[Math] the probability of winning Casino War

gamblingprobability

For a class assignment I need to calculate the probability of winning the game called Casino War. I've never done anything with chance before (only had algebra and calculus-like stuff). Basically there are 6 decks of cards. If you get a higher card than the dealer, you win. If you and the dealer have the same value, you can either get half your money back or continue playing where the player and the dealer get a new card. A simple 1 minute video explanation can be found here: https://www.youtube.com/watch?v=uimPY3I1aw4

I need to calculate the probability of winning. On a website (in Dutch unfortunately) I found that the probability of winning should be 0.502735, but there's no explanation why. My teacher says this is indeed correct but due to the holidays I can't ask him why anymore. This is my calculation so far:

$$\frac{6*52/2-1}{6*52-1} + (\frac{6*4-1}{6*52-1} * \frac{6*4-3}{6*52-3}) \approx 0.503418 \ne 0.502735$$

Explanation of calculation

Here's my guess; There are 6 decks (hence the * 6 every time). The probability of getting higher than the opponent equals half of the decks minus the first picked card. This is the first part of the equation.
Then I add the probability of getting the same value as the opponent and the probability of winning after that. There are two extra cards removed from the deck so in the calculation I add the -3 to compensate for that.

My calculation adds up to 0.503418 while the answer should be 0.502735. I don't see what part of my equation isn't right. I hope someone can help!

Edit

Alright, keeping the chance of getting the same card in mind, I came up with the following formula:

$$\frac{6*52/2-4}{6*52-4} + (\frac{6*4-1}{6*52-1} * (\frac{6*4-2}{6*52-1-2} + \frac{6*52/2-2}{6*52-1-2} )) \approx 0.53563 \ne 0.502735$$
Breaking this up in 4 segments:

  1. the probability of getting a higher card the first time

  2. the probability of getting the same value, which continues to getting two new cards;

  3. the probability of getting the same value another time (keeping in mind the two cards that are already out of the game + a new card) and;

  4. the probability of getting a higher value than the opponent (keeping in mind the two cards that are already out of the game + new card)

    I've never had to do anything with probability and I've only had a single one hour lecture on this subject so I hope I'm heading in the right direction.

Best Answer

First round

In the first round the player must beat the dealer. The dealer's card is excluded, leaving $311$ possible cards for the player. On average

$23$ cards tie (there are $24$ cards in each rank, but the dealer has one of them.)
$144$ cards win
$144$ cards lose.

Second round

The second round takes place if the first round was a tie, so the probability it happens is $23/311$.

In the second round the player wins the bet if the card values relative to the dealer are either a win or a tie. There are now $310\cdot309=95790$ possible deals.

As before, non-ties are split equally on average.

For $12$ of the ranks there are $24\cdot23=552$ ways to tie.
For the rank that occurred in the first round tie there are $22\cdot21=462$ ways to tie.

So in the second round there are

$7086$ ways to tie $(12\cdot552+1\cdot462)$
$44352$ ways for the player's card to be higher.
$44352$ ways for the player's card to be lower.

Conclusion

The overall probability of a win is$$\frac{144}{311}+\frac{23}{311}\cdot\frac{7086+44352}{95790}\approx0.502375$$

Incidentally, the house advantage comes from the way the betting works, and happens in a way that is not relevant to the question asked here.

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