pays | probability |
---|---|

13 | 27.1% |

7 | 22.2% |

5 | 11.1% |

2 | 11.1% |

1 | 13.3% |

pays | probability | expected payout |
---|---|---|

13 | 27.1% | 3.51919 |

7 | 22.2% | 1.55556 |

5 | 11.1% | 0.55556 |

2 | 11.1% | 0.22222 |

1 | 13.3% | 0.13333 |

total: | 5.98586 | |

based on 6 coins in: | ||

return: | 0.99764 | |

house edge: | 0.24% |

Imagine further that several of these slot machines existed and we are lucky to have a peek inside them.

Type A.

Some of the boxes have a single board computer inside. The sixth coin inserted triggers the random number generator (RNG). Based on this random result, the amount payed out is selected from the paytable given above.

Type B.

The other boxes have gaming consoles inside. They are set to play passline bets with maximum odds in a 3x4x5x odds craps game. The sixth coin inserted in the box triggers a single hand of passline to be played. After this hand, the payout is derived from the coins inserted (6) plus the winnings or minus the losses of the passline outcome.

The payout amounts of the box and their corresponding probabilities are the same as those given in the paytable above.

The payout of 13 coins is produced by three different possible outcomes in the case of the passline bet. Their probabilities add up to the probability given in the original paytable above. The player outside the box is not able to distinguish which of the random outcomes caused cashing out 13 coins.

pays | probability | expected payout | outcome | win(initial bet) | win(odds bet) | |
---|---|---|---|---|---|---|

13 | 5.6% | 0.72222 | point is 4 or 10: won | 1 | 6 | |

13 | 8.9% | 1.15556 | point is 5 or 9: won | 1 | 6 | |

13 | 12.6% | 1.64141 | point is 6 or 8: won | 1 | 6 | |

7 | 22.2% | 1.55556 | 7, 11: frontline winner | 1 | 0 | |

5 | 11.1% | 0.55556 | 2, 3, 12: craps | -1 | 0 | |

2 | 11.1% | 0.22222 | point is 4 or 10: 7 out | -1 | -3 | |

1 | 13.3% | 0.13333 | point is 5 or 9: 7 out | -1 | -4 | |

0 (not in paytable) | 15.2% | 0.00000 | point is 6 or 8: 7 out | -1 | -5 | |

total: | 100.0% | 5.98586 | ||||

based on 6 coins in: | ||||||

return: | 0.99764 | |||||

house edge: | 0.24% |

Initial bet plus free odds bet.

It could be argued that on the passline bet, in general, not all six coins are wagered. Some coins would remain "unused", and are cashed out together with the actual resulting amount.

pays = cashes out | probability | outcome | initial bet | odds bet | expected wager | win(initial bet) | win(odds bet) | payout | expected payout | |
---|---|---|---|---|---|---|---|---|---|---|

13 | 5.6% | point is 4 or 10: won | 1 | 3 | 0.22222 | 1 | 6 | 11 | 0.61111 | |

13 | 8.9% | point is 5 or 9: won | 1 | 4 | 0.44444 | 1 | 6 | 12 | 1.06667 | |

13 | 12.6% | point is 6 or 8: won | 1 | 5 | 0.75758 | 1 | 6 | 13 | 1.64141 | |

7 | 22.2% | 7, 11: frontline winner | 1 | 0 | 0.22222 | 1 | 0 | 2 | 0.44444 | |

5 | 11.1% | 2, 3, 12: craps | 1 | 0 | 0.11111 | -1 | 0 | 0 | 0.00000 | |

2 | 11.1% | point is 4 or 10: 7 out | 1 | 3 | 0.44444 | -1 | -3 | 0 | 0.00000 | |

1 | 13.3% | point is 5 or 9: 7 out | 1 | 4 | 0.66667 | -1 | -4 | 0 | 0.00000 | |

0 (not in paytable) | 15.2% | point is 6 or 8: 7 out | 1 | 5 | 0.90909 | -1 | -5 | 0 | 0.00000 | |

total: | 100.0% | 3.77778 | 3.76364 | |||||||

based on expected wager: | ||||||||||

return: | 0.99626 | |||||||||

house edge: | 0.37% |

Question 3: How could this higher house edge be mathematically justified for the slot machines of type A, which are behaving identically on the outside?

Initial bet only.

