I understand when you have a FD on the flop you have a 9/47 (9 diamond cards left on the deck / by 47 unseen cards) = 19.1 chance of hitting a diamond OTT and a 9/46 - 19.5 chance of hitting a diamond OTR. Therefore a total of 35% chance of hitting by the river.
The example above would make perfect sense to me if:
1. It was heads-up match (2 players)
2. Zero "burn cards"
3. Opponent has zero diamonds in his/ her hand
Therefore, all the diamonds would be available and we would be able to use the 9/47 and 9/46 method.
However, in the games we play in, its usually a full ring game 10 handed so that means that there are approximately 18 cards ( 9 players) + 1 "burn card" = 19 cards that will not to be seen OTT.
Assuming Full ring game 10 handed and we have a FD on the flop. 19 cards that will not be seen OTT out of 47 cards.
1. What is the probability that there are zero diamonds in the 19 cards that are not in play? I bet it is pretty minuscule.
2. What is the probability that there is 1, 2, 3, etc diamonds?
3. What is the average number of diamonds in 19 random cards?
I did a small sample size experiment based on the above. I took out 4 diamond cards and a non diamond card. I shuffled 47 cards and took out 19 random cards to see how many diamonds in 19 cards. My results were are as follow:
Test 1: 2 diamonds
Test 2: 5 diamonds
Test 3: 5 diamonds
Test 4: 3 diamonds
Test 5: 3 diamonds
Test 6: 5 diamonds
Test 7: 2 diamonds
Test 8: 4 diamonds
Test 9: 5 diamonds
Test 10: 3 diamonds
Average: 37 diamonds/ # of tests = 3.7 diamonds
Therefore, based on data above should we really figure 9 diamonds still available when we are on a FD or a lesser number (9 diamonds - 3.7 diamonds (Average) = 5.3 diamonds) since there will be 19 random cards that will not be seen OTT?
****Correction it should be 20 cards that will not be seen i didnt take into account the burn card before the turn. So even less diamonds available with 20 cards not to be seen OTT.