Jul 1, 2019 at 16:01
  • 1
    $\begingroup$ Here I was thinking we just needed better than even odds, not thinking about the actual stakes. This also explains what Evargalo was getting at, and why I don't gamble. $\endgroup$
    – Skosh
    Jul 1, 2019 at 16:59
  • 1
    $\begingroup$ This is the intended answer. I admit the wording could have been better -- maybe "can't expect to lose" would make more sense considering @Evargalo's point, which hadn't occurred to me at all. $\endgroup$
    – Jafe
    Jul 1, 2019 at 17:53

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