I am trying to figure out the optimal times to double up my poker game reward in the casino in Danmachi Memoria Freese. Here is the details to figure it out.

This poker version uses joker as wild card.

Current Tokens: 7,666,474 tokens Max bet: 1000 tokens

Without doubling up:

  • Royal Straight Flush: 500000 tokens
  • Five of a Kind: 300000 tokens
  • Straight Flush: 100000 tokens
  • Four of a Kind: 20000 tokens
  • Full House: 12000
  • Flush: 10000
  • Straight: 8000
  • Three of a Kind: 3000
  • Two Pair: 2000

I can double up reward to a maximum of 10 times (both success and draw cause counter to increment).

If I win the double up the reward becomes: $totalpayout = payoutBeforeDoubleUp * 2$ and I can chose to collect reward or gamble it on another double up which the then the payout from winning last double up gets doubled if I win.

If there is a draw, I get the amount won back.

The double up works as follows:

  • Computer puts 1 face up card which can have any value except joker.
  • We will assume for face down cards that deck has only one joker in it.
  • The Computer will put 4 face down cards which can be any value including a joker.
  • The Computer will seed the value so there is at least a minimum of 1 card that is higher than face up card so its possible to win.
    • We will assume the computer will check if 1 or more cards is higher than face up and if not then it will reshuffle face down into deck and then redraw 4 face down and repeat the check until at least 1 or more cards is higher than face up.

Based on these facts how much double ups should be risked for optimal token gain if you are not worried about time and assume there is enough current tokens that you won't run out?

Sorry for asking this complex question. I tried to figure it out myself but my expertise is not in probability as I don't have to use that math at my job. Also I am asking this so I can get the token rewards in my game called DanMachi Memoria Freese and as such it is not for class homework so there is no need to phrase it in way as to help but not give away answer.

  • $\begingroup$ I forgot to add this but assume there is only one joker in the deck since I have never pulled a hand with 2+ jokers and due to how much I have played the chances of there being more than one is so slim that I can say with confidence there is only 1 joker. $\endgroup$ – 8bitedge Oct 13 '18 at 22:51

For simplification let’s say every time you challenge it’s a 1/4 chance. Yes there are times that it is a 1/2 or 100% but sense we don’t get to know the face up card before you get to try. We are guaranteed a 1/4 minimal. The risk inceses the more times in a row you go. Here is the order, 1/4, 1/16, 1/64, 1/256, 1/1,024, 1/4,096, 1/16,384, 1/65,536, 1/262,144, 1/1,048,576.

  • $\begingroup$ Thanks for the help. $\endgroup$ – 8bitedge Nov 9 '18 at 12:59

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