Some might argue that the free odds bets shouldn't be included in the calculation because they have no house edge.

pays = cashes out | probability | outcome | initial bet | win(initial bet) | payout | expected payout | |
---|---|---|---|---|---|---|---|

13 | 5.6% | point is 4 or 10: won | 1 | 1 | 2 | 0.11111 | |

13 | 8.9% | point is 5 or 9: won | 1 | 1 | 2 | 0.17778 | |

13 | 12.6% | point is 6 or 8: won | 1 | 1 | 2 | 0.25253 | |

7 | 22.2% | 7, 11: frontline winner | 1 | 1 | 2 | 0.44444 | |

5 | 11.1% | 2, 3, 12: craps | 1 | -1 | 0 | 0.00000 | |

2 | 11.1% | point is 4 or 10: 7 out | 1 | -1 | 0 | 0.00000 | |

1 | 13.3% | point is 5 or 9: 7 out | 1 | -1 | 0 | 0.00000 | |

0 (not in paytable) | point is 6 or 8: 7 out | 1 | -1 | 0 | 0.00000 | ||

total: | 100.0% | 0.98586 | |||||

based on initial bet: | |||||||

return: | 0.98586 | ||||||

house edge: | 1.41% |

Question 5: How could this rather high house edge be mathematically justified for the slot machines of type A, which are behaving identically on the outside?

Experiment 1.

Take a friend to the craps tables. Choose a table with 3x4x5x odds. Let your friend do the betting at the table. You won't bet yourself but instead you will watch over the bankroll and do the statistics. Follow this procedure:

1) Hand your friend six chips and ask him to play a single hand of passline with full odds.

2) After this one hand your friend returns back to you the number of chips resulting from this hand.

3) Keep records of the number of chips your friend returns and how often this happens. Don't forget to also keep track of those times your friend returns nothing, i.e. zero chips.

4) Repeat from step 1).

After a large number of hands played, for each amount of chips returned, calculate the percentage of hands played.

Question 6: Would it be wrong to expect that your statistics will approximate the paytable given above for the slot machine?

Question 7: What would be the house edge of your play?

Experiment 2.

If your friend is not available for some craps try the following procedure at a table with 3x4x5x odds:

1) Define a bankroll for your current session as a multiple of six chips and keep records of its total. Put the chips in one rail of your chip rack.

2) Separate six chips by moving them a little on the side. From these six chips bet one hand of passline with full odds.

3) Collect the chips of the result of this one hand and add them to the second rail. Also move any remaining of the six separated chips to the second rail.

4) Repeat from step 2) until your bankroll from the first rail is used up.

5) Add to your records the number of chips you accumulated in the second rail and take your winnings home.

After a large enough number of sessions played, add up all the chips you recorded for the second rail and devide it by the sum of chips you recorded for the first rail. Divide that value by six.

Question 8: Would it be wrong to expect for your play a house edge of 0.24%?

Final question: From a pure mathematical point of view, which game would you prefer? Playing the hypothetical slot machine given above or playing the passline with full odds on a 3x4x5x odds craps table?

Therefore, I only care about the house edge based on coins put into play. The two games are not identical for my purposes.

Quote:ThomasKIf your friend is not available for some craps try the following procedure at a table with 3x4x5x odds:

1) Define a bankroll for your current session as a multiple of six chips and keep records of its total. Put the chips in one rail of your chip rack.

2) Separate six chips by moving them a little on the side. From these six chips bet one hand of passline with full odds.

3) Collect the chips of the result of this one hand and add them to the second rail. Also move any remaining of the six separated chips to the second rail.

4) Repeat from step 2) until your bankroll from the first rail is used up.

5) Add to your records the number of chips you accumulated in the second rail and take your winnings home.

After a large enough number of sessions played, add up all the chips you recorded for the second rail and devide it by the sum of chips you recorded for the first rail. Divide that value by six.

Question 8: Would it be wrong to expect for your play a house edge of 0.24%?

link to original post

Yes, and here's why:

Instead of setting aside 6 coins, you set aside 100. In effect, your return will be 94 coins higher; the ER is 5.98586 + 94 = 99.98586. On a bet of 100, the house is expected to gain 100 - 99.98586 = 0.01414; that is a house edge of 0.014%.

Quote:unJonWhat a great post! Love the thought experiment. Have not yet looked over the math.

link to original post

Thanks for liking this one.

I already posted the same idea in a different context earlier but there it seems to be rejected as "nonsense".

Quote:MentalEvery casino I play at only rates me or comps me or gives me cashback based on coins put into play. The extra coins that are never used on an odds bet are not rated and do not earn me cashback or satisfy play-through requirements for bonuses.

Therefore, I only care about the house edge based on coins put into play. The two games are not identical for my purposes.

link to original post

Thanks for your reply.

I already have thought of this drawback, that the idea does not work for comps.

Maybe in the future I will work out the math of how comps would effect the results.

Quote:ThatDonGuyQuote:ThomasKIf your friend is not available for some craps try the following procedure at a table with 3x4x5x odds:

1) Define a bankroll for your current session as a multiple of six chips and keep records of its total. Put the chips in one rail of your chip rack.

