# What are the odds of winning \$25,000 in the Catch 21 end game?

In order to meaningfully talk about probabilities you need to deal with the element of skill.  You can deal with the number of power chips by splitting the problem into 3 (you get 2, 3, or 4 power chips, at least in season 4, except for a few games where you get Burger King power chips instead of regular ones, but there's really no difference).  You can deal with the choice of when to stop by insisting that players go for \$25,000 (no stopping).  Dealing with skill at placing the cards in hands / using or not using power chips is harder.  There are conflicting goals of maximizing the chance of 21 on the next card and keeping an unbustable hand.

You can run stats on the individual games.  Since they play out the games you can tell the percentage who would have won if they had to go for \$25,000.  The sample I took gives an answer of 6 wins / 53 games = 11% (1 actual \$25,000 win; 5 "would haves").  Broken down by power chips: 2: 1/12 = 8%, 3: 0/26 = 0%, 4: 5/15 = 33%.  This says I really need more data, and I'm not sure enough games have been played to get enough data.

I did a computer simulation with a stop strategy of never stop, and a card placement strategy of maximizing the number of cards in the deck that will get you a 21 on the next draw – ties between placing a card and using a chip are broken in favor of playing a card (this is likely not optimal, but I'm not sure what is.  Anyway, it's something that is fairly easy for a person to compute on the fly, and the host talks about the number of cards in the deck a lot).  I ran 100,000 games for each case.  The results for each number of power chips are 2: 13%, 3: 18%, and 4: 24%.   Considering \$ won / game, an extra power chip is worth maybe \$1,300 on average.

Incidentally, the distribution of power chips for my sample of 53 games is 2: 23%, 3: 49%, 4: 28% with the average number of chips 3.06 .  The prize won is \$0: 21%, \$1000: 25%, \$5000: 53%, and \$25000: 2% (totals do not add to 100% due to rounding).