Probablity of Winning on a Slot Machine at Online Casino

Question

A Player from India, Rahul, decides to play at a slot machine at an Online Casino in india. The Slot machine has only digits from 0 to 9 spread across 3 columns with 1 digit in each row. 

You Win under the following circumstances.

• If you get same numbers in all 3 columns like 777 or 888 or 999 (except 000)
• If you get all 3 adjacent numbers like (123, 456, 234).
• You can continue to spin the machine at the casino as many times as you want until you are Flagged.
• If you win, you have to stop.

You will be Flagged under following circumstance.

• The number 0 appears in one of the columns. 
• So if you get any of the following - 000, 012, 023, 540, etc - You will have to leave the game and can no longer spin.

Questions - 

• At which spin is Rahul most likely to be Flagged?
• What is the actual probablity of Rahul winning something at the online casino?

Answer 1

First, we need to find out how many digit combinations there are which contain zero. The digit combinations give us 1000 combinations (000-999).

There is only 1 combination containing three zeroes: 000

For 00X where X is 1-9, there are 9, but we need to consider 0X0 and X00 combinations, so there are 27 of these.

For 0XX there are 81 combinations and we need to consider X0X and XX0, making 243 combinations.

So the number of combinations containing at least one zero is 1+27+243=271.

Another way of doing this is to consider probabilities for each digit place. For the first digit there is a probability of 9/10 for choosing X. The same applies for the second and third digits so the total probability is (9/10)³=729/1000, that is 729 combinations out of 1000. The remaining 271 therefore must contain zeroes. The probability of spinning one of these is 271/1000 or 0.271. So the probability of being flagged on the first spin is 0.271. The probability of not being flagged on the first spin is 0.729. However, to proceed to the next spin, Rahul must neither win nor be flagged.

The probability of winning three in a row on the first spin is 9/1000=0.009.

The sequential combinations are: 123, 234, 345, 456, 567, 678, 789 so the probability is 7/1000=0.007.

Combining these we get 0.009+0.007=0.016, the probability of a winning row. The probability of a non-winning row is 0.984.

The game ends on a win (W) or a flag (F). A non-win, without being flagged, (L) allows the game to continue.

On spin 1 we have the following outcomes: W, F, or L with related probabilities 0.016, 0.271, 1-(0.016+0.271)=0.713. So there is a 71.3% chance that Rahul can continue to play, a 1.6% chance that he will win, and a 27.1% chance that he will be flagged (and win nothing).

To get to spin 2 he must lose spin 1 but not be flagged.

Taking the two spins together we get:

0.713×0.016=0.011408 (W)

0.713×0.271=0.193223 (F)

0.713²=0.508369 (L)

He can only continue if the second spin is L. The probability of 3 L's is about 0.36, so about 0.64 that he will either win or be flagged.

The game pattern emerges: L^n-1W or L^n-1F, where n is the number of spins, ending with winning or being flagged. In terms of probabilities we have 0.713^n-1×0.016 or 0.713^n-1×0.271.

• He is most likely to be flagged on the first spin, because the probability of being flagged is 0.271 (27.1%). When n=2, this reduces to LF=0.713×0.271=0.193223 (about 19.3%), and continues to reduce with each successive spin.
• His chances of winning are 0.016 (1.6%) on the first spin. This reduces to LW, which is about 0.011 (1.1%) on the second spin, and continues to reduce. His best chance to win is on the first spin.

Comments

Thanks so much. You explained it really well.