2) Separate six chips by moving them a little on the side. From these six chips bet one hand of passline with full odds.

3) Collect the chips of the result of this one hand and add them to the second rail. Also move any remaining of the six separated chips to the second rail.

4) Repeat from step 2) until your bankroll from the first rail is used up.

5) Add to your records the number of chips you accumulated in the second rail and take your winnings home.

After a large enough number of sessions played, add up all the chips you recorded for the second rail and devide it by the sum of chips you recorded for the first rail. Divide that value by six.

Question 8: Would it be wrong to expect for your play a house edge of 0.24%?

link to original post

Yes, and here's why:

Instead of setting aside 6 coins, you set aside 100. In effect, your return will be 94 coins higher; the ER is 5.98586 + 94 = 99.98586. On a bet of 100, the house is expected to gain 100 - 99.98586 = 0.01414; that is a house edge of 0.014%.

link to original post

Thanks for this one.

I alreay expected someone to come up with this suggestion: The higher the coin in the lower the house edge.

However, the six coins will be lost on a seven out of a six or an eight. So the six coins actually are at stake in the game.

The 94 coins, on the other hand, will be returned in each possible outcome and therefore have a probability of 100% of being payed.

(The slot machine would only warm them up.)

This brings up another interesting question:

0 < p(X) < 1 ... is considered a gamble.

The sure things p(X) = 0 and p(X) = 1 are handled mathematically like any 0 < p(X) < 1.

Are both sure probabilities a gamble, anyway?

Quote:unJonWhat a great post! Love the thought experiment. Have not yet looked over the math.

link to original post

unJon,

Thanks again for your appreciation of Schroedinger's craps.

My ideas about progressive betting seem to lack some substance, if I'm not mistaken by your reply given in another related thread:

Quote:unJonI thought the Schrodinger craps game was brilliant. But this thread is just math semantics misunderstanding. You are doing a bait and switch on denominators.

You lose 5.3% of every bet you place at risk on a single spin of the wheel. Full stop and incontrovertible. It doesn’t matter if net sizing changes for independent or dependent reasons.

link to original post

I decided to re-phrase the latest of my examples using the structure of Schroedinger's craps and I am keenly interested in your general opinion and in your specific arguments in how far this thought experiment is inferior to Schroedinger's craps.

Thank you very much in advance for your constructive criticism:

"As good or as bad

as double zero roulette:

The 'Tip your Dealers" bet."

Another notice explains "Each winning hand triggers the donation of $1 to the dealers' tip pool."

Inserting the seventh coin triggers the random mechanism which then cashes out the amount given on this very rudimentary paytable:

pays | probability |
---|---|

$39 | 85.42% |

pays | probability | expected payout |
---|---|---|

39 | 85.42% | 33.31404 |

total: | 33.31404 | |

based on 7 x $5 coin in: | ||

return: | 0.95183 | |

house edge: | 4.82% |

Imagine further that the casino distributed several of these slot machines all over the casino floor but we are lucky to have a peek inside them.

Type A.

Some of the boxes have a single board computer inside. The seventh coin inserted triggers the random number generator (RNG). Based on this random result, the amount payed out is selected from the paytable given above.

Type B.

The other boxes have gaming consoles inside. They are set to play one attempt of a three step Martingale on red in a double zero roulette game. The seventh coin inserted in the box triggers the Martingale to be played. The payout is derived from the coins inserted (7) plus the winnings or minus the losses of the steps of the Martingale.

The payout amount of the box and its corresponding probability is the same as that given in the paytable above.

The payout of $39 is produced by three different possible outcomes in the case of the Martingale. Their probabilities add up to the probability given in the original paytable above. The player outside the box is not able to distinguish which of the random outcomes caused cashing out $39.

pays | probability | expected payout | outcome | win | tip | |
---|---|---|---|---|---|---|

39 | 47.37% | 18.47368 | won on first step | 5 | -1 | |

39 | 24.93% | 9.72299 | won on second step | 5 | -1 | |

39 | 13.12% | 5.11736 | won on third step | 5 | -1 | |

0 (not in paytable) | 14.58% | 0.00000 | lost on third step | -35 | 0 | |

total: | 100.00% | 33.31404 | ||||

based on 7 x $5 coin in: | ||||||

return: | 0,95182971278612 | |||||

house edge: | 4.82% |

Winning the Martingale before the third step.

It could be argued that, in general, not all seven coins are wagered. Some coins would remain "unused", and are cashed out together with the actual resulting amount.

