Probability Question Help

I’ve only posted a few time but had a question come up the other day that I just can’t get out of my head. I was playing a baccarat table with a buddy that had the dragon-7 bet (if the banker wins with a 3-card-7 it pays 40:1, yes I know its a high HE). Player had 1, Banker 0, player drew an ace for 2 my buddy gets the final banker card. He looks and sees a 3-side (6, 7, or 8) and says “I’m 50/50 to hit the Dragon” he ultimately made an 8 and won the hand but not the dragon. Afterward I told him he was 1 in 3 as he could be holding a 6, 7, or 8 and only a 7 won the dragon. Here was his explanation:

I look and see a 3-side so I know it’s a 6, 7 or 8. Then I turn the card long-ways and see either a pip or a blank, if it’s a pip it’s 50/50 between a 7 and 8, if it’s a blank it’s 50/50 between a 6 and a 7 since either a pip or a blank gives me a 50/50 shot to draw a 7 I’m 50/50 and a 3-side has a 50% chance of being a 7.

Now this explanation creates a decision tree showing 4 outcomes (PP-8, PB-7, BB-6, BP-7) and shows: 6-25%, 7-50% and 8-25% but this doesn’t reconcile with my immediate reaction that knowing it’s a 3-side each has a 1/3 chance. At first I thought this was a case of a Monty Hall situation, where additional information alters your odds, allowing either 6 or 8 to be eliminated on the first peek however under that reasoning a 7 is a 50% chance and I don’t think that’s right. Trying to engineer a solution where a 7 is 1/3 I realize that if you had a 1/3 chance, 100% of the time, this would equal my 1/3 gut reaction. This leads to a conclusion that when you see a pip the odds are: 7 -1/3 and 8- 2/3 and when you see a blank its 7- 1/3 and 6- 2/3 but I don’t know if that’s right or why.

Help please this is racking my brain!

Quote: CrazyCanuck

Now this explanation creates a decision tree showing 4 outcomes (PP-8, PB-7, BB-6, BP-7) and shows: 6-25%, 7-50% and 8-25% but this doesn’t reconcile with my immediate reaction that knowing it’s a 3-side each has a 1/3 chance. At first I thought this was a case of a Monty Hall situation, where additional information alters your odds, allowing either 6 or 8 to be eliminated on the first peek however under that reasoning a 7 is a 50% chance and I don’t think that’s right. Trying to engineer a solution where a 7 is 1/3 I realize that if you had a 1/3 chance, 100% of the time, this would equal my 1/3 gut reaction. This leads to a conclusion that when you see a pip the odds are: 7 -1/3 and 8- 2/3 and when you see a blank its 7- 1/3 and 6- 2/3 but I don’t know if that’s right or why.

Help please this is racking my brain!

You are absolutely correct. Let's look at the following:

6 end 1 (no pip)
6 end 2 (no pip)
7 end 1 (no pip)
7 end 2 (pip)
8 end 1 (pip)
8 end 2 (pip)

If you don't see a pip, then you know it's 1 of the 1st 3 scenarios. 2 of those are 6's, and 1 is 7. So you have a 1/3 chance of picking the 7. Similarly, if you see a pip, 2 of those scenarios are 8's, and only 1 is 7. So you have a 1/3 chance of picking 7, and 2/3 chance of picking 8.

I believe somewhere in the Wizard's math problems, there is one where you have a 2-sided heads coin and a normal coin inside a bag. You pull 1 coin out of the bag and see a head. What are the odds that the other side is a tail? It is 1/3, for the same reasoning.

edit: yes here it is (#16), though he asks the chance of the other side also being heads. His solution is also much more explanatory: http://mathproblems.info/working.php#16

Your buddy is making just one error: if you see the pip or blank, the chance of the card being a 7 is NOT 50/50. This problem falls under the realm of conditional probability and Bayes' Theorem. What you need to consider is the probability of the card being a 7 given you see a pip. This is typically written as: Prob(card=7 | pip). Now, using Bayes' Theorem, you can calculate: P(7|pip) = P(pip|7)*P(7) / P(pip).

P(pip|7), or probability of seeing a pip given the card is 7, is 0.50. P(7), probability of the card being a 7, is 0.33 (given you saw the 3-side). P(pip) is 0.5--kind of skipping a step here. Therefore, P(7|pip) = 0.5*0.33 / 0.5 = 0.33.

This probably makes it clear as mud. But what's important to remember is that because the probability of a pip when the card is 8 is 1 and the probability of a pip when the card is 7 is 0.5, it is more likely to be a 8 when you see the pip. This is known as the prior probability.

I welcome the math experts to clarify/correct my explanation. Hope this helps!

Quote: jc2286

If you don't see a pip, then you know it's 1 of the 1st 3 scenarios. 2 of those are 6's, and 1 is 7. So you have a 1/3 chance of picking the 7. Similarly, if you see a pip, 2 of those scenarios are 8's, and only 1 is 7. So you have a 1/3 chance of picking 7, and 2/3 chance of picking 8.

Beat me to it! Good explanation.