Expected Value, Variance, and Bankroll Risk in a Card Betting Game

# How would you approach this positive EV and high variance betting problem?


# How would you approach this positive EV and high variance betting problem?












My friend was asked this question and I’m curious as to how people would play. There are 15 cards face down on a table. You can draw any number, n, of them at random. You only see the cards you have drawn after drawing all n cards. 14 of the cards will double your money, and one of the cards will multiply your money by 2^-14. Lets call the double cards ‘good cards’ and the 2-^14 card the ‘bad card’. (I.e. suppose you start with $1 and you draw 5 cards, if all 5 of them are good, then your bankroll goes from 1 to 32. However, if one of those 5 is the bad card, then you have 4 good cards and one bad card and thus your bank roll goes from 1 to 2^-10). The question is how many cards do you want to draw if: a) you can play many times, b) you can play just once. And how the answers would change if you were playing with all of your personal wealth or playing with a weekly trading allowance that will get topped up at some point?

In terms of expected gain, it will be 0 if you draw 0 cards and 0 if you draw all 15 cards. In between, the expected value is always positive, with the peak being at 14. In terms of standard deviation, this would increase from 0 to 14. ‘Sharpe’ would be maximised at around 4.

What would people’s approach be in deciding how many cards to draw. Even though drawing 14 cards has very high variance, most of this variance is to the upside. When you draw 14 cards your loss is limited to 1/2 as the worst case scenario is you draw 13 doubles and one 2^-14. And you have a 1/15 chance of 2^14 multiplying your money. Even though drawing 4 cards has less variance and higher sharpe, why would you do this over drawing 11 cards? Because when you draw 4 cards your worst case is that you have a 4/15 probability of drawing three doubles and one 2^-14 in which case you 2^-11 your wealth. This seems much worse than just taking the chance on drawing 14 cards,but my friend argues it has a better sharpe so is better.

Thoughts?

## Answer by THATS MY QUANT MY QUANTITATIVE (score 3)

https://quant.stackexchange.com/a/77597

You always choose more cards because the game is biased (unless I misunderstood your question). The issue lies in the fact that you keep the winnings from the $n-1$ cards. You can see this from the EV of only choosing 1 card vs choosing 14 cards.

$$EV(n=1) = 2\left(\frac{14}{15} \right)+2^{-14}\left(\frac{1}{15}\right)\approx 1.87$$ $$EV(n=14) = 2^{14}\left(\frac{1}{15} \right) + 2^{13}\cdot 2^{-14}\left(\frac{14}{15}\right)\approx 1092.73$$

For every $k$ card, the potential loss amount decreases by $2^{k-1}$:

$$EV(n=k) = 2^{k}\left(\frac{15-k}{15} \right) + 2^{k-1}\cdot 2^{-14}\left(\frac{k}{15}\right)$$

The standard deviation is irrelevant because you can never bust (there's no stopping time, $\tau$). So the more games you play with maximum cards, the more money you will win. Sharpe ratio only matters if you can go bust.

## Answer by Pranav Poduval (score 0)

https://quant.stackexchange.com/a/77605

The whole problem reminds me of Poker, especially when I was working on building GTO bots

The question is how many cards do you want to draw if: a) you can play many times, b) you can play just once

The expectation is the same and you cannot go bust so the only right answer is one with highest expectations. Only when you can go broke like in Poker bad hands does variance matter (even then the right answer is the manage bank roll rather than chose sub optimal low variance actions)

So 14 it is