# Positive Expectancy with Random Chance

A conversation came up at work about positive expectancy. I am having difficulty getting the same answer as the guys so I am throwing the question out to you folks... Any help is appreciated.

A game costs €50 to play. If you win you get €100. You get 52 chances to pick the Ace of hearts from a deck of cards. Every time you pick a card it is then returned to the deck and the deck is shuffled again. You pay the €50 once and you then have up to and including 52 tries to get the Ace of Hearts. Is there a positive expectancy on this game?

I worked it that you have a 1/52 chance of winning every time you pick a card. While you do get to pick 52 times your chance of winning is still 1/52 each time you play.

( 1/52 * 100 ) - ( 51/52 * 50) = (200 - 4750) {I used rounded 2% for 1/52 and rounded 98% for 51/52}

What do you folks think?

• I'm not sure the rules are clear. You can draw up to $52$ times without paying again? – lulu Apr 26 '16 at 17:13
• Edited for clarity. You pay the €50 once and you then have up to and including 52 tries to get the Ace of Hearts. – Ciaran O'S Apr 26 '16 at 17:15
• Fair enough. I can't follow your calculation. There's a $\frac {51}{52}$ probability of missing on any given trial, so there is a $\left( \frac {51}{52}\right)^{52}\sim .3643$ probability of missing on all $52$ trials. Thus the expected payout is about $63.57$ which is, indeed, greater than $50$. – lulu Apr 26 '16 at 17:18
• I have copy and pasted this from below: That is exactly the argument I have heard, but I am unable to work out the maths to get a percentage for (1/52)^52 which would be how may times you win. It should be ≈0.64 – Ciaran O'S Apr 26 '16 at 17:20
• $\left(\frac 1{52}\right)^{52}$ represents the probability of getting the desired Ace $52$ times in a row. That is effectively $0$ and has nothing to do with the question. – lulu Apr 26 '16 at 17:22

You fail each time with probability $51/52$. Since you return the card to the deck, the draws are independent so you fail all and lose the game with probability $(51/52)^{52} \approx 0.36$. Since you win with probability greater than one half, the odds are good for you.

• That is exactly the argument I have heard, but I am unable to work out the maths to get a percentage for (1/52)^52 which would be how may times you win. It should be ≈0.64 – Ciaran O'S Apr 26 '16 at 17:17
• $(1/52)^{52}$ is not the "number of times you win", it's the probability that you pick the Ace of hearts every time. What you care about is picking the Ace at least once. The $0.64$ is $1-0.36$. – Ethan Bolker Apr 26 '16 at 17:21
• Thank you Ethan. I understand now. Looked at it all wrong. Cheers – Ciaran O'S Apr 26 '16 at 17:26

To clarify for OP. The expected return of the game is

$E[R] = 100 \cdot P(\text{win}) - 50$

Where the probability that you lose is

$P(\text{lose}) = (51/52)^{52}$

and the probability that you win is

$P(\text{win}) = 1 - P(\text{lose}) = 1 - (51/52)^{52}$

Hence

$E[R] = 100 \cdot P(\text{win}) - 50$

$E[R] = 100 \cdot (1 - (51/52)^{52}) - 50 \approx \$13.7\$

So this is a fantastic game :)