Without tip (for comparison only):

pays = cashes out | probability | outcome | wager | expected wager | win | payout | expected payout | |
---|---|---|---|---|---|---|---|---|

39 | 47.37% | won on first step | 5 | 2.36842 | 5 | 10 | 4.73684 | |

39 | 24.93% | won on second step | 15 | 3.73961 | 5 | 20 | 4.98615 | |

39 | 13.12% | won on third step | 35 | 4.59251 | 5 | 40 | 5.24858 | |

0 (not in paytable) | 14.58% | lost on third step | 35 | 5.10278 | -35 | 0 | 0.00000 | |

total: | 100.00% | 15.80332 | 14.97157 | |||||

based on expected wager: | ||||||||

return: | 0.94737 | |||||||

house edge: | 5.26% |

With tip:

pays = cashes out | probability | outcome | wager | expected wager | win | tip | payout | expected payout | |
---|---|---|---|---|---|---|---|---|---|

39 | 47.37% | won on first step | 5 | 2.36842 | 5 | -1 | 9 | 4.26316 | |

39 | 24.93% | won on second step | 15 | 3.73961 | 5 | -1 | 19 | 4.73684 | |

39 | 13.12% | won on third step | 35 | 4.59251 | 5 | -1 | 39 | 5.11736 | |

0 (not in paytable) | 14.58% | lost on third step | 35 | 5.10278 | -35 | 0 | 0 | 0.00000 | |

total: | 100.00% | 15.80332 | 14.11736 | ||||||

based on expected wager: | |||||||||

return: | 0.89332 | ||||||||

house edge: | 10.67% |

Question 3: How could this horrible house edge be mathematically justified for the slot machines of type A, which are behaving identically on the outside?

Experiment 1.

Take a friend to the roulette tables. Let your friend do the betting at the table. You won't bet yourself but instead you will watch over the bankroll and do the statistics. Follow this procedure:

1) Hand your friend seven $5 chips and ask him to play a single attempt of a three step Martingale.

2) After this one attempt, if your friend won, he tips the dealer $1 from the amount he now holds. In any case he eventually returns to you either with nothing or with what he still holds from the winnings.

3) Keep records of the amount your friend returns and how often this happens. Don't forget to also keep track of those times your friend returns nothing, i.e. zero chips.

4) Repeat from step 1).

After a large number of hands played, for each amount of chips returned, calculate the percentage of hands played.

Question 6: Would it be wrong to expect that your statistics will approximate the paytable given above for the slot machine?

Question 7: What would be the house edge of your play?

Experiment 2.

If your friend is not available for some roulette try the following procedure at a double zero roulette table:

1) Define a bankroll for your current session as a multiple of seven $5 chips and keep records of its total. Try to get hold of two coin buckets and put your chips in one of them as the bankroll bucket. The second empty bucket will be your payout bucket.

2) Take seven chips out of your bankroll bucket and play one attempt of a three step Martingale. If you won the attempt, tip the dealer $1 of the resulting amount.

3) Collect the chips of the result of this one attempt plus any remaining of the seven chips and put them all in the payout bucket.

4) Repeat from step 2) until your bankroll bucket is empty.

5) Add to your records the number of chips you accumulated in the payout bucket and take your winnings home.

After a large enough number of sessions played, add up all the chips you recorded for the payout bucket and devide it by the sum of chips you recorded for the bankroll bucket. Divide that value by seven.

Question 8: Would it be wrong to expect for your play a house edge of 4.82%?

Final question: From a pure mathematical point of view, which game would you prefer? Playing the hypothetical slot machine given above or playing the Martingale with tipping the dealer?

Quote:ThomasKI

Experiment 2.

If your friend is not available for some roulette try the following procedure at a double zero roulette table:

1) Define a bankroll for your current session as a multiple of seven $5 chips and keep records of its total. Try to get hold of two coin buckets and put your chips in one of them as the bankroll bucket. The second empty bucket will be your payout bucket.

2) Take seven chips out of your bankroll bucket and play one attempt of a three step Martingale. If you won the attempt, tip the dealer $1 of the resulting amount.

3) Collect the chips of the result of this one attempt plus any remaining of the seven chips and put them all in the payout bucket.

4) Repeat from step 2) until your bankroll bucket is empty.

5) Add to your records the number of chips you accumulated in the payout bucket and take your winnings home.

After a large enough number of sessions played, add up all the chips you recorded for the payout bucket and devide it by the sum of chips you recorded for the bankroll bucket. Divide that value by seven.

Question 8: Would it be wrong to expect for your play a house edge of 4.82%?

original post

Leave the game unchanged, but put 7,000 chips in the bucket. Take 7,000 chips out each play, but only use a max of 7 chips to play the same 3-step sequence using one-chip to start.

Reductio ad absurdum question: Would it be wrong to expect for your play a house edge of 0.